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# Algebra Notes in Newsletter Form

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This is a resource I found that is probably the most creative idea that I have seen in a long time. This teacher presented Algebra Notes in the form of newsletters. Awesome idea!

This is a resource I found that is probably the most creative idea that I have seen in a long time. This teacher presented Algebra Notes in the form of newsletters. Awesome idea!

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• 1. Algebra Newsletter NotesA creative way to presentconcepts, notes & study guidesin a newsletter format.Excellent Resource!www.clubtnt.org/my_algebra
• 2. In many ways, working withrational expressions is like workingwith fractions. Recall that:, equalnotdoesbbcabcba +=+0, equalnotdoesbbcabcba −=−As long as we have commondenominators, adding andsubtracting rational expressions isfairly straightforward – simplyrewrite the addition or subtractionon the common denominator.0equalnotdoesRRQPRQRPRQPRQRP−=−+=+To add or subtract rationalexpressions with differentdenominators, we need to find theLowest Common Denominator forthe polynomials that represent thedenominators or our rationalexpressions.To find the LCD forpolynomials:1. Factor each denominatorcompletely. Use exponentnotation for repeated factors.2. Write the product of all thedifferent factors that appear inthe denominators.3. On each factor, use the highestpower that appears on thatfactor in any or thedenominators.Example: LCD of 20, 501. Factor completely: 20=22*5,50=2*522. Product of all different factors:2 and 53. Use highest power: 22*52=100Example: LCD x3yz2, x5y2z, xyz51. Factor completely, eachexpression already is a factor:x3yz2, x5y2z, xyz5.2. Product of all the differentfactors (x, y, z)3. Use highest power: x5y2z5Example: LCD a2+5a+6, a2+4a+41. Factor completely: (a+2)(a+3)and (a+2)22. Product of all different factors:(a+2), (a+3), and (a+2)3. Use highest power:(a+3)(a+2)2Adding and Subtracting RationalExpressions with DifferentDenominatorsWhen we add or subtractrational expressions with differentdenominators, we must convertthem to contain identicaldenominators. The easiest way todo this is to use the LCD that wehave just reviewed. Once the LCDis determined, rewrite each rationalexpression as an equivalent rationalexpression with the LCD. Theprocess is just like working withfractions.Example:127203+1. Factor completely: 20=22*5,50=2*522. Product of all different factors:2 and 53. Use highest power: 22*52=100151115*411*4604460356095512733203===+=+=Academic Support ServicesFREE Tutoring And Academic Support Services!!!Basement of McCutchan Hall, Rm. 1Mon-Thurs: 9 a.m. – 9 p.m.Fri: 9 a.m. – 3 p.m. and Sun 5 p.m. – 9 p.m.Mr. Breitsprecher’s Edition April 5, 2005 Web: www.clubtnt.org/my_algebra
• 4. Mr. Breitsprecher’s Edition January 18, 2005 FREE!Decimals are a method ofwriting fractional numbers withoutwriting a fraction having anumerator and denominator.The fraction 7/10 could bewritten as the decimal 0.7. Theperiod or decimal point indicatesthat this is a decimal.If a decimal is less than 1, placea zero before the decimal point.Write 0.7 not .7There are other decimals suchas hundredths or thousandths. Theyall are based on the number ten justlike our number system. A decimalmay be greater than one, 3.7 is anexample.The fraction 37/100 could bewritten as the decimal 0.37. Theperiod or decimal point indicatesthat this is a decimal. Of course, adecimal with one hundredthsdecimal may be greater than 1.Decimal numbers, such asO.6495, have four digits after thedecimal point. Each digit is adifferent place value.The first digit after the decimalpoint is called the tenths place value.There are six tenths in the numberO.6495.The second digit tells you howmany hundredths there are in thenumber. The number O.6495 hasfour hundredths.The third digit is thethousandths place.The fourth digit is the tenthousandths place which is five inthis example.Therefore, there are six tenths,four hundredths, nine thousandths,and five ten thousandths in thenumber 0.6495.If there are two decimalnumbers we can compare them.One number is either greater than,less than or equal to the othernumber.A decimal number is just afractional number. Comparing 0.7and 0.07 is clearer if we compared7/10 to 7/100. The fraction 7/10 isequivalent to 70/100 which isclearly larger than 7/100.Therefore, when decimals arecompared start with tenths place andthen hundredths place, etc. If onedecimal has a higher number in thetenths place then it is larger than adecimal with fewer tenths. If thetenths are equal compare thehundredths, then the thousandths etc.until one decimal is larger or there areno more places to compare. If eachdecimal place value is the same thenthe decimals are equal.Sound simple – great! This is thefoundation of percentages, a veryimportant tool for analysis.Source: www.aaamath.comTo write decimals as fractions,use the “placeholder” as thedenominator and write the number(without the decimal) as thenumerator.To write fractions as decimals,divide the numerator by thedenominator.To write a percent as a decimal,drop the % symbol and move thedecimal points two places to the left.To write a decimal as a percent,move the decimal point two places tothe right and attach the % symbol.To write fraction as a percent,write the decimal as a fraction(above), perform that division, andmultiply by 100.To Add or Subtract Decimals1. Write the decimals so thatthe decimal points line upvertically.2. Add or subtract for wholenumbers.3. Place the decimal point in thesum or difference so that it linesup vertically with the others.To Multiply Decimals1. Multiply the decimals as thoughthey are whole numbers.2. The decimal point in the productis placed so that the number ofdecimal places is equal to theSUM of the number of decimalplaces in the factors.To Divide Decimals1. Move the decimal point in thedivisor to the right until thedivisor is a whole number.2. Move the decimal point in thedividend to the right the SAMENUMBER OF PLACES as thedecimal point was moved in step1.3. Divide. The decimal point in thequotient is directly over themoved decimal point in thedividend
• 6. howMr. Breitsprecher’s Edition March 9, 2005 Web: www.clubtnt.org/my_algebraEasy does it! Lets look at howto divide polynomials by startingwith the simplest case, dividing amonomial by a monomial. This isreally just an application of theQuotient Rule that was coveredwhen we reviewed exponents.Next, well look at dividing apolynomial by a monomial. Lastly,we will see how the same conceptsare used to divide a polynomial by apolynomial. Are you ready? Let’sbegin!Quotient of a Monomial by aMonomialTo divide a monomial by amonomial, divide numericalcoefficient by numerical coefficient.Divide powers of same variableusing the Quotient Rule ofExponents – when dividingexponentials with the same base wesubtract the exponent on thedenominator from the exponent onthe numerator to obtain the exponenton the answer.Examples:• (35x³ )/(7x) = 5x²• (16x² y² )/(8xy² ) = 2xRemember: Any nonzero numberdivided by itself is one (y²/y² = 1),and any nonzero number to the zeropower is defined to be one (ZeroExponent Rule).• 42x/(7x³ ) = 6/x²Remember: The fraction x/x³simplifies to 1/x². The NegativeExponent Rule says that anynonzero number to a negativeexponent is defined to be onedivided by the nonzero number tothe positive exponent obtained. Wecould write that answer as 6x-2.Quotient of a Polynomial by aMonomialTo divide a polynomial by amonomial, divide each of the termsof the polynomial by a monomial.This is really an example of how weadd fractions, recall:cbcacba+=+Think of dividing a polynomial by amonomial as rewriting each term ofthe polynomial over thedenominator, just like we did whenworking with fractions. Then, wesimplify each term.Example:• (16x³ - 12x² + 4x)/(2x)= (16x³)/(2x)-(12x² )/(2x)+(4x)/(2x)= 8x² - 6x + 2Review: Exponents and Polynomialsanmeans the product of n factors, each of which is a (i.e. 32= 3*3, or 9)Exponent Rules• Product Role: am*an=am+n• Power Rule: (am)n=amn• Power of a Product Rule: (ab)n= anbn• Power of a Quotient Rule: (a/b)n=an/bn• Quotient Rule: am/an=am-n• Zero Exponent: a0=1, a ≠0Working with Polynomials• Adding Polynomials. Remove parenthesis and combine like terms.• Subtracting Two Polynomials. Change the signs of the terms in the secondpolynomial, remove parenthesis, and combine like terms.• Multiply Polynomials. Multiply EACH term of one polynomial by EACHterm of the other polynomial, and then combine like terms.Special Products• FOIL When multiplying binomials, multiply EACH term in the firstbinomial by EACH term in the second binomial. This multiplication ofterms results in the pattern of First, Outside, Inside, and Last.• Difference Of Squares: (a+b)(a-b) = a2-b2• Perfect Square Trinomials:(a+b)2= a2+2ab+b2(a-b)2= a2– 2ab + b2
• 7. Quotient of a Polynomial by aPolynomialTo divide a polynomial by a polynomial,use long division, similar to the longdivision technique used in arithmetic.Remember in starting the long divisionprocess:1. Write dividend and divisor in termsof descending powers of variable,leaving space for any missingpowers of the variable or writing inthe missing powers with coefficientzero. If there is more than onevariable, arrange dividend anddivisor in terms of descending order(from the term with the highest“degree” to the lowest “degree”).2. Divide first term of divisor into firstterm of dividend (On subsequentiterations, into first term of previousdifference). Place this answer abovelong division symbol.3. Multiply divisor by the expressionjust written above division symboland align like terms.4. Subtract line just written from lineimmediately above it. Remember tosubtract we change the sign of theMr. Breitsprecher’s Edition March 9, 2005 Algebra Connections, Page 2subtrahend and add (add theopposite).5. Repeat steps 2 through 4 untilthe difference you obtain is apolynomial of degree less thanthe degree of the divisor.6. If the final difference is zero,the division is exact. Thequotient is the polynomial givenacross the top. If the differencein nonzero division is not exact,the quotient is the polynomialgiven across the top plus theremainder (polynomial in lastline) divided by the divisor.Example:)12/()1(32450412223−−++−+++xRxxxxxx2324 xx −xx −223646−−xx1−Source: http://home.sprynet.comAcademic Support ServicesFREE Tutoring And Academic Support Services!!!Basement of McCutchan Hall, Rm. 1Mon-Thurs: 9 a.m. – 9 p.m.Fri: 9 a.m. – 3 p.m. and Sun 5 p.m. – 9 p.m.OnlineResourcesHow to Add, Subtract, Multiply,and Divide Polynomialshttp://faculty.ed.umuc.edu/~swalsh/Math%20Articles/Polynomial.htmlDividing Polynomialshttp://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut35_div.htmhttp://tutorial.math.lamar.edu/AllBrowsers/1314/DividingPolynomials.aspDr. Math: Dividing Polynomials.http://mathforum.org/library/drmath/view/52901.htmlLong Division of Polynomials.http://www.sosmath.com/algebra/factor/fac01/fac01.htmlhttp://www.purplemath.com/modules/polydiv2.htmhttp://www.mathwords.com/p/polynomial_long_division.htmhttp://www.math.utah.edu/online/1010/euclid/http://tutorial.math.lamar.edu/AllBrowsers/1314/DividingPolynomials.aspPolynomial Division andFactoringhttp://campus.northpark.edu/math/PreCalculus/Algebraic/Polynomial/FactoringOnline Solutions. Scroll down thissite to the form identified asDivision of polynomials: p(x) /f(x). It will divide polynomials todegree 6. Just set the coefficients,and click "Divide". Be sure that thedegree of p(x) >= degree of f(x).Leading zero coefficients areignored – use zeros for terms in theform that are not present in theproblem you want to solve.http://www.egwald.com/linearalgebra/polynomials.phpDefinitionsMonomial has one term: 5y or -8x2or 3.Binomial has two terms: -3x2+ 2, or 9y- 2y2Trinomial has 3 terms: -3x2+ 2 +3x, or9y - 2y2+ yDegree Of The Term is the exponent ofthe variable: 3x2has a degree of 2.Degree of a Polynomial is the highestdegree of any of its terms.When the variable does not have anexponent -understand that theres a 1.One thing you will do whensolving polynomials is combine liketerms. Understanding this is the keyto accurately working withpolynomials. Let’s look at someexamples:• Like terms: 6x + 3x - 3x• NOT like terms: 6xy + 2x - 4The first two terms are like and theycan be combined:5x2+ 2x2– 3Combining like terms, we get:7x2– 3
• 9. 7272727Working with exponents is animportant part of algebra and otherclasses. Lets go over theprocedures of exponent rules indetail and review some examples.Rules of 1There are two simple "rules of1" to remember.First, any number raised to thepower of "one" equals itself. Thismakes sense, because the powershows how many times the base ismultiplied by itself. If its onlymultiplied one time, then its logicalthat it equals itself.Secondly, one raised to anypower is one. This, too, is logical,because one times one times one, asmany times as you multiply it, isalways equal to one.x1= x31= 31m= 114= 1 * 1 * 1 * 1 = 1Product RuleThe exponent "product rule"tells us that, when multiplying twopowers that have the same base, youcan add the exponents. In thisexample, you can see how it works.Adding the exponents is just a shortcut!xm* xn= xm+n42* 43= 4 * 4 * 4 * 4 * 442+5= 45Power RuleThe "power rule" tells us that toraise a power to a power, justmultiply the exponents. Here yousee that 52raised to the 3rd power isequal to 56.(xm)n= xm*n(52)3= 56Note: This is really where thePower of a Quotient rule comesfrom. Numbers without exponentsare assumed to be to the power of 1.(a/b)m= am/bm, b ≠ 0Quotient RuleThe quotient rule tells us thatwe can divide two powers with thesame base by subtracting theexponents. You can see why thisworks if you study the exampleshown.xm÷ xn= xm-n, x ≠ 045÷ 42= 4 * 4 * 4 * 4 * 4 / 4 * 44 5-2= 4 3Zero RuleAccording to the "zero rule," anynonzero number raised to the powerof zero equals 1.x0= 1, x ≠ 0Negative ExponentsThe last rule in this lesson tells usthat any nonzero number raised to anegative power equals its reciprocalraised to the opposite positivepower.x-n= 1/xn4-2= 1/42= 1/16Source: www.math.comMr. Breitsprecher’s Edition October 4, 2005 Web: www.clubtnt.org/my_algebraCommon Errors With Exponent RulesWith practice, we can all apply and work with these rules. Sometimes, we learn howto do things by making mistakes. Here are some common errors with exponents.Please review the following examples of how NOT to work with exponents.1. The exponent next to a number applies ONLY to that number unless there areparenthesis (grouping) that indicate another number or sign is actually part ofthe base.-52≠ (-5)2-(5)(5) ≠ (-5)(-5)-25 ≠ 252. The product rule (xm* xn= xm+n) only applies to expressions with the samebase.42* 23≠ 82+3(4)(4)(2)(2)(2) ≠ 85128 ≠ 32,7683. The product rule ((xm* xn= xm+n) applies to the product, not the sum of 2numbers22+ 23≠ 22+3(2)(2) + (2)(2)(2) ≠ 254+8 ≠ (2)(2)(2)(2)(2)12 ≠ 32Math.com has a variety of online lessons and interactive tutorials to help studentsmaster many of the concepts in our math. Check out the “workout” of 10 interactiveexponent examples with solution and helpful tips at:http://www.math.com/school/subject2/practice/S2U2L2/S2U2L2Pract.html
• 10. Everything we have done up tothis point in Beginning Algebra hasbeen to get ready to apply basicprocedures to more involvedalgebraic concepts such as factoringpolynomials.As a quick review, let’s startout by reviewing what factoring is.Factoring is a process to determinewhat we can multiply to get thegiven quantity. If means to write anas a product – the reverse ofmultiplication. In this example, wewant to rewrite a polynomial as aproduct.Examples: Factors of 12Possible Solutions: 2*6 or 3*4or 2*2*3 or [1/2(12)] or (-2*-6) or(-2*2*-3) Note: There are manymore possible ways to factor 12, butthese are representative of many ofthem.A useful method of factoringnumbers is to completely factorthem into positive prime factors. Aprime number is a number whoseonly positive factors are 1 and itself.For example 2, 3, 5, and 7 are allexamples of prime numbers.Examples of numbers that aren’tprime include 4, 6, and 12.If we completely factor anumber into positive prime factors,there will only be one solution.This is why prime factorization isuseful. That is the reason forfactoring things in this way. For ourexample above with, the completefactorization of 12 is: (2*2*3).Factoring PolynomialsFactoring polynomials issimilar, determine all the terms thatwere multiplied together to get theoriginal polynomial. For many of us,it is probably easiest to factor insteps – start with the first factors wesee and continue until we can’t factoranymore. When we cannot find anymore factors we will say that thepolynomial is completely factored.Example: x2-16 = (x+4)(x-4).This is completely factored, wecannot find another way that thefactors on the right can be furtherfactored (think: write as product).Example: x4-16 = (x2+4)(x2-4).This is not completely factored,because the second factor on theright can be further factored. Do yousee that it is one of our specialproducts, a difference of squares?Breaking that example down further,we see: x4-16 = (x2+4)(x+2)(x-2).Greatest Common FactorOne way to factor polynomials,probably the easiest one whenapplicable, is to factor out thegreatest common factor. In generalthis should ALWAYS be the firstmethod we consider wheneverfactoring.Start by looking at each termsand determine if there is a factorthat is in common to all the terms.If there is, factor it out of thepolynomial. Let’s take a look atsome examples.Example: 8x4-4x3-10x2Notice that we can factor out a2 from every term, we can alsofactor out an x2. Some of us willsee that we can factor out a 2x2immediately, but we can also factorthis in steps if that helps us getstarted. Our solution to thisproblem is: 2x2(4x2-2x+5).Example: x3y2+3x4y+5x5y3On closer inspection, we cansee that each term has a commonfactor has x3y. Of course, we mighthave seen this in steps, that eachterm has a factor of x3and then thateach has a factor of y. It doesn’tmatter, as long as we keep lookingfor factors. Our solution is:x3y(y+3x+5x2y2)Example: 3x6-6x2+3xDo you see that each term has acommon factor (greatest commonfactor) of 3x? Please remember,when we factor a 3x out of the lastterm (3x), we are left with +1. BeAcademic Support ServicesFREE Tutoring And Academic Support Services!!!Basement of McCutchan Hall, Rm. 1Mon-Thurs: 9 a.m. – 9 p.m.Fri: 9 a.m. – 3 p.m. and Sun 5 p.m. – 9 p.m.Mr. Breitsprecher’s Edition March 10, 2005 Web: www.clubtnt.org/my_algebra
• 11. Mr. Breitsprecher’s Edition March 10, 2005 Algebra Connections, Page 2careful, forgetting this 1 is acommon error when factoring! Oursolution is: 3x(x5-2x+1)Example: 9x2(2x+7)-12x(2x+7)At first glance, this one looksstrange. It is factored like the otherexamples, however. There is a 3x ineach term and there is also a (2x+7)in each term. Both can be factoredout of this polynomial. This leavesa 3x in the first term (because3x*3x=9x2) and a -4 in the second(3x*-4 gives us our original -12x).The solution is: 3x(2x+7)(3x-4)Factoring By GroupingThis is another way to factorpolynomials that have 4 terms. Itwill not work with every 4 termpolynomial, but when it can be used,it is a simple and direct approach. Itis also a great review of important,basic algebraic procedures. Here isa step-by-step procedure forfactoring 4 term polynomials bygrouping:1. Group terms into 2 groups of 2terms.2. Factor out the greatest commonfactor from each of these twogroups.3. If we are left with a commonbinomial factor, factor it out –then we are done!4. If not, rearrange the terms andtry steps 1-3 again.Example: 3x2-2x+12x-81. Group terms into 2 groups of 2terms: (3x2-2x) + (12x-8)2. Factor out the greatest commonfactor from each term:x(3x-2) + 4(3x-2)3. Factor out a common binomialfactor. Our solution is:(3x-2)(x+4).That’s it! We do not need todo step 4 because we had a commonbinomial term of (3x-2). Once wefactor it out, we are done.Example: x4+x-2x3-21. Group terms into 2 groups of 2terms. BE CAREFUL WITHPOLYNOMIALS WITH A “-” SIGN IN FRONT OF THE3RDTERM! The process is thesame, but notice that we have acommon factor in the 3rdand 4thterms of -1 or just “-”. If wefactor out this “-” and group thefirst 2 and last 2 terms together,we get: (x4+x)-(2x3+2). THISIS AN IMPORTANT STEPWHEN WE HAVE A “-”SIGN IN THE THIRDTERM.2. Factor out the greatest commonfactor from each term:x(x3+1)-2(x+1)3. Factor out a common binomialfactor. Our solution is:(x3+1)(x-2).Example: x5-3x3-2x2+61. Group terms into 2 groups of 2terms. Again, NOTICE THATWE HAVE A “-” IN FRONTOF THE THIRD TERM.Also note the “+” in front of the4thterm. We will still factor outthe ‘-’, so that we do not losetrack of it. This gives us:(x5-3x3) – (2x2-6).2. Factor out the greatest commonfactor from each term:x3(x2-3)-2(x2-3)3. Factor out a common binomialfactor. Our solution is:(x2-3)(x3-2).Example: 5x-10+x3-x21. Group terms into 2 groups of 2terms. Note that in this case,the 3rdterm is positive. Our 2groups are: (5x-10) + (x3-x2)2. Factor out the greatest commonfactor from each term:5(x-2)+(x2-1)3. Note that there is no commonbinomial factor. No groupingwill lead to a common factor –We cannot find a solution byfactoring by grouping.Example: 3xy+2-3x-2y1. Note that the first 2 terms haveno common factors other thanone. If we rearrange the terms,however, we can create 2groupings with commonfactors: (3xy-3x)+(-2y+2).NOTICE THAT WE HAVEA “-” IN FRONT OF THETHIRD TERM. Also note the“+” in front of the 4thterm. Wewill still factor out the ‘-’, sothat we do not lose track of it.This gives us: (3xy-3x)-(2y-2)2. Factor out the greatest commonfactor from each term: 3x(y-1)-2(y-1)3. Factor out a common binomialfactor. Our solution is:(y-1)(3x-2)Remember, factoring bygrouping can be efficient, but itdoesn’t work all that often. It is angood review of algebraicprocedures. BE CAREFUL whenthere is a “-” in front of the thirdterm. ALWAYS factor that out ofthe third and fourth terms whengrouping.Online ResourcesGCF & Factoring by Groupinghttp://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut27_gcf.htmInteract Tutorial: GCG &Factoring by Groupinghttp://www.mathnotes.com/intermediate/Mchapter05/aw_MInterAct5_7.htmlGCF & Factoring by GroupingGuide (.pdf file)http://online.math.uh.edu/Math1300/ch4/s41/ex41.pdfGCF & Factoring by Groupinghttp://online.math.uh.edu/Math1300/ch4/s41/GCF/Lesson/Lesson.htmlFactoring & Polynomialshttp://www.okc.cc.ok.us/maustin/Factoring/Factoring.htmlFactoring Strategieshttp://hhh.gavilan.cc.ca.us/ybutterworth/intermediate/ch5Angel.docShouldn’t your baby be a Gerbor baby?
• 13. Because the first term andcoefficient were perfect squares andthe middle term was positive, if thistrinomial is factorable, we have aperfect square trinomial that is the sumof two terms.Example: 4m2-4m = -1Step 1. Rewrite in standard form:4m2-4m+1 = 0. NOTE: This is aperfect square trinomial with anegative term; a2-2ab+b2= (a-b)2Step 2. Factor completely: (2x-1)2= 0Step 3. Set each factor containing avariable equal to 0: 2x-1 = 0. Note:In a perfect square trinomial, there isonly 1 factor.Step 4. Solve the resulting equation:x-1 = 0x = 1/2Because the first term andcoefficient were perfect squares andthe middle term was negative, if thistrinomial is factorable, we have aperfect square trinomial that is thedifference of two terms.Mr. Breitsprecher’s Edition March 29, 2005 Algebra Connections, Page 2Example: 12x3+5x2= 2xStep 1. Rewrite in standard form:12x3+5x2-2x = 0.Step 2. Factor completely. Note:first, factor out the greatest commonfactor (GCF) x(12x2+5x-2). Thenfactor the trinomial:x(4x-1)(3x+2)=0Step 3. Set each factor containing avariable equal to 0: x = 0, 4x-1 = 0,and 3x+2 = 0.Step 4. Solve the resulting equations:x = 0 (4x-1) = 0 (3x+2) = 04x = 1 3x=-2x = 1/4 x = - 2/3Don’t forget your GCF! THIS ISALWAYS THE FIRST STEPWHEN FACTORING. If is containsa variable (or is a variable as in thiscase), be sure to set it equal to zerotoo.Example: x(4x-11) = 3Step 1. Rewrite in standard form:4x2-11x-3 = 0. NOTE: Before werewrite this polynomial in standardform, we must simplify the equation.Remove the parenthesis by distributingthe x through the factor (4x-11). Then,rewrite the resulting equation to beequal to 0.Step 2. Factor completely (4x+1)(x-3).NOTE: This is not a “perfect squaretrinomial” the first term cannot befactored as (2x)(2x).Step 3. Set each factor containing avariable equal to 0:4x+1 = 0 and x-3 = 0Step 4. Solve the resulting equations:4x+1 = 0 x-3 = 04x = -1 x = 3x = -1/4The Fundamental Theorem ofAlgebraLook at the examples given.Compare the number of solutions withthe degree of the polynomial. Thenumber of solutions to any polynomialequation is ALWAYS less than orequal to the degree of the polynomial.This fact is known as the fundamentaltheorem of algebra.Additional Key Concepts: Factoring PolynomialsThe Greatest Common FactorFactoring is the process of writing an expression as aproduct.The GCF of a list of common variables raised topowers is the variable raised to the smallest exponent inthe list.The GCF of a list of terms is the product of all commonfactors.Factoring by Grouping:1. Group the terms into two groups of two terms.2. Factor out the GCF from each group.3. If there is a common binomial, factor it out.4. If not, rearrange the terms and try steps 1-3 again.Factoring Trinomials in the Form x2+bx+cx2+bx+c = (x+?)(x+?), where the numbers indicated by the“?” sum to “b” and the product of the numbers indicatedby the “?” is “c”Factoring Trinomials in the Form ax2+bx+cTo factor ax2+bx+c, try various combinations of factors ofax2and c until the middle term of bx is obtained whenchecking (the product of the outside term and the productof the inside terms sum to equal the middle term).Factoring Trinomials in the Form ax2+bx+c by Grouping1. Find two numbers whose product is a*c and whosesum is b2. Rewrite bhx, using the factors found in step 1.3. Factor by grouping.Factoring Perfect Square Trinomials (Trinomials that arethe square of some binomial)a2+2ab+b2= (a+b)2a2-2ab+b2= (a-b)2Difference of Two Squaresa2-b2= (a+b)(a-b)Online ResourcesQuadratic Equationshttp://www.mathpower.com/tut99.htmQuadratic Equations: Solutions by Factoringhttp://www.sosmath.com/algebra/quadraticeq/sobyfactor/sobyfactor.htmlhttp://www.mathpower.com/tut105.htmhttp://www.mathpower.com/tut110.htm
• 17. Let’s look at factoringtrinomials with a “leadingcoefficient other than 1” We canwrite that form as ax2+ bx + c,where a, b, and c are integers.Like factoring any otherpolynomial, the first thing to do is tofactor out all constants which evenlydivide all three terms (GreatestCommon Factor). If “a” is negative,factor out -1.This will leave an expression inthe form: d(ax2+ bx + c), where a,b, c, and d are integers, and a > 0.Now we have a factor (GCF) times atrinomial in the form ax2+ bx + c.The next step is to factor theremaining trinomial. We willpresent 2 methods.The first method, which will bereferred to as “Old SchoolAlgebra.” NOTE THAT THIS ISNOT THE METHOD THAT ISPRESENTED IN OURTEXTBOOK. It is similar to thethe method presented in ourtextbook (unit 4.3), but uses adifferent set of “tests” utilizingabsolute values when “c” is positiveand when “c” is negative.It is presented here as analternative perspective – studentsneed not study this method. Likethe method presented in our text,unit 4.3, it comes down to “trial-by-error.”Our text presents a second,similar method, which I call “Guessand By Golly” (Unit 4.3). Like the“Old School” method, it involvesdetermining possible factors andchecking them until one works.Our text’s method differs in how itlooks at positive and negative valuesfor “c.” Rather than creating a set ofrules about when “c” is positive andwhen “c” is negative, “Guess and ByGolly” relies on an understanding ofpatterns with signs.The third method, “Factoringby Grouping,” (Unit 4.4 in our text)works by rewriting our trinomialinto a 4 term polynomial that can befactored by groups.Old School AlgebraHere is how to factor anexpression ax2+ bx + c, where a > 0(“a” is positive):1. Write out all the pairs ofnumbers that, when multiplied,produce “a” (i.e. a1*a2=a,a3*a4=a, etc.).2. Write out all the pairs ofnumbers that, when multiplied,produce “c” (i.e. c1*c2=c,c3*c4=c, etc.).3. Pick one of the a pairs, (a1, a2),and one of the c pairs, (c1, c2).4. If c > 0 (“c” is positive):Compute a1c1 + a2c2. If | a1c1 +a2c2 | = b, then the factored formof the quadratic is:(a1x + c2)(a2x + c1) if b > 0(“b” is positive).(a1x - c2)(a2x - c1) if b < 0(“b” is negative).5. If a1c1 + a2c2 ≠ b, compute a1c2+ a2c1. If a1c2 + a2c1 = b, thenthe factored form of thequadratic is (a1x + c1)(a2x + c2)or (a1x + c1)(a2x + c2). If a1c2 +a2c1 ≠ b, pick another set ofpairs.6. If c < 0 (is negative). Computea1c1 -a2c2. If | a1c1 - a2c2 | = b,then the factored form of thequadratic is (a1x - c2)(a2x + c1)where a1c1 > a2c2 if b > 0 (ispositive) and a1c1 < a2c2 if b <0 (is negative).7. Using FOIL, the outside pairplus (or minus) the inside pairmust equal b. This is a check.Example 1: 3x2- 8x + 4Numbers that produce 3: (1, 3).Numbers that produce 4: (1, 4),(2, 2).Continuing:(1, 3) and (1, 4): 1(1) + 3(4) =11≠8. 1(4) + 3(1) = 7≠ 8.(1, 3) and (2, 2): 1(2) + 3(2) = 8.(x - 2)(3x - 2).Check: (x - 2)(3x - 2) = 3x2 -2x -6x + 4 = 3x2 - 8x + 4.Example 2: Factor 12x2+ 17x + 6Numbers that produce 12: (1, 12),(2, 6), (3, 4).Numbers that produce 6: (1, 6),(2, 3).Continuing:(1,12) and (1,6): 1(1)+12(6) = 72.Academic Support ServicesFREE Tutoring And Academic Support Services!!!Basement of McCutchan Hall, Rm. 1Mon-Thurs: 9 a.m. – 9 p.m.Fri: 9 a.m. – 3 p.m. and Sun 5 p.m. – 9 p.m.Mr. Breitsprecher’s Edition March 16, 2005 Web: www.clubtnt.org/my_algebra3Methods,YouChoose!
• 19. Factoring Trinomials in theform, where “b” and “c” are integersis a good place to start factoringtrinomials. Notice that the firstcoefficient (x`) is 1. Let’s look atthe patterns and procedures we willneed to solve these problems. Thiswill help us when we will looktrinomials where the leadingcoefficient is other than 1.When we factor trinomials witha lead coefficient of 1, we will endup with a factorization in the formof 2 binomials: (x+?)(x+?). If wecan determine the numbers thatreplace the “?” in each binomial, wehave factored the trinomial.Remember, IF THETRINOMIAL IS FACTORABLE(x2+bx+c), there are 2 numberswho’s SUM is “b” and who’sPRODUCT is “c” Just look for apair of numbers that will sum to “b”and then check their product to seeif it is “c.” This should determine ifthe trinomial is factorable and, if so,will fill in the missing terms of our 2binomials: (x+?)(x+?).Alternatively, find 2 numberswho’s product is “c” and then checktheir sum to see if it is “b.” Eitherway, once we see these “patterns,”we can use them to help identify themissing terms when we factor atrinomial in the form of of x2+bx+cinto (x+?)(x=?). We will focus onthis alternative perspective.The following steps can be usedto factor trinomials in the formx2+bx+c:1. If possible, factor out anygreatest common factor (otherthan 1 or –1)2. Look at each possible sets of 2factors that will result in “c”3. Add each of those 2 factors andidentify which set sums to “b”4. Write our binomial factors byfilling in the missing terms:(x+?)(x+?)Patterns in the SignsIn many ways, algebra is allabout seeing patterns in numbers.When factoring trinomials in theform of x2+bx+c, look at the signsof “b” and “c” and notice:• If “c” is positive, then thenumbers we need to complete ourfactorization [(x=?)(x+?)] havethe same sign – they are bothpositive or negative. We see:If “b” is positive, then bothfactors of “c” must be positive.If “b” is negative, then bothfactors of “c” must be negative.• If “c” is negative, then one of thenumbers we need to complete ourfactorization [(x=?)(x+?)] must bepositive and the other must benegative (two opposite signs).Factor Out GCF FIRST!Don’t forget, wheneverfactoring a polynomial – ALWAYSFACTOR OUT ANYGREATEST COMMONFACTOR before attempting anyother factorization.Example: x2+6x+8Note that “c” is positive (8).This tells us (if the trinomial isfactorable) that the numbers neededto complete our factorization[(x+?)(x+?)] must have the samesign. Because “b” is positive, weknow that we are looking at twofactors of “c” (or 8) that have to bepositive.We now know our choices are(2*4) and (1*8). We need to lookat which of these factors (if any)will sum to “b,” in this case, 6.Note that (2+4=6) and (1+8=9).We now know that this trinomial isfactorable and our solution is(x+2)(x+4).Example: x2-11x+10Again, “c” is positive (10).This tells us (if the trinomial isfactorable) that the numbers neededto complete our factorization[(x+?)(x+?)] must have the samesign. This time, because “b” isAcademic Support ServicesFREE Tutoring And Academic Support Services!!!Basement of McCutchan Hall, Rm. 1Mon-Thurs: 9 a.m. – 9 p.m.Fri: 9 a.m. – 3 p.m. and Sun 5 p.m. – 9 p.m.Mr. Breitsprecher’s Edition March 15, 2005 Web: www.clubtnt.org/my_algebra
• 20. Mr. Breitsprecher’s Edition March 15, 2005 Algebra Connections, Page 2negative (-11), we know that we arelooking at two factors of “c” (or 10)that have to be negative.We now know our choices are(-2*-5) or (-1*-10). Now we needto look at which of these factors (ifany) will sum to “b,” in this case, -11. Note that (-2 + -5 = -7) and (-1+ -10 = -11). We now know thatthis trinomial is factorable and oursolution is (x + -1)(x + -10). Wewould rewrite this. Our solution is(x-1)(x-10)Example: x2-3x-10Here we see that “c” is negative(-10). This tells us (if the trinomialis factorable) that the numbersneeded to complete our factorization[(x+?)(x+?)] must have different oropposite signs (one positive, theother negative). We now know thatour choices are (-1 * 10), ( 1*- 10),(-2*-5), or (-2*5). Now we look ateach of these factors to see which (ifany) will sum to “b,” in this case –3.The only pair of these factors thatwill sum to –3 is (2+ -5). Therefore,this trinomial is factorable as(x+2)(x+ -5). We would rewritethis. Our solution is (x+2)(x-5).Example: 2x3-24x2+64xNote that this one, at firstglance, does not look like atrinomial in the form x2+bx+c; but itis! Remember to always look for agreatest common factor BEFOREattempting any other factorization ofa polynomial. In this case, we havea GCF of 2x. We would start byrewriting this polynomial as 2x(x2-12x+32). Now we can factor it likethe other examples.Note that “c” is positive (32).This tells us (if the trinomial isfactorable) that the numbers neededto complete our factorization[(x+?)(x+?)] must have the samesign. Because “b” is negative (-12),we know that we are looking at twofactors of “c” (or 32) that have to benegative.Our choices are (-1 * -32),(-2* -16) or (-4 * -8). Only the lastof these three choices will sum toour “b” term (-4 + -8 = -12)Therefore, this trinomial isfactorable (DON’T FORGET OURGCF) as 2x(x+ -4)(x+ -8). Wewould rewrite this. Our solutionis 2x(x-4)(x-8).Example: 4x2+36x+80Again, at first glance, this doesnot look like a trinomial in the formx2+bx+c; but it is! Remember toalways look for a greatest commonfactor BEFORE attempting anyother factorization of a polynomial.In this case, we have a GCF of 4.We would start by rewriting thispolynomial as 4(x2+9x+20). Nowwe can factor it like the otherexamples.Note that “c” is positive (20).This tells us (if the trinomial isfactorable) that the numbers neededto complete our factorization[(x+?)(x+?)] must have the samesign. Because “b” is positive (9),we know that we are looking at 2factors of “c” (or 20) that have to bepositive.We know our choices are(1*20), (2*10) or (4*5). Only thelast of these three choices will sumto our “b” term (4+5=9) Therefore,this trinomial is factorable (don’tforget our GCF) and our solution is4(x+4)(x+5).Not Just “x”Just because we illustrate theform of a trinomial with a degree of2 and “leading coefficient of 1” as:x2+bx+c, any variable can bereplace the “x,” (i.e. “w” or “z”) justas any number can be the coefficientof our 2ndterm (we called it “b”) andany number can be our constant (wecalled it “c”).Online ResourcesIn most cases, you can go to the“domain” (i.e. http://domainname.ext)and follow links to the pagesidentified. ASK ME TO SENDTHEM TO YOU VIA EMAIL ASLINKS IF YOU HAVE ANYPROBLEMS!Factoring Power Pointshttp://students.loyola.ca/classes/powellt/Math4_5/factring.ppthttp://www.tcc.edu/vml/Mth03/Trinom/documents/FactoringTrinomials.ppsSimple Trinomials as Product ofBinomials (Many examples, printable)http://www.math.bcit.ca/competency_testing/testinfo/testsyll11/basicalg/basops/factoring/trinom/trinom.docFactoring Trinomials (Somepresentations do not treat trinomialsin the form of x2+bx+c, or leadingcoefficient of 1, as different fromax2+bx+c. We have started with thesimplest form, before introducingother trinomials.http://www.mathmax.com/introalg/chapter/bk3c5im.htmlhttp://www.coolmathalgebra.com/Algebra1/10FactDivPolys/04_undoingFOIL.htmhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut28_facttri.htmhttp://www.jamesbrennan.org/algebra/polynomials/factoring_polynomials.htmOnline Factoring Solutions (Enterexponents as x^2)http://www.webmath.com/factortri.htmlhttp://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=algebra&s2=factor&s3=basic (Note: Domain forthis one is http://www.quickmath.com/)Factoring. Process of writing an expression as a product.Greatest Common Factor. Product of ALL commonfactors from a set of terms. When there are commonvariable factors in terms, use the common variable(s) raisedto the smallest exponent(s) of the in the terms.Special Products(a+b)2= a2+ 2ab + b2(a-b)2= a2– 2ab + b2(a-b)(a+b) = a2– b2Common Errors(a+b)2≠ a2+ b2(a-b)2≠ a2– b2We forgot middle terms!
• 21. Before working with formulas,let’s look at some units ofmeasurement – many formulas willbe based on units. In the US, we usefeet, yards, and miles.12 inches = 1 foot; 3 feet = 1yard; 5,280 feet = 1 mile. Unitsused in formulas must be the same –so changing from one unit intoanother equivalency is important.Here are the conversions:Feet to Inches. Number offeet * 12Inches to Feet. Number ofinches/12Yards to Feet. Number ofyards * 3Miles to Feet. Numbermiles * 5,280Metrics are used throughout theworld and are universally acceptedin the sciences. These measures areall based on a system of 10’s – wewill look at them in another editionof Mr. Breitsprecher’s AlgebraConnections.Often, proportions are the bestway to solve measurementproblems. Recall that a proportionis a statement that 2 ratios are equal.The key is to use the same units ineach ratio, identify 3 of the figuresin the proportion, and solve for thefourth. The cross product is animportant shortcut to help us set upthese problems (see previous editionMr. Breitsprecher’s AlgebraConnections for more on this)Once we are comfortableworking with units and performingconversions, we are ready to workwith formulas. There is nothingnew here – identify the knownquantities and solve for theunknown using the algebraicprocedures reviewed in class. Hereare some common formulas that willbe used in many practical situationsand in Algebra.PerimeterThe distance around the outsideof a given area, or 2 dimensionalshape, is called the perimeter.Think of it as the total length of theborder around that shape. Here aresome common perimeter formulas.Perimeter of a Triangle. Aclosed figure with three straightsides is called a triangle. We canrefer to each side as side 1, side 2,and side 3. The perimeter is the sumof these sides. The formula is:P = s1+s2+s3Perimeter of a Square. A 4-sided, closed figure with each cornermeasuring 90 degrees (right angle)is a square. While we could expressthe formula for the perimeter as thesum of the 4 sides, this would lookawkward – it is easier to express itin terms of multiplication. Theformula is: P = 4sPerimeter of a Rectangle. A4-sided, closed figure with eachcorner measuring 90 degrees (rightangle) is a rectangle. We could alsoexpress its perimeter as the sum ofthe sides, but that would also lookawkward – it is easier to express itin terms of multiplication. Theformula is: P = 2l (length)+2w(width).Perimeter of a Circle. Acircle has a perimeter, but we call itthe circumference. This wordderives from circumstance – thinkof the situations around you. Thereare no sides to measure, so we usethe diameter (D) – a line drawnthrough the center point of a circlethat touches both sides of thefigure. Sometimes, we are onlygiven the radius (r), which is halfthe diameter (D/2). We need aspecial mathematical figure for thisformula, π. For most purposes, wecan approximate this with 3.14 –though it really will go on and onforever without repeating. Whenconvenient, we can also use thefraction 22/7. The formula for thecircumference of a circle is: C=πDAreaA measurement of how many2-dimensional units (squares) aparticular object or surface coversis called the area. Think of it as theflat space a shape occupies. It ismeasured in square units, usuallysquare inches, square centimeters,square feet, or square miles, and soforth. Imagine a floor covered withsquare tiles that are each 1-foot x 1foot. If we count the tiles, we havethe area in square feet.Academic Support ServicesFREE Tutoring And Academic Support Services!!!Basement of McCutchan Hall, Rm. 1Mon-Thurs: 9 a.m. – 9 p.m.Fri: 9 a.m. – 3 p.m. and Sun 5 p.m. – 9 p.m.Mr. Breitsprecher’s Edition October 3, 2005 Web: www.clubtnt.org/my_algebra
• 23. A fraction has 2 main parts, thenumerator (top) and denominator(bottom). The line between the twois called the “fraction bar.” Need alittle “trick” to keep this straight?Just think “N” for North (up),numerator; “D” for Down,denominator.DN=43The denominator (down) orlower part of a fraction indicateshow many equal portions or piecesof the whole there are. Note thatwhen we divide a pizza into ½, eachpiece or half is of equal size.The same is true if we cut achunk of fudge into 4 pieces or ¼slices. Each piece of fudge will beequal. The numerator (north or up)refers to how many of those equalpieces the fraction represents.Types of FractionsWhen we start dividing wholethings into pieces, we have threesituations: proper, improper, andmixed fractions. Let’s review each.Proper Fractions. Thenumerator is always smaller than thedenominator. Most of us think ofthis type of fraction when thinkingof parts of a whole. This type offraction always represents less than1. This is the simplest type offraction to work with.Examples: Proper Fractions17612587ororImproper Fractions. Thenumerator is always larger than thedenominator. This fraction seems“top-heavy” or an odd way toexpress parts of a whole, becausethe fraction is actually greater than1. The denominator (down) stilltells us what size the equal partsare. The numerator (north or up)tells us that we have more thanenough pieces to make a whole –this fraction always representssomething greater than 1. This typeof fraction can be useful to workwith in many situations.Examples: Improper Fractions72435712ororMixed Fractions. Looking atimproper fractions can beconfusing, that’s why we havemixed fractions. They clearlyindicate how many parts of thewhole there are with proper fractionand use a whole number to indicatehow many complete “wholes” are.This is much easier for most of usto interpret than when looking atMr. Breitsprecher’s Edition March 3, 2005 Web: www.clubtnt.org/my_algebraAcademic Support ServicesFREE Tutoring And Academic Support Services!!!Basement of McCutchan Hall, Rm. 1Mon-Thurs: 9 a.m. – 9 p.m.Fri: 9 a.m. – 3 p.m. and Sun 5 p.m. – 9 p.m.½¼Lowest Common Multiple = Lowest Common DenominatorThe Lowest Common Multiple (LCM) is the smallest number that is a commonmultiple of two or more numbers. When working with fractions, this is also theLowest Common Denominator (LCD). For example, the L.C.M of 3 and 5 is 15. Thesimple method of finding the L.C.M of smaller numbers is to write down the multiplesof the larger number until one of them is also a multiple of the smaller number.Example: Find the Lowest Common Multiple of 8 and 12.Solution: Multiples of 12 are 12, 24... 24 is also a multiple of 8, so the L.C.M of 8and 12 is 24.Example: LCM/LCD of Big Numbers. Find all the prime factors of both numbers.Multiply all the prime factors of the larger number by those prime factors of the smallernumber that are not already included. (Note: this sounds different than our textbook;but if you think about it, it is really the same thing. We end up with prime factors fromeach number and use each prime factor the number of times it appears the most).Example: Find the Lowest Common Multiple of 240 and 924.Solution: Write the prime factorization of each. 924 = 2 x 2 x 3 x 7 x 11 and240 = 2 x 2 x 2 x 2 x 3 x 5, therefore the lowest common multiple is:(2 x 2 x 3 x 7 x 11) x (2 x 2 x 5) = 924 x 20 = 18,480
• 24. improper fractions.Examples: Mixed Fractions81937167321 ororSimplifying Fractions`The key to working with fractionsis to express them as simply aspossible. It is easiest to work with thesmallest possible numerator (north –up) and denominator (down). Thismakes it easier to visualize andperform mathematical operations. Afraction is in its lowest terms whenboth the numerator and denominatorcannot be divided evenly by anynumber except 1.Equivalent Fractions. If thenumerator and denominator of anyfraction are multiplied by the samenumber, the fraction looks differentbut really represents the same part ofthe whole. We know that any singlenumber multiplied by 1 is equal to thenumber we started with.Note that any fraction that has thesame numerator and denominator isequal to one – this is what we reallyhave when we multiply the numeratorand denominator by the same number.We have equivalent fractionswhen we express a fraction in 2different ways, both of which can besimplified to the same lowest terms.We also create an equivalent fractionwhen we divide the numerator anddenominator by the same number – inthis case, we are expressing the samefractions with smaller numbers.Examples: Equivalent Fractions10584634221====Simplest TermsKeeping fractions easy to look at,visualize, and work with is important.That is what we are doing when wesimplify fractions. If we can divideboth the numerator and denominatorby the same number (just another wayof expressing 1), we reduce thenumbers used to express that fraction.This makes it simpler to work with.When we reduce the numeratorand denominator by dividing each bythe same number and cannot reduce itMr. Breitsprecher’s Edition March 3, 2005 Algebra Connections, Page 2any further, we have expressed thatfraction in simplest terms.Examples: Simplest Terms4333129=÷5412126048=÷Common DenominatorsIf we are comparing 2 differentfractions (which is larger and which issmaller), we need to think of eachfraction in terms of the same, equalparts of the whole (denominator).When we add or subtract fractions, weneed to work with fractions that areexpressed with the same denominator– a common denominator.Rewriting fractions with thelowest common denominator is just aspecial example of writing anequivalent fraction – in this case, wefind equivalent fractions that share thesame denominator.Finding the Common DenominatorHere is a step-by-step method forfinding common denominators thatwill work every time:1. Examine the fractions – can wesee by observation what thecommon denominator is? If so –skip steps 2 and 3, goimmediately to step 4. The morewe practice working withfractions, the easier this stepbecomes.2. If we did not see anythingobvious in step 1, determinewhich fraction has the largestdenominator.3. Check to see if the smallerdenominator divides into thelarger one evenly. If so, move onto step 4. If not, check multiplesof the larger denominator untilyou can find one that the smallerdenominator can divide intoevenly. If you remember ourearlier lesson about lowestcommon multiples – that processwill take you immediately to thelowest common denominator(LMC=LCD).4. Write the 2 fractions asequivalent fractions with thecommon denominator.Note that we can always find acommon denominator by multiplyingthe denominators together – while thisworks, it can result in working withlarge numbers. Once this number isfound, go to step 4. If you cancomfortably work with these largernumbers, this method can be efficient.Examples: Common Denominators146&1472273&772173&21××=3015&3016151521&2215821&158××=Improper Fractions & MixedFractionsWhile working with an improperfraction is no different than workingwith a mixed fraction, most of us willfind it easier to interpret our results ifwe convert the improper fraction to amixed fraction (whole number &proper fraction).Divide the numerator (north = up)by the denominator (down) – thenumber of times the denominator goesevenly into the numerator is our“whole” number. The remainder, orparts that are left-over, are thenumerator. Now we have are left witha fraction that represents less than oneand a whole number for everythingelse.921911=Note: 11 divided by 9 is 1with a remainder of 3.1024010402=Note: 402 divided by 10 is 40with a remainder of 2.744732=Note: 32 divided by 7 is 4with a remainder of 4.
• 25. Graph of (3, 4)Quadrant IQuadrant IIQuadrant III Quadrant IVMr. Breitsprecher’s Edition November 10, 2005 Web: www.clubtnt.org/my_algebraA rectangular coordinatesystem is a plane with vertical andhorizontal number lines thatintersect at their 0 coordinate. Thevertical line is referred to as they-axis; the horizontal line, the x-axis. We call the point ofintersection the origin (0, 0).We graph or plot an orderedpair by locating the corresponding xand y values on our rectangularcoordinate system. To plot x=3 andy=4 (3, 4); start at the origin andmove right 3 and then up 4.The rectangular coordinatesystem is divided into quadrants.The quadrants are:Quadrant x-axis y-axisI Positive PositiveII Negative PositiveIII Negative NegativeIV Positive NegativeAn ordered pair is a solution ofan equation in 2 variables ifreplacing the variables with thecoordinates of the ordered pairresults in a true statement.If we are given one coordinateof an ordered pair solution, the othervalue can be determined bysubstitution. For example: x-4y=16Start by assuming x equalssome convenient number towork with, say 0 (0, y). Bysubstitution, we have:0-4y = 16-4y=16(-4y)/-4 = 16/-4y = (-4)Our ordered pair is (0, -4)A linear equation in twovariables is an equation that can bewritten in the form Ax+By=C,where A and B are are not both 0.We call the form Ax+By=Cstandard form.To graph a linear equation intwo variables, find three orderedpairs that are solutions for theequation. Two points (each anordered pair) determine the line.We use the third point as our“check.” When we plot the threepoints, a straight line should connectall three points.Graphing Linear Equations: InterceptsAn intercept point of a graph is the point where the line crosses an axis. The x-intercept is the point where a line crosses the x-axis. If that point is some number, let’scall it “a,” then the x-intercept is “a” and the corresponding intercept point is (a, 0). Ifa graph intersects the y-axis at a point we call “b,” then b is the y-intercept and thecorresponding intercept point is (0, b). To find intercept points:x-intercept point is determined by letting y=0 and solving for xy-intercept point is determined by letting x=0 and solving for xExample: 5x + 2y = 10We find the x-intercept bysetting y = 0 and solvingfor x.5x + 2(0) = 105x=10(5x)/5 = 10/5x = 2The x-intercept is (2, 0)Next, we will find they=intercept by settingx = 0 and solving for y.5(0) + 2y = 102y = 10(2y)/2 = 10/2y = 5The y-intercept is (0, 5)
• 26. Mr. Breitsprecher’s Edition November 10, 2005 Algebra Connections, Page 2Academic Support ServicesFREE Tutoring And Academic Support Services!!!Basement of McCutchan Hall, Rm. 1Mon-Thurs: 9 a.m. – 9 p.m.Fri: 9 a.m. – 3 p.m. and Sun 5 p.m. – 9 p.m.Linear Equations: SlopeThe slant or steepness of a line is refered to as slope. Theslope (m) of a line passing through points (x1, y1) and (x2, y2)can determined by:rise change in y y2 – y1m = run change in x x2 – x1Example: Assume a line passes through (2, 1) and(4, 4).Solving for m, we get:4 - 1m = 4 - 23m = 2Graphing this line, we can see how the slope indicates howmany units the line’s rise (y2 – y1) and run (x2 – x1)It makes no difference what 2 points of a line we choose tofind its slope. The slope of a line is the same everywhere onthe line. Note that when we calculate the slope, it makes nodifference what points we assign as (x1, y1) and (x2, y2) as longas we make sure that when we call one point x1, we use itscorresponding y-coordinate as point y1. A positive slope goesup; a negative slope goes down (from left to right).A horizontal line has a slope of 0, because the “rise” iszero and its run is infinite – by definition, performing thisdivision results in 0.A vertical line has a slope that is undefined, because the“rise” is infinite and its run is 0 – by definition, division by 0 isundefined.Nonvertical parallel lines have the same slope. Twononvertical lines are perpendicular if the slope of one is thenegative reciprocal of the slope of the other.Linear Equations: Intercepts(Continued from page 1)Special Case: Graph of x = c is a vertical line withx-intercept of “c.” In this case, c = 3.Special Case: Graph of y = c is a vertical line with x-intercept of “c.” In this case, c = 3.
• 27. Mr. Breitsprecher’s Edition November 14, 2005 Web: www.clubtnt.org/my_algebraA graph of a straight line is calleda linear equation in 2 variables.They can always be written in the formAx+By = C where A and B are notBOTH 0. Note that a linear equationmay appear to have only one variable(x or y). This means that the othervariable has a coefficient of 0. We callthe form Ax+By = C standard form.A linear equation defines arelationship between 2 variables, x andy, and is called a relation. The set ofall x-coordinates is called the domainof the relation. The set of all y-coordinates is called the range of therelation.A function is a set of orderedpairs that assigns each x-valueprecisely one y-variable. In otherwords, each x-value predicts a y-value.When setting up linearequations, it is important establishthe relationship so that each value ofy is dependent on the value of x.This results in each value of y beinga “function” of x.Function notation uses thesymbol f(x) means function of x. forexample: f(x) = 3x-7.f(-1) = 3(-1)-7 = -10In other words, f(-1) means tosubstitute that value (-1) for x anddetermine the corresponding y-value.Because function notation meansthat each x-value has precisely 1 y-value, there can only be 1 value of yfor each value of x. In other words, avertical line can always be drawnthrough and it will only intersect alinear equation 1 time – this is calledthe vertical line test. A functioncannot curve back on itself or containany type of angle that results in a valueof x to “predict” more than 1 y.Three Useful Forms for LinearEquation in 2 VariablesWe will use 3 forms to representlinear equations in 3 variables:1. Standard Form: Ax+By = C,where A and B are not both 0.2. Slope-Intercept Form: y = mx+b,where m is the slope of the line andb is the y-intercept (0, b).3. Point-Slope Form: y-y1 = m(x-x1),where m is the slope and x1, y1 is apoint on the line.Note that standard form is themost general equation – recall that themultiplication property of equalitytells us we can multiply BOTH SIDESof an equation in standard from bysome number and not change theidentity of the equation (line).In other words, A, B, and C willchange and each will still be valid.For this reason, standard form haslimited practical applications.The slope-intercept form,however, will always have uniquevalues of m (slope) and b (x-intercept)for each line.What Does Slope MeanSlope tells us the slant or tilt of a line – it is defined as rise over run and can beexpressed as:rise change in y y2 – y1m = run change in x x2 – x1Note that when we take an equation in standard form (Ax+By = C) and solve it fory, the resulting coefficient of x is the slope – THIS IS NOT THE SAME AS SAYINGTHAT” A” IS THE SLOPE – IT CERTAINLY IS NOT!Please see the examplebelow:10x-5y = (-5)-10x+10x-5y = (-5)–10x-5y = (-10)-10x(-5y)/(-5) = [(-5)-2x]/(-5)y = 1+2xy = 2x+1Note that we nowhave a slope of 2, whichis not the coefficient Afrom our standard form.When we solve any equation in standard form, the resulting coefficient of x isactually the slope of the line. Please see the examples above.
• 28. Important DefinitionsRectangular Coordinate System. Plane with vertical and horizontal number lines thatintersect at their 0 coordinate.X-axis. The number horizontal lineY-axis. The number vertical lineOrigin. Where the lines cross, represented by the point x=0, y=0Ordered Pair. A pair of coordinates, one representing “x” and one representing “y.” Byconvention, written in the format of (x,y).Relation. Set of ordered pairsDomain. Set of all x-coordinatesRange. Set of all y-coordinatesFunction. Set of ordered pairs that assigns to each x-value exactly one y-valueVertical Line Test. If a vertical line intersects a graph more than once, the graph is not afunctionOnline ResourcesThe Coordinate Plane and Graphing Linear Equations. In this unit well be learningabout equations in two variables. A coordinate plane is an important tool for workingwith these equations. http://www.math.com/school/subject2/lessons/S2U4L1GL.htmlTutorial: Graphing Linear Equations. When you graph linear equations, you will endup with a straight line. Lets see what you can do with these linear equations.http://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut21_graph.htmGraph of a Line. We can graph the linear equation defined by y = x + 1 by findingseveral ordered pairs. For example, if x = 2 then y = 2 + 1 = 3, giving the ordered pair (2,3). Also, (0, 1), (4, 5), (-2, -1), (-5, -4), (-3, -2), among many others, are ordered pairs thatsatisfy the equation. http://www.algebra-online.com/graph-lines-1.htmGraphs of Linear Equations: Lines and Slope. Lessons, some practice setshttp://www.math.com/school/subject2/lessons/S2U4L1GL.htmlGraphing Inequalities in 2 Variables. The solution set for an inequality in twovariables contains ordered pairs whose graphs fill an area on the coordinate plane called ahalf-plane. An equation defines the boundary or edge of the half-plane.http://www.algebra-online.com/graphing-inequalities-1.htmWorksheets: Linear Equations and Inequalities. Need more practice? Theseworksheets will help. http://www.edhelper.com/LinearEquations.htmThinkQuest: Graphine Linear Equations and Inequities. Let’s review basics ofgraphic equations and inequalities. http://library.thinkquest.org/10030/6gleai.htmCreate Graphs ofEquations and InequalitiesAutomatic Graph Solutions: Equations. Enter the equation you want to plot, in termsof the variables x and y, set the limits and click the Plot button.http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=graphs&s2=equations&s3=basicAutomatic Graph Solutions: Inequalities. Enter the polynomial inequality you wantto plot, in terms of the variables x and y, set the limits and click the Plot button.http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=graphs&s2=inequalities&s3=basicLikewise, the point-slope formwill always have a unique value for m(slope) and x1, y1 must be points onthe line.Graphing Linear Inequalities inTwo VariablesA linear inequality is two variables isan inequality that can be written inone of the following forms:Ax+By < CAx+By ≤ CAx+By > CAx+By ≥CThe solution set for an inequalityin two variables contains orderedpairs whose graphs fill an area on thecoordinate plane called a half-plane.An equation defines the boundary oredge of the half-plane.Graphing Linear Inequalities1. Graph the boundary line bygraphing the related equation.Draw the line solid if theinequality symbol is ≤ or ≥.Draw the line dashed if theinequality is < or >.2. Choose a test point not on theline. Substitute its coordinatesinto the original inequality.3. If the resulting inequality is true,shade the half-plane that containsthe test point. If the inequality isnot true, shade the half-pane thatdoes not contain the test point.Example: y-2x ≤ 1The boundary would be y =2x+1. Choose easy to work withvalues for x and find corresponding y-values, i.e. (0, 3), (1, 3), and (3, 7).Remember, 2 points determine a line.The third point is a “check.” Wewould graph this as a solid line; ourboundary is included in our half-plane.To determine which half-plane,represents our solution, choose a testpoint; say (0, 0). Substituting thisback in our original inequality:y-2x ≤ 1, we see 0-2(0) ≤ 1 resultsin 0 ≤ 1. This is a true statement,so we shade that half-plane.Mr. Breitsprecher’s Edition November 14, 2005 Algebra Connections, Page 2
• 30. In order to understand linearequations in 1 variable, recall that avariable is a number that is notidentified. It is often represented by"x" or "y," any letter can be used. Alinear expression is a mathematicalstatement that performs functions ofaddition, subtraction, multiplication,and division, but has no exponents(or powers) and no variables thatmultiply or divide each other. Someexamples of linear expressionsinclude:• x + 4• 2x = 4• 2x + yThe following examples ARENOT linear expressions:• x2• 2xy + 4• 2x / 4yA linear equation is amathematical statement that has anequal sign and linear expressions.Examples include:• 2x + 4 = 10 (linearequation in 1 variable)• 3x - 4 = -10 (linearequation in 1 variable)• 4x - 4y = 8 (linearequation in 2 variables)Solving linear equationsinvolves applying logical steps tosimplify the expressions while stillmaintaining the equality andoriginal identity or solution. Eachstep creates an equivalent equation,not a new one. Like mostprocedures, this will be easiest tolearn if we establish a sequence ofsteps and practice using them untilwe are comfortable.Most of us would agree thatmath problems are easier to workwith if we ELIMINATE ORCLEAR ALL FRACTIONS first.To SOLVE LINEAREQUATIONS IN ONEVARIABLE, practice the following6-step process:1. Multiply both sides of theequation to CLEARFRACTIONS if they occur –multiply both sides of theequation by the lowest commondenominator.2. Use the distributive property toREMOVE PARENTHESES ifthey occur. This will oftencreate “like terms.”3. Simplify both sides of theequation, COMBINE LIKETERMS.4. Get all terms containingvariables, all VARIABLETERMS ON ONE SIDE OFTHE EQUATION and allnumbers on the other side byusing the addition property ofequality.5. GET THE VARIABLEALONE by using themultiplication property ofequality to remove thecoefficients of any term with avariable.6. CHECK YOUR SOLUTIONby substituting it into theoriginal equation.Let’s look at how the examplesof linear equations in one variable,identified earlier, would be solvedby applying our set of logicalprocedures.We will not look at our 3rdexample of a linear equationbecause it is a linear equation in 2variables.Example 1: 2x + 4 = 101. Note that there are no fractions,so we will go to step 2.2. There are no parentheses toremove, so we will go to step3.3. There are no like terms tocombine, so we will go to thenext step, 4.4. Isolate "x" to one side of theequation with the additionproperty of equality. Add theopposite, a –4, on both sides ofthe equation to move that termto the other side: 2x + 4 - 4 =10 – 4. This simplifies to 2x =6. We now have only the termwith the variable on the left,but it has a coefficient.Mr. Breitsprecher’s Edition February 9, 2005 Web: www.clubtnt.org/my_algebraAcademic Support ServicesFREE Tutoring And Academic Support Services!!!Basement of McCutchan Hall, Rm. 1Mon-Thurs: 9 a.m. – 9 p.m.Fri: 9 a.m. – 3 p.m. and Sun 5 p.m. – 9 p.m.
• 32. The following are examples ofinequalities:a < b a is less than ba ≤ b a is less than or equal to ba > b a is greater than ba ≥ b a is greater than or equal to bGraphing InequalitiesWhen we draw equalities on anumber line, we merely have to place adot or point on the location indicated.Graphing inequalities is not hard, but weneed to indicate the entire set of numbersthat are part of that inequality – it is notjust one point, there are infinite points.We will do this by using a “o” forthe point that is NOT included in ourgraph of an inequality. We will thendraw an arrow from that point to showwhich way the set of numbers thatrepresents that inequality extends.Example: x < 1|--|--|--|--o--|--|-3 -2 -1 0 1 2 3Example: x > -1|--|--o--|--|--|--|-3 -2 -1 0 1 2 3Remember, EVERY inequalityrepresents a SET of numbers, to the leftor right of where we start.When we graph inequalities thatinclude “equal to” such as “greater thanor equal to” or “less than or equal to,”we need to INCLUDE the starting pointin our graph. We do this by using “ ”and drawing our arrow to represent theset from that point.Example: x ≤ 1|--|--|--|-- --|--|-3 -2 -1 0 1 2 3Example: x ≥ -1|--|-- --|--|--|--|-3 -2 -1 0 1 2 3In order to understand linearinequalities, recall that a variable is anumber that is not identified. It is oftenrepresented by "x" or "y," any letter canbe used.A linear expression is amathematical statement that performsfunctions of addition, subtraction,multiplication, and division, but has noexponents (or powers) and no variablesthat multiply or divide each other.A linear inequality is amathematical statement that has one ofthe inequality signs above and linearexpressions.The examples of linear equalitiesdiscussed so far are fairly straight-forward. The graphs will becomeintuitively obvious with practice. Wheninequalities are not so obvious, we canuse a similar process to that which isused to solve linear equations. Theproperties of equalities are similar to theproperties of inequality.Addition Property for InequalitiesIf a < b, then a + c < b + cIn other words, adding orsubtracting the same expression toboth sides of an inequality does notchange the inequality.Example: x – 5 ≤ 1x – 5 ≤ 1x – 5 + 5 ≤ 1 + 5x ≤ 6|--|--|--|-- --|--|2 3 4 5 6 7 8Recall that the product of 2 realnumbers with the same sign isALWAYS positive. Also recall thatthe product of 2 numbers withdifferent signs is ALWAYS anegative number.Multiplication Properties forInequalitiesWhen multiplying by a positivevalue, If a < b AND c ispositive, then ac < bcWhen multiplying by a negativevalue, If a < b AND c isnegative, then ac > bc.In other words, multiplying thesame POSITIVE number to bothsides of an inequality DOES NOTCHANGE the inequality.Multiplying the same NEGATIVEnumber to both sides of aninequality REVERSES the sign ofthe inequality.Academic Support ServicesFREE Tutoring And Academic Support Services!!!Basement of McCutchan Hall, Rm. 1Mon-Thurs: 9 a.m. – 9 p.m.Fri: 9 a.m. – 3 p.m. and Sun 5 p.m. – 9 p.m.Mr. Breitsprecher’s Edition February 14, 2005 Web: www.clubtnt.org/my_algebra
• 33. Multiplication Properties forInequalities (continued)Example: 5x < -10(1/5) * 5x < -10 * (1/5)x < (10/5)x < 5|--|--|--|--|--|--o--|-1 0 1 2 3 4 5 7Example: (x / 3) > 93 * (x / 3) > 9 * 3[(3x) / 3] > 27x > 27|--o--|--|--|--|--|--|26 27 28 29 30 31 32 33In the next examples, we willmultiply BOTH sides of the equationby a negative number. T ISALWAYS MEANS we need tochange the direction of the inequalitysign.Example: (-3x) ≥ 9 Note: Thecoefficient of the first term is (-3).Therefore, we will multiply BOTHSIDES by (-1/3). The inequalityneeds to be reversed to make thestatement true.(-1/3) * (-3x) ≤ 9 * (-1/3)x ≤ -9/3x ≤ -3|--|-- --|--|--|--|–5 –4 –3 -2 -1 0 1If you are unsure about this, pleasechoose any number from oursolution and check it in our originalequation. You will get a “true”statement.Example: (-x/3) > 7. Note: Thecoefficient of the first term is (-1/3).Therefore, we will multiply BOTHSIDES by (-3). The inequalityneeds to be reversed to make thestatement true.-3 * (-x / 3) < 7 * -3x > -27|---o---|---|---|---|-28 -27 -26 -25 -24 -23Solving Linear Inequalities is the Same asSolving Linear Equalities, EXCEPT…Yes, if you have mastered keeping track of positive and negative signs when performingmultiplication (we have defined division in terms of multiplication), solving linearequations is the same WITH ONE VERY IMPORTANT EXCEPTION.Let’s look at an example and see if we can make this clear. Look at this statement, 6 > 3.Certainly, we would all agree that 5 is greater than 3.Now we will multiply both sides by a convenient number, let’s use –1; however, anynegative number would do. Is (-1)(6) still greater than (3)(-1)? NO!!!!!-6 is not greater than –1!Notice that it is left than –1 on the number line:|-- --|--|--|--|-- --|--|-7 –6 –5 –4 –3 –2 –1 0 1 2Notice that multiplying by a negative number changes the signA positive number multiplied by a negative number results in a negative number.A negative number multiplied by a negative number is a positive number.Multiplying BOTH the left and right side of an equation by a negative number is never anissue (equations contain an “=” sign). IT IS ALWAYS AN ISSUE when we multiplyBOTH the left and right side of an INEQUALITY by a negative number. WE HAVETO “FLIP,” (or change the direction ) OF THE SIGN! This is also true if you divideby a negative.Online ResourcesNote: These URL’s can be long and frustrating to enter. You should be able to go to thedomain name (domain_name.com) and find links to that will take you to these pages.Algebra: Solving Linear Inequalitieshttp://en.wikibooks.org/wiki/Algebra:Solving_linear_inequalitiesS.O.S. Math: Inqualitieshttp://www.sosmath.com/algebra/inequalities/ineq01/ineq01.htmlMath.com: Inequalitieshttp://www.math.com/school/subject2/lessons/S2U3L4DP.htmlMathematics Help Central: Solving Linear Inequalitieshttp://www.mathematicshelpcentral.com/lecture_notes/intermediate_college_algebra/solving_linear_inequalities.htmThinkQuest. Solving Linear Equationshttp://library.thinkquest.org/C0110248/algebra/ielinearsolving.htmMr. Breitsprecher’s Edition February 14, 2005 Algebra Connections, Page 2Express
• 34. Multiplying and dividingrational expressions is just likemultiplying and dividing fractions –they just look more complicated.Recall that:0≠=× bdwherebdacdcba≠=×=÷ bdwherebcadcdbadcbaWhen working with fractions,we can perform the arithmetic andthen look at our results to see if it isin “lowest terms.” Many of us, withpractice, can intuitively see how tosimplify fractions because werecognize when numbers containcommon factors. We are able to“see” the common factors withoutfactoring the numerators anddenominators.Also notice that dividingfractions is simply multiplying bythe inverse. Once we arecomfortable multiplying fractions,we will quickly learn to divide themif we are able to remember torewrite the division asmultiplication. Instead of dividing,multiply by the inverse.Multiplying Rational ExpressionsMultiplying rationalexpressions works exactly likefractions, but the expressions aremore complicated. Not only are weworking with numerators anddenominators, each are apolynomial. We need a way to keepthe polynomials in the numeratorsand denominators manageable.Few of us will intuitively seeall common factors in rationalexpressions. The key will be tocompletely factor the numeratorsand denominators FIRST, thencancel out common factors (which isapplying the FundamentalPrincipal of RationalExpressions). The factors that areleft represent our answer and thisanswer is expressed in its simplestform.Do you see that we DO NOTACTUALLY PERFORM ANYMULTIPLICATION WHEN WEMULTIPLY RATIONALEXPRESSIONS?We factor, which means writerational expression as a product, andthen cancel out all common factorsthat appear in BOTH the numeratorand denominator (FundamentalPrincipal of Rational Expressions).What is left after factoring andcanceling out common factorsbetween the numerator anddenominator IS OUR PRODUCT!Steps in Multiplying RationalExpressionsStated formally, as in a typicalAlgebra Textbook, the process formultiplying rational expressionsthat we have been discussing can bewritten as:1. Factor numerators anddenominators.2. Multiply Numerators andmultiply denominators.3. Write the product inlowest terms (applyFundamental Principal ofRational Expression,PR/QR = P/Q, where R isa nonzero polynomial)Note that we can apply thesesteps BY COMBINING 1 & 2 byfactoring the numerators andnominators and rewriting each ofthese sets of factors as 1 rationalexpression that contains ALLfactors from each numerator anddenominator.This is how AlgebraConnections will present itsexamples. We will completelyfactor each rational expression,writing the factors as one rationalexpression, keeping factors of thenumerators on top and factors of thedenominators on the bottom.Academic Support ServicesFREE Tutoring And Academic Support Services!!!Basement of McCutchan Hall, Rm. 1Mon-Thurs: 9 a.m. – 9 p.m.Fri: 9 a.m. – 3 p.m. and Sun 5 p.m. – 9 p.m.Mr. Breitsprecher’s Edition April 4, 2005 Web: www.clubtnt.org/my_algebra
• 35. Mr. Breitsprecher’s Edition April 4, 2005 Algebra Connections, Page 2Example 1:xyyyx31059×yxyyyx635)52()33(=××××=Example 2: 353421538yxzzxy×−35342)3()53()222(yxzzxy×××××−=Example 3:166212722−×+++xxxxx82)4(2)4)(4()]3(2[)]4()3[(−=−=+−×+×+××+=xxxxxxxxxDividing Rational ExpressionsJust like fractions – once we arecomfortable with the process ofmultiplying rational expressions, weare ready to divide if we rememberto rewrite our division asmultiplication.Instead of dividing, multiply bythe reciprocal. As we have justseen, that means to factorcompletely, cancel out commonterms that appear in BOTH thenumerator and denominator, andwrite down the remaining factors.Note that we can write divisionof rational expressions with the “÷”sign or with a fraction bar.Example 4:xxxx65356535÷=Either way, we invert thedivisor and multiply.Example 4:xx 6535÷21253)32(55635==×××=×=xxxxExample 5: )272(2xx÷422121252727xxxxx=××=×=Example 6:252651222−−−÷++xxxxx235)23)(12[()5()]5)(5[()12(262551222−−=−+×+−+×+=−−−×++=xxxxxxxxxxxxxIndicated Division with FractionBarMany beginning algebra books,such as K. Elayn Martin-Gay’sIntroductory Algebra, do notpresent division of rationalexpressions with fraction bars – theyintroduce this concept as “complexfractions.”Remember, fractions alwaysindicate division. Thedenominator is the divisor. Theresult of division is called thequotient. Regardless of how wewrite division (fraction bar or “÷”),we need to invert the divisor (itsreciprocal) and rewrite the problemas multiplication.Example 7:613ba +babababa222)(13)32()(163+=+=×××+=×+=Example 8:31212−−xx)1(23)]1)(1[(13212−××+−=−×−=xxxxxExample 9:2352+a652135 22+=×+=aaOnline ResourcesMultiplying Rational Expressionshttp://www.purplemath.com/modules/rtnlmult.htmhttp://www.algebra-online.com/multiplying-rational-expressions-1.htmDividing Rational Expressionshttp://www.purplemath.com/modules/rtnlmult2.htmhttp://www.algebra-online.com/dividing-rational-expressions-1.htmMultiplying & Dividing RationalExpressionshttp://www.sci.wsu.edu/~kentler/Fall97_101/nojs/Chapter7/section1.htmlhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut9_mulrat.htmhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut32_multrat.htmhttp://faculty.ed.umuc.edu/~swalsh/Math%20Articles/RationalE.htmlMultiplying, Dividing, Adding,and Subtracting RationalExpressionshttp://tutorial.math.lamar.edu/AllBrowsers/1314/RationalExpressions.aspFactoring a Polynomial:3. Are there any common factors? If so,factor them out.4. How many terms are in thepolynomial:a. Two terms: Is it the difference of2 squares? a2-b2= (a-b)(a+b)1. See if any factors can befactored further. Watch fordifference of 2 squares!Quick Review: Two ImportantExponent Rulesam*an= am+nam/an= am-na. Three terms: Try one of thefollowing patternsi. a2+2ab+b2= (a+b)2ii. a2-2ab+b2= (a-b)2iii. Otherwise, try to use anothermethod.b. Four terms: Try factoring bygrouping.
• 36. howMr. Breitsprecher’s Edition February 22, 2005 Web: www.clubtnt.org/my_algebraMultiplying polynomials is justanother application of thecommutative and associativeproperties of multiplication.Combined with the multiplicationrules for exponents, we have all thetools necessary to practice andmaster this algebraic skill.The key will be to apply theseprocedures in an organized,simplified fashion. Understandingthis process is important – timespent becoming familiar with theseprocedures of multiplyingpolynomials will repay itself.Like many things – it isprobably best to look at and practicesimpler examples before we move tothose that require a greaterunderstanding. Recall that apolynomial is an expression withone or more terms added to orsubtracted from each other. Forexample, x³ + 5x² - 8 is apolynomial.The simplest form of apolynomial only has 1 term – it iscalled a monomial. Let’s start bylooking at multiplying monomialsby monomials.Example 1: MultiplyingMonomials:(4x3)(3x4)• Group coefficients and likebases:(4*3)*( x2x4)• Add exponents and simplify:12x7Example 2: MultiplyingMonomials:(-3c4d)(2c2d3e)• Group coefficients and likebases:(-3*2)(c4c2)(dd3)(e)• Simplify:-6c6d4eMultiplying a Polynomial by aMonomialThis isn’t new; we looked atthis earlier without using theseterms. Recall that distributiveproperty: a(b+c) = ab + bc. Do yousee that this is really multiplying apolynomial by a monomial?Example 1: Multiplying aBinomial by a Monomial:4t(2t-3)• Multiply each term of thebinomial by 4t:4t(2t-3)• Apply the distributive property:(4t)(2t) + 2t(-3)• Simplify each term:8t2-6tSpecial Case Product FormulasThere are patterns in numbers and in algebra – it is important to see theserelationships. In some cases – multiplying 2 binomials takes on a special pattern. Wewill not discuss why they work or “prove” them; you can apply foil to see for yourself.Please practice these patterns so that you can recognize and apply them – they areextremely important when factoring polynomials.• Difference Of Squares: (a+b)(a-b) = a2-b2• Perfect Square Trinomials:(a+b)2= a2+2ab+b2(a-b)2= a2– 2ab + b2More Links for Polynomial Expressionshttp://www.ifigure.com/math/algebra/algebra.htmSolve Your Math Problem (you supply the problem, it provides the solution)http://www.webmath.com/polymult.htmlHow to Add, Subtract, Multiply, and Divide Polynomialshttp://faculty.ed.umuc.edu/~swalsh/Math%20Articles/Polynomial.htmlInteractive Exponents and Polynomial Review (requires plug-in, available at site)http://www.mathnotes.com/Intro/aw_introchap3.htmlPractice Test: Polynomial Concepts and Operations (interactive)http://www.mccc.edu/~kelld/polynomials/polynomials.htm
• 37. Example 2: Multiplying a Trinomialby a Monomial:-4a2(-3a2+ 2a – 3)• Multiply each term of the trinomialby –4a2:–4a2(-3a2+ 2a – 3)• Apply the distributive property:(-4a2)(-3a2) + (-4a2)(2a) + (-4a2)(-4)• Simplify:12a4– 8a3+ 16a2So we have seen how to multiplypolynomials by monomials – this wasjust an application of the distributiveproperty. Apply the same principleswhen multiplying polynomials withmore than 1 term (i.e. binomial,trinomials, etc.).THE KEY IS TO REMEMBERTO MULTIPLY EACH TERM OFTHE FIRST POLYNOMIAL BYEACH TERM OF THE SECONDPOLYNOMIAL. We can draw arrowsto help us keep track – this is especiallyuseful when multiplying polynomialsthat contain many terms.Mr. Breitsprecher’s Edition February 22, 2005 Algebra Connections, Page 2Example: MultiplyingPolynomials With More Than 1Term:(x+5)(x+3)• Multiply each term in thesecond polynomial by eachterm in the first:(x+5)(x+3)• Apply the distributionproperty:(x*x)+(x*3)+(5*x)+(5*3)• Simplify:x2+3x+5x+15)• Combine like terms:x2+8x+15Note how the product of twobinomials (example above) equalsthe sum of the products of the firstterms, the outer terms, the innerterms and the last terms.The acronym FOIL (FirstOuter Inner Last) can be used tohelp memorize how to multiply 2binomials. In this author’s opinion– that is just an extra rule tomemorize; we don’t need it. Pleaseuse it if you find it helpful and beaware that you will probably see itin another math class.We can accomplish the samething be remembering to multiplyeach term of the first polynomial byeach term of the second polynomial.We can easily keep track of thisif we draw arrows for each productwhile we work out the solution.This approach works whenmultiplying any polynomial withmore than one term by any otherpolynomialUsing FOIL will ONLY workwhen multiplying two binomials. ItWILL NOT work whenmultiplying a binomial by atrinomial or with any other type ofpolynomial. If you have beensuccessful accurately using FOIL inthe past – please continue to use it.Sometimes, new approaches arehelpful – drawing arrows to keeptrack of each term as you multiplypolynomials, can be a convenientmethod. This author recommendskeeping memorization to aminimum when learning algebra.DefinitionsMonomial has one term: 5y or -8x2or 3.Binomial has two terms: -3x2+ 2, or 9y- 2y2Trinomial has 3 terms: -3x2+ 2 +3x, or9y - 2y2+ yDegree Of The Term is the exponent ofthe variable: 3x2has a degree of 2.Degree of a Polynomial is the highestdegree of any of its terms.When the variable does not have anexponent -understand that theres a 1 .One thing you will do whensolving polynomials is combine liketerms. Understanding this is the keyto accurately working withpolynomials. Let’s look at someexamples:• Like terms: 6x + 3x - 3x• NOT like terms: 6xy + 2x - 4The first two terms are like and theycan be combined:5x2+ 2x2– 3Combining like terms, we get:7x2– 3Academic Support ServicesFREE Tutoring And Academic Support Services!!!Basement of McCutchan Hall, Rm. 1Mon-Thurs: 9 a.m. – 9 p.m.Fri: 9 a.m. – 3 p.m. and Sun 5 p.m. – 9 p.m.Report• Helium was up, feathers weredown.• Paper was stationery.• Fluorescent tubing was dimmedin light trading.• Knives were up sharply.• Pencils lost a few points.• Hiking equipment was trailing.• Elevators rose, while escalatorscontinued their slow decline.• Light switches were off.• Mining equipment hit rockbottom.• Diapers remain unchanged.• Shipping lines stayed at an evenkeel.• The market for raisins dried up.• Caterpillar stock inched up a bit.• Sun peaked at midday.• Balloon prices were inflated.• Kleenex nosed up.
• 39. Adding Integers (from page 1)Examples:2 + 5 = 7(-7) + (-2) = -(7 + 2) = -9(-80) + (-34) = -(80 + 34) = -114When adding integers of theopposite signs, we take theirabsolute values, subtract the smallerfrom the larger, and give the resultthe sign of the integer with thelarger absolute value.Example:8 + (-3) = ?The absolute values of 8 and -3are 8 and 3. Subtracting the smallerfrom the larger gives 8 - 3 = 5, andsince the larger absolute value was8, we give the result the same signas 8, so 8 + (-3) = 5.Example:8 + (-17) = ?The absolute values of 8 and -17 are 8 and 17. Subtracting thesmaller from the larger gives 17 - 8= 9, and since the larger absolutevalue was 17, we give the result thesame sign as -17, so 8 + (-17) = -9.Example:-22 + 11 = ?The absolute values of -22 and11 are 22 and 11. Subtracting thesmaller from the larger gives 22 - 11= 11, and since the larger absolutevalue was 22, we give the result thesame sign as -22, so -22 + 11 = -11.Example:53 + (-53) = ?The absolute values of 53 and -53 are 53 and 53. Subtracting thesmaller from the larger gives 53 - 53=0. The sign in this case does notmatter, since 0 and -0 are the same.Note that 53 and -53 are oppositeintegers. All opposite integers havethis property that their sum is equalto zero. Two integers that add up tozero are also called additiveinverses.Subtracting IntegersSubtracting an integer is thesame as adding its opposite.Examples:In the following examples, weconvert the subtracted integer to itsopposite, and add the two integers.7 - 4 = 7 + (-4) = 312 - (-5) = 12 + (5) = 17-8 - 7 = -8 + (-7) = -15-22 - (-40) = -22 + (40) = 18Note that the result ofsubtracting two integers could bepositive or negative.Multiplying IntegersTo multiply a pair of integers ifboth numbers have the same sign,their product is the product of theirabsolute values (their product ispositive). If the numbers haveopposite signs, their product is theopposite of the product of theirabsolute values (their product isnegative). If one or both of theintegers is 0, the product is 0.Examples:In the product below, bothnumbers are positive, so we just taketheir product.4 × 3 = 12In the product below, bothnumbers are negative, so we take theproduct of their absolute values.(-4) × (-5) = |-4| × |-5| = 4 × 5 = 20In the product of (-7) × 6, thefirst number is negative and thesecond is positive, so we take theproduct of their absolute values,which is |-7| × |6| = 7 × 6 = 42, andgive this result a negative sign: -42,so (-7) × 6 = -42.In the product of 12 × (-2), thefirst number is positive and thesecond is negative, so we take theproduct of their absolute values,which is |12| × |-2| = 12 × 2 = 24,and give this result a negative sign: -24, so 12 × (-2) = -24.To multiply any number ofintegers:1. Count the number of negativenumbers in the product.2. Take the product of theirabsolute values.3. If the number of negativeintegers counted in step 1 iseven, the product is just theproduct from step 2, if thenumber of negative integers isodd, the product is the oppositeof the product in step 2 (givethe product in step 2 a negativesign). If any of the integers inthe product is 0, the product is0.Example:4 × (-2) × 3 × (-11) × (-5) = ?Counting the number ofnegative integers in the product, wesee that there are 3 negativeintegers: -2, -11, and -5. Next, wetake the product of the absolutevalues of each number:4 × |-2| × 3 × |-11| × |-5| = 1320.Since there were an odd numberof integers, the product is theopposite of 1320, which is -1320,so:4 × (-2) × 3 × (-11) × (-5) = -1320.Dividing IntegersTo divide a pair of integers ifboth integers have the same sign,divide the absolute value of the firstinteger by the absolute value of thesecond integer.To divide a pair of integers ifboth integers have different signs,divide the absolute value of the firstinteger by the absolute value of thesecond integer, and give this result anegative sign.Examples:In the division below, bothnumbers are positive, so we justdivide as usual.4 ÷ 2 = 2.In the division below, bothnumbers are negative, so we dividethe absolute value of the first by theabsolute value of the second.(-24) ÷ (-3) = |-24| ÷ |-3|=24÷3=8.In the division (-100) ÷ 25, bothnumber have different signs, so wedivide the absolute value of the firstnumber by the absolute value of thesecond, which is |-100| ÷ |25| = 100÷ 25 = 4, and give this result anegative sign: -4, so (-100) ÷ 25 = -4.In the division 98 ÷ (-7), bothnumber have different signs, so wedivide the absolute value of the firstnumber by the absolute value of thesecond, which is |98| ÷ |-7| = 98 ÷ 7= 14, and give this result a negativesign: -14, so 98 ÷ (-7) = -14.Source: www.mathleague.comMr. Breitsprecher’s Edition January 26, 2005 Algebra Connections, Page 2
• 40. Percents, like fractions anddecimals, are way to express how apart of something relates to a whole.If we talk about 1/2, .5, or 50% of apizza, we are talking about one of 2equal parts – half the pizza.Percentages describe therelationship to a whole divided into100 parts. The word “percent”means “out of 100.” The whole partis considered 100%. Do you seethat the percent sign (%) consists of2 zeros, as in 100?Percentages can also be used torepresent more than the wholeamount (100%). Assume we have100 pieces of candy. That wholerepresents 100%. Let’s assume webuy 50 more pieces of candy – 50parts of 100 equals 50%. Noticethat our original 100% + 50% more= 150% of what we started with.The advantage of working withpercentages is that we express eachpart of the whole in terms of 1/100.Percentages allow us to makemeaningful comparisons betweenparts of a whole, because they alwaysrefer to parts out of 100. If weconvert fractions to percentages (byperforming the indicated division andmultiplying by 100), we can noweasily see the relationship betweenfractions – even when thedenominators are different.A percentage is really just adecimal multiplied by 100. Most ofus will find it easier to work withpercentages, because we don’t have todeal with as many decimal places (i.e.12.5% = .125).The Three Percentage CasesTo explain the cases that arise inproblems involving percents, it isnecessary to define the terms that willbe used. Rate (r) is the number ofhundredths parts taken. This is thenumber followed by the percent sign.The base (b) is the whole on whichthe rate operates. Percentage (p) isthe part of the base determined by therate. In the example:5% of 40 = 25% is the rate (r)40 is the base (b)2 is the percentage. (p)There are three cases that usuallyarise in dealing with percentage, asfollows:Case I. Find the percentagewhen the base and rate are known, i.e.What number (p) is 6% (r) of 50 (b)?Case II. Find the rate when thebase and percentage are known, i.e. 20(p) is what percent (r) of 60 (b)?Case III. Find the base when thepercentage and rate are known, i.e. thenumber 5(p) is 25% (r) of whatnumber (b)?Source: http://www.tpub.com/math1/7a.htmHere are some common equivalents. Do you see the patterns when you relate onecolumn to another?Fraction Decimal PercentageTwentieths.........................................1/20......................05................................5%Tenths................................................1/10......................10..............................10%3/10......................30..............................30%7/10......................70..............................70%9/10......................90..............................90%Eighths.................................................1/8......................125......................... 12.5%3/8......................375......................... 37.5%5/8......................625......................... 62.5%7/8......................875......................... 87.5%Sixths ...................................................1/6......................16666..................16 2/3%5/6......................83333..................83 1/3%Fifths....................................................1/5......................2................................20%2/5......................4................................40%3/5......................6................................60%4/5......................8................................80%Quarters ..............................................1/4 .....................25..............................25%3/4 .................75.......................75%Thirds ..................................................1/3......................33333..................33 1/3%2/3......................66666..................66 2/3%Half ......................................................1/2 .....................5................................50%Mr. Breitsprecher’s Edition January 26, 2005 Web: www.clubtnt.org/my_algebra
• 41. Case I. 6% of 50 = ? The "of"has the same meaning as it does infractional examples, such as1/4 of 16 = 4In other words, "of" means tomultiply. Thus, to find thepercentage, multiply the base by therate. Of course the rate must bechanged from a percent to a decimalbefore multiplying can be done. Tofind the percentage, multiply theRate times base and then divideby 100.[6% (r) of 50 (b)]/100 = p(6 * 50)/100 = 3The number that is 6% of 50 is 3.To explain Case II and CaseIII, we notice in the foregoingexample that the base corresponds tothe multiplicand, the ratecorresponds to the multiplier, andthe percentage corresponds to theproduct.50 (b) * .06 (r) = 3.00 (p)Case II, ?% of 60 = 20.Recalling that the product dividedby one of its factors gives the otherfactor, we can solve the followingproblem:?% of 60 = 20We are given the base (60) andpercentage (20).60 (b) * ? (r) = 20 (p)We then divide the product(percentage) by the multiplicand(base) to get the other factor (rate).Percentage divided by base times100 equals rate. The rate is found asfollows:20/60 * 100 = 33 1/3 %Case III, 25% of ? = 5. Therule for Case II, as illustrated in theforegoing problem, is as follows: Tofind the rate when the percentageand base are known, divide thepercentage by the base and multiplyby 100.The unknown factor in Case IIIis the base, and the rate andpercentage are known.25% of ? = 5We are given the rate (25) andpercentage (5). To find the basewhen the percentage and rate areknown, divide the percentage by therate and multiply by 100.5/25 * 100 = 20Mr. Breitsprecher’s Edition January 26, 2005 Algebra Connections, Page 2Another Perspectives: Ratios, Proportions & PercentsCross-Products to the Rescue!!!Sometimes a different point of view is helpful – it gives us choices. Analternative way to think of percentages is to understand the relationshipbetween ratios, proportions, and percentages.Ratios tell how one number is related to another number. A ratio may bewritten as A:B or A/B or by the phrase "A to B". A ratio of 1:5 says that thesecond number is five times as large as the first. Percentages are really allratios – just remove the “%” sign and write the remaining number over 100.For example, 25% equals “25 to 100.” or the ratio 25/100.A proportion is a mathematical statement that two ratios are equal. If the tworatios are not equal, then it is not a proportion. An example would be theproportion that compares 3/6 = 4/8. Do you see that each is an equivalentfraction (in this example, 1/2). Please look at the proportion above andnotice that the product of the 2 outside terms equals the product of the 2inside terms, i.e. (3*8)=(4*6). This will always be true of any proportion.This is called a “cross-product.”If we remember that all percentages can be expressed as a ratio by removingthe “%” and placing the remaining number over 100, then we can use cross-products to solve the 3 types of percentage problems we have beendiscussion. Many will find this approach easier; because if we set up ourproportions correctly and understand how to algebraically solve for themissing variable, then we do not need to consider each of the three casesdiscussed in the preceding article. Cross-products will work for all three!Case I, 6% of 50 = ? Realize that we are solving for the percentage. 6%(rate or “r”) can ALWAYS be expressed as the ratio 6/100. The base is 50.We can use a variable (p) to represent the question mark (percentage) andwrite that ratio as p/50. ALWAYS express this ratio as the percentage overthe base. Now we have the following proportion: 6/100 = p/50. The crossproducts are (6*50) = (100*p), the product of the outside terms equals theproduct of the inside terms. This will simplify to 300= 100p. Solving for“p”, we get p=300/100 or 3. This is the same percentage that we calculatedearlier.Case II, ?% of 60 = 20. Realize that we are solving for the rate (r). It can beexpressed as the ratio as r/100 (always rate/100). We are given thepercentage (20) and the base (60). These are expressed as the ratio 20/60(always percentage over base). Now we have the following proportion:r/100=20/60. The cross products are (r* 60) = (100*20) When proportion isexpressed in this format, the cross-product is always the product of theoutside terms equals the product of the inside terms. This can be simplified,r(60) = 2000. Solving for rate (r), we get r = 2000/60 or 33 1/3%. This is thesame rate we calculated earlier.Case III, 25% of ? = 5. Realize that we are solving for the base. We canexpress the rate as the ratio 25/100 (always as rate/100). We are given thepercentage (and are solving for the base). This can be expressed as the ratio5/b (always percentage/base). Now we have the proportion of (25/100) =(5/b). The cross product is (25*b) = (100*5). When proportion is expressedin this format, the cross-product is always the product of the outside termsequals the product of the inside terms. Solving for b, we get b = (100*5)/25or 20. This is the same base we calculated earlier.
• 42. Mr. Breitsprecher’s Edition March 2, 2005 Web: www.clubtnt.org/my_algebraPolynomials are algebraicexpressions. They can contain realnumbers and variables. There cannotbe any division and square roots thatinvolve the variables. Only addition,subtraction and multiplication ofvariable terms are allowed.Polynomials contain more thanone term – that’s the meaning of theroot-word “poly” Think ofpolynomials as the sums ofmonomials.A monomial has one term:5y or -8x2or 3.A binomial has two terms:-3x2+ 2, or 9y - 2y2A trinomial has 3 terms:-3x2+ 2 +3x, or 9y - 2y2+ yThe degree of the term is theexponent of the variable: 3x2has adegree of 2.When the variable does nothave an exponent, it is assumed tobe 1. (i.e. 1x has a degree of 1).Example:x2- 7x - 6Each part, or “chunk” ofmathematical information is a termand x2is referred to as the leadingterm.CoefficientsA constant used to multiply anotherquantity or series is called acoefficient. Examples include 3x andax. In these 2 cases, 3 and a arecoefficients of x.Examples:Term Coefficientx21-7x -7-6 -6Examples of Expressions andPolynomials• 8x2+ 3x -2 is a polynomial withthree terms. It is a trinomial.• 8x-3+ 7y -2 is NOT a Polynomial,because the exponent is negative.• 9x2+ 8x -2/3 is NOT aPolynomial; it cannot havedivision.• 7xy is a Monomial. There is only1 term.Polynomials are usually written indecreasing order of terms. The largestterm or the term with the highestexponent in the polynomial is usuallywritten first. The first term in apolynomial is called a leading term.When a term contains an exponent, ittells you the degree of the term.DegreesDegree of a Term. Each term in apolynomial has a “degree,” or theexponent of that term. A term writtenPolynomial Definitions of Terms:• A monomial has one term: 5y or -8x2 or 3.• A binomial has two terms: -3x2 + 2, or 9y - 2y2• A trinomial has 3 terms: -3x2 + 2 +3x, or 9y - 2y2 + y• The degree of the term is the exponent of the variable: 3x2 has a degree of 2.• When the variable does not have an exponent -understand that theres a 1 e.g., 3xWorking With Polynomials: Just Collect Like Terms!One thing you will do when solving polynomials is combine like terms.Understanding this is the key to adding and subtracting polynomials. Let’s look atsome examples:• Like terms: 6x + 3x - 3x• NOT like terms: 6xy + 2x - 4The first two terms are like and they can be combined: 5x2+ 2x2– 3, combining liketerms, we get: 7x2– 3.Adding and Subtracting PolynomialsJust like working with polynomials, they key is to identify like terms. To addpolynomials, you must clear the parenthesis, combine and add the like terms. In somecases you will need to remember the order of operations. Remember, when adding andsubtracting like parts, the variable never changes.