Algebra Newsletter NotesA creative way to presentconcepts, notes & study guidesin a newsletter format.Excellent Resource!www.clubtnt.org/my_algebra
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
Mr. Breitsprecher’s Edition April 5, 2005 Algebra Connections, Page 2Example:15461−1. Factor completely: 6=2*3,15=3*52. Product of all different factors:2, 3, and 53. Use highest power: 2*3*5=301013*103*1303308305221545561−=−=−=−=−Rational expressions are addedand subtracted just like rationalnumbers (quotients of rationalnumber or fractions). Whennecessary, rewrite rationalexpressions with a commondenominator.Follow these steps to add orsubtract rational expressions whenthe denominators differ.1. Find the Lowest CommonDenominator (LCD)2. Rewrite each rationalexpression as an equivalentexpression whose denominatoris the LCD.3. Add or subtract numerators andplace the sum or difference overthe common denominator.4. Write the result in lowest terms.Example:3225+x+=xxx 22323325xx6415 +=Example:xxx 329122++−)3(2)3)(3(1+++−=xxxx)3)(3(63)3)(3(6233)3(2)3)(3(1+−−=+−−+=−−+++−=xxxxxxxxxxxxxxxxxExample:5254−−− aaNote: because (5-a) and (a-5) differ onlyin signs, we can obtain identicaldenominators by multiplying only the firstexpression by -1 in BOTH the numeratorand denominator (maintain and equivalentexpression).565652545211454−−=−−=−−−−=−−−−−=aaaaaExample:6112612212−+++++xxxxxNote: The LCD for these threedenominators is 6x(x+2)xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx63)2(6)2)(3()2(6652)2(6222266)2(622)2(622)2(666)2()2(61)2(61266)2(161)2(612)2(1+=+++=+++=+−−+++=++−+++++=++−+++++=−+++++=Online ResourcesFinding the Lowest CommonDenominatorhttp://www.csun.edu/~ayk38384/Math093-Rational%20Expression.htmhttp://library.thinkquest.org/20991/textonly/alg/frac.htmlAdding and Subtracting RationalExpressionshttp://www.mathsteacher.com.au/year10/ch14_rational/05_addition_and_subtraction/addsub.htmhttp://a-s.clayton.edu/garrison/Math%200099/sect44.htmhttp://faculty.ed.umuc.edu/~swalsh/Math%20Articles/RationalE.htmlhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut33_addrat.htmhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut33_addrat.htmhttp://www.algebra-online.com/adding-substracting-rational-expressions-like-denominators-1.htmhttp://tutorial.math.lamar.edu/AllBrowsers/1314/RationalExpressions.asphttp://www.sparknotes.com/math/algebra2/rationalexpressions/section2.rhtmlPurple Math’s Tutorial: Addingand Subtracting RationalExpressions (2 parts)http://www.purplemath.com/modules/rtnladd.htmhttp://www.purplemath.com/modules/rtnladd2.htmFactoring a Polynomial:1. Are there any common factors? If so,factor them out.2. How many terms are in thepolynomial:a. Two terms: Is it the difference of2 squares? a2-b2= (a-b)(a+b)3. 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.
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
Widgets by Plant As FractionFraction As% of Whole As % of TotalPant 1 342.166 342 1/6 16.66666667 24.64449928Plant 2 30.9 30 9/10Plant 3 7.9558 7 65/68Plant 4 61.619Plant 5 0.13209Plant 6 38.87Plant 7 45.4Plant 8 75Plant 9 37.39Plant 10 94.728Plant 11 56.2Plant 12 57.16Plant 13 6.251Plant 14 52.6Plant 15 32.14Plant 16 99.34Plant 17 26.2Plant 18 99.3302Plant 19 15.229Plant 20 53.7Plant 21 8.268Plant 22 44.5Plant 23 56.5801Plant 24 46.748Total 1388.40719What’s A Widget?A “widget” is an imaginary unit ofproduction – it is often used in Economicclasses to illustrate ideas (it is easy to createdata for imaginary units – virtually anythingworks!)Please create the spreadsheet on the left,using ALL HEADINGS including PLANTNumbers. ONLY ENTER THE DATA FORTHE UNITS OF PRODUCTION (secondcolumn) when you start. We will enter theinformation under: As a Fraction, Fraction as% of Whole, as % of Total, and Total later byapplying our Decimals & Fraction Review,using formulas or functions as appropriate.Please apply BOLD as illustrated to theleft. If you have any problems entering thenumbers, it is usually because EXCEL hasdefaulted to one format (for example, DATE),when we are looking for another format (i.e.number). ONLY USE A “GENERAL”NUMBER FORMAT.If you have any questions or problems,PLEASE ASK A CLASSMATE OR YOURINSTRUCTOR. This activity is written toassume that students have worked with EXCELenough that specific directions about thesoftware are not needed. Please do not feel shyor hesitate to ask for assistance if necessary.PROCEDUREOnce you have entered the information per the instructionsunder the section, WHAT’S A WIDGET, let’s review ourconcepts from our Review Chapter, Decimals and Percents.Please follow the directions carefully and do not use any otherformats, functions, or MS EXCEL features. You will Emailthis to your instructor and no credit will be given for projectsthat have not applied the review principals in this project.1. The column, AS A FRACTION will be used to re-writethe decimal portion (partial unit) of each plant’sproduction as a fraction (including the whole number ofunits). Simply RE-ENTER THE WHOLE NUMBERFROM EACH DECIMAL and then ENTER A SPACE.2. Express the partial unit, after that space, by RE-ENTERING THE DECIMAL PART OF THENUMBER (without the decimal), FOLLOWED BY /(for the fraction bar) and then the DECIMALPLACEHOLDER (10, 100, 1000, or 10000).3. Note how EXCEL will re-calculate the fraction in itssimplest term. (Note examples above). For this class,ALL FRACTIONS IN FINAL ANSWERS MUST BESIMPLIFIED!4. Use AUTOSUM (or SUM FUNCTION) to total the number ofwidgets for ALL plants (cell B26 in my example, if you havedifferent cells, adjust accordingly).5. Next, in COLUMN C, create a formula that will convert thefraction in COLUMN B (DO NOT INCLUDE THE WHOLENUMBER) to a PERCENT OF WHOLE. Note that fromour review, this involves writing the decimal as a fraction,performing the indicated division, and the results multiply by100. Please also note that we always started with a fraction thatwas less than 1 – our calculated percentage, therefore, has to beless than 100. (In example shown, =1/6*100)6. Next, we will calculate each plants output as a percentage of thetotal. Note that each plant only represents a small fraction ofthe total. The percentage of each plant’s output to the totalMUST BE LESS THAN 100. The sum of these percentagesMUST EQUAL 100. Therefore, we will create a formula thatwill divide each plants output by the total of all plants. Tosimplify things, use an ABSOLUTE CELL REFERENCEand FILL this formula down. In this example, I used=B2/$B$26*1007. Use AUTOSUM to verify that the percentages in this columntotal to 100 and add the three borders that are shown in theexample. Be sure all headings and data are displayed.8. Save your work to your stustorage account. Your instructorwill give directions to Email this assignment in to earn credit.
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
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
There are several methods fortesting the divisibility of a numberwithout actually performing thedivision. A great interactive tutorial isat:http://www.oswego.org/mtestprep/math8/a/divisibilityrulesl.cfm (interactivetutorial)Dividing by 2• All even numbers are divisibleby 2. E.g., all numbers ending in0,2,4,6 or 8.Dividing by 3• Add up all the digits in the number.• Find out what the sum is. If thesum is divisible by 3, so is thenumber• For example: 12123(1+2+1+2+3=9) 9 is divisible by 3,therefore 12123 is too!Dividing by 4• Are the last two digits in yournumber divisible by 4?• If so, the number is too!• For example: 358912 ends in 12which is divisible by 4, thus so is358912.Dividing by 5• Numbers ending in a 5 or a 0 arealways divisible by 5.Dividing by 6• If the Number is divisible by 2 and 3 itis divisible by 6 also.Dividing by 7 (2 Tests)• Take the last digit in a number.• Double and subtract the last digit inyour number from the rest of the digits.• Repeat the process for larger numbers.• Example: 357 (Double the 7 to get 14.Subtract 14 from 35 to get 21 which isdivisible by 7 and we can now say that357 is divisible by 7.OR• Take the number and multiply eachdigit beginning on the right hand side(ones) by 1, 3, 2, 6, 4, 5. Repeat thissequence as necessary• Add the products.• If the sum is divisible by 7 - so is yournumber.• Example: Is 2016 divisible by 7?• 6(1) + 1(3) + 0(2) + 2(6) = 21• 21 is divisible by 7 and we can now saythat 2016 is also divisible by 7.Dividing by 8• This ones not as easy, if the last 3digits are divisible by 8, so is the entirenumber.• Example: 6008 - The last 3 digits aredivisible by one, therefore, so is 6008.Dividing by 9• Almost the same rule and dividing by 3.Add up all the digits in the number.• Find out what the sum is. If the sum isdivisible by 9, so is the number.• For example: 43785 (4+3+7+8+5=27)27 is divisible by 9, therefore 43785 istoo!Dividing by 10• If the number ends in a 0, it is divisibleby 10.Seem like too much? Easy does it –work on mastering the rules for the primenumbers: 2, 3, 5, and 7 (Note: 7’s arecommon football scores).1. A scholarship fund has $213,198.Can the money be awarded equallyamong 8 students?2. 2465 girls signed up to live in adormitory. If each room can hold 3girls, will each room be completelyfilled after all the girls have beenassigned a room?3. Nine teachers receive 190 filefolders. Can each teacher have thesame number of folders?4. Professor Fields spilled in on amemo from the math department.Find the missing Digit:TO: All Math Faculty9 classes have been scheduled forthis year. Each class has an equalnumber of students. The totalnumber of students is 4__25.5. There are 128 students enrolled in amath course. Can the students bedivided into 4 equal classrooms?6. The headwaiter at the restaurantforgot the last digit of the numberof people in a banquet party. Theparty will be seated in tables of 9.Find the missing digit. It will be aparty of 167__7. The English department has 190pens. Can 8 teachers receive thesame number of pens?8. The school health center has 124boxes of gauze. Can each of the 4wards receive an equal number ofboxes?9. A table seats 4 people. Can 1,5024people sit at these tables and havethem all be full?10. Five students washed cars all daySaturday. They made $108.00.Can they divide this amountevenly?11. Could three students divide$108.00 evenly?12. Could 4 students divide $108.00evenly?Mr. Breitsprecher’s Edition January 20, 2005 Web: www.clubtnt.org/my_algebra
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
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
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?
Solving equations is anessential part of Algebra. Factoringequations breaks mathematicalstatements that look complex intosmaller parts. Often, what lookslarge, complex and difficultbecomes clear when we can look atit in smaller or simpler terms.A polynomial is a finite sum ofterms of the form axnwhere a is areal number and n is a whilenumber. Example: 5x3-6x2+3x-6A quadratic equation is anequation that can be written in theform ax2+bx+c=0 with a not equalto 0. When a quadratic equation iswritten in the form ax2+bx+c=0, itis called standard form.If we can write a polynomial inthis standard form, it is a quadraticequation; 2x2+5x = -3 is an exampleof a polynomial equation. Here, thesame equation is written in standardform:2x2+5x+3=0While the polynomial westarted with did not look like aquadratic equation, it is. Note thatthe degree of the above a quadraticequation is 2. This is a commontype of equation.Zero Factor PropertyIf ab = 0, then a = 0 or b = 0.Zero is the key, because the onlyway a product can become 0 is if atleast one of its factors is 0. Wewould not be able to make a generalstatement about the factors if theproduct was set equal to any othernumber. For example, if ab = 1,then a = 7 and b = 1/7 or a = 4 and b= 1/4, etc. But with the product setequal to 0, we can guarantee findingthe solution by setting each factorequal to 0. This is why standardform (ax2+bx+c=0) is so important.Solving a Quadratic Equations byFactoringStep 1: Write the equation instandard form. First, if necessary,simplify the equation (i.e. clear anyfractions or parenthesis () andcombine like terms BEFORErewriting in standard form)Step 2: Factor completelyStep 3: Set each factor containing avariable equal to 0.Step 4: Solve the resultingequations.Example: 3x2= 13x-4Step 1. Rewrite in standard form:3x2-13x+4 = 0Step 2. Factor completely:(3x-1)(x-4) = 0Step 3. Set each factor containing avariable equal to 0: 3x-1 = 0 andx+4 = 0Step 4. Solve the resultingequations:3x-1 = 0 x-4 = 03x = 0+1 x = 0+4x = 1/3 x = 4In this problem, there is nosimplifying – we were able toimmediately rewrite in standardform by using the addition propertyof equality.Example: x2= 121Step 1. Rewrite in standard form:x2-121 = 0. NOTE: This is adifference of squares;a2-b2= (a+b)(a-b)Step 2. Factor completely:(x-11)(x+11) = 0Step 3. Set each factor containing avariable equal to 0: x-11 = 0 andx+11 = 0Step 4. Solve the resultingequations:x-11 = 0 x+11=0x = 11 x = -11being able to recognizedifference of squares and perfectsquare trinomials is important.Example: x2+12x = -36Step 1. Rewrite in standard form:x2+12+36 = 0. NOTE: This is aperfect square trinomial with apositive middle term;a2+2ab+b2= (a+b)2Step 2. Factor completely:(x+6)2= 0Step 3. Set each factor containing avariable equal to 0: x+6 = 0. Note:In a perfect square trinomial, thereis only 1 factorStep 4. Solve the resultingequations:x + 6 = 0x = -6Mr. Breitsprecher’s Edition March 29, 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.
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
Factoring is an idea you might befamiliar with from multiplication.Numbers that can be multiplied togetherto get another number are its factors. Forexample, 4*3 = 12, so 3 and 4 arefactors of 12. However, theyre not itsonly factors; 1, 2, 6, and 12 are otherfactors of 12. (Another way of defininga factor is a number that goes evenlyinto the number youre factoring.)A number is prime if it can not bedivided evenly by anything except itselfand 1. For example, 5 is a primenumber, because the only factors of 5are 1 * 5 = 5. However, 12 is not aprime number, because 1 * 12 = 12, 2 *6 = 12, and 3 * 4 = 12. Primefactorization means finding all the primenumbers that are factors of a number.Why is this important – it is thefoundation of what we will do the rest ofthe semester! We need to be able to findLowest Common Multiples (LCM). Inour “Review,” we saw that the LowestCommon Multiple is the LowestCommon Denominator (R-2).It will not be possible to work withfractions without this understanding.Please NEVER attempt to add orsubtract fractions unless they have acommon denominator!Suppose you want to find the LeastCommon Multiple (Please rememberLCM = LCD) of 6 and 10. If we start byfactoring them into primes, we get this:6: 2 * 310: 2 * 5Now the challenge is to find thesmallest possible set of prime factorsthat contains all the factors of eachoriginal number.In other words, we need a 2 and a 3from the 6. So far we have 2 * 3. Butwe dont need to include another 2 fromthe 10 since we already have the 2 fromthe 6.All we need to add to our set is the5, so now we have 2 * 3 * 5. As acheck, can you find 2 * 3 in our set of 2* 3* 5? How about 2 * 5? Yes, bothoriginal factor sets are included andthere is nothing extra. So, our LCM is 2* 3 * 5 or 30.Lets try another one. Suppose youare looking for the LCM of 8, 10 and 12.Start by factoring each one out:8: 2 * 2 * 210: 2 * 512: 2 * 2 * 3Now lets put together the smallestset that contains each of those three sets.We start with three 2s from the 8. Thatgives us: 2 * 2 * 2For the 10, we certainly dont needanother 2 since we already have three ofthem, but we do need to add in a 5.Now we have: 2 * 2 * 2 * 5For the 12, we dont need to doanything with the two 2s because wealready have three of them, but we doneed to toss in the 3. Now we have: 2 *2 * 2 * 5 * 3Again, can you find each of thethree smaller factor sets within thatbigger one? Yes, so were in business.The LCM is 2 * 2 * 2 * 5 * 3 or 120.Source: mathforum.orgBeing able to think about numbersis the first step towards being able towork with them. Fundamental to numbertheory are numbers themselves, and thebasic building blocks for numbers areprime numbers.A prime number is a countingnumber that only has two factors, itselfand one. Counting numbers which havemore than two factors (such as six,whose factors are 1, 2, 3 and 6), are saidto be composite numbers.The number one only has onefactor and is considered to be neitherprime nor composite.You can learn EVERYTHING youcould ever imagine about primenumbers at:http://www.utm.edu/research/primes/Here are some prime numbers startingwith 2:2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37,41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83,89, 97, 101, 103, 107, 109, 113, 127,131, 137, 139, 149, 151, 157, 163, 167,173, 179, 181, 191, 193, 197, 199, 211In this class, you will benefit if youcan recognize prime numbers and factorsup to at least 59.MS EXCEL and Primes?Can you generate these lists from MSExcel? Well, no – not without special“plug-ins.” Software to generate primes isavailable FREE as shareware.Prime Derivatives is one suchprogram. It is available at:http://www.sharewareconnection.com/prime-derivatives.htmThis attractive, handy little primenumber calculator takes any number up totwenty digits, and then factorizes all theunique prime numbers that evenly divideinto the number that was entered into thefield. Its small enough to neatly fit in thecorner of your monitor while you easilyfactorize the numbers you need for thatschool or office assignment.A great interactive site to check primefactorizations is at:http://www.gomath.com/algebra/factor.phpMr. Breitsprecher’s Edition January 20, 2005 Web: www.clubtnt.org/my_algebra
Factoring a polynomial is theopposite process of multiplyingpolynomials. Factor means write asa product. Recall that when wefactor a number, we are looking forprime factors that multiply togetherto give the number; for example6 = 2 * 3, or 12 = 2 * 2 * 3.When we factor a polynomial,we are looking for simplerpolynomials that can be multipliedtogether to give us the polynomialthat we started with. Understandinghow to multiply polynomials is thekey to understanding how to factorthem.When we factor a polynomial,we are usually only interested inbreaking it down into polynomialsthat have integer coefficients andconstants.Simplest Case: RemovingCommon FactorsThe simplest type of factoring iswhen there is a factor common toevery term. In that case, you canfactor out that common factor. Whatyou are doing is using thedistributive property in reverse.Recall that the distributive law saysa(b + c) = ab + ac.Source: www.jamesbrennan.org/algebraThinking about it in reverse meansthat if you see ab + ac, you can writeit as a(b + c).Example: 2x2+ 4xNotice that each term has a factor of2x, so we can rewrite it as:2x2+ 4x = 2x(x + 2)Difference of Two SquaresIf you see something of the form(a2- b2), you should remember theformula(a+b)(a-b) = a2-b2Example: x2– 4 = (x – 2)(x + 2)This only holds for aDIFFERENCE of two squares.There is no way to factor a sum oftwo squares such as a2+ b2intofactors with real numbers. Tryingto do so is a common mistake inalgebra – please avoid this one!Trinomials (Quadratic)A quadratic trinomial has the formax2+ bx + c,where the coefficients a, b, and c,are real numbers (for simplicity wewill only use integers, but in real lifethey could be any real number). Weare interested here in factoringquadratic trinomials with integercoefficients into factors that haveinteger coefficients.Not all such quadraticpolynomials can be factored usingthe real numbers, and even fewerinto integers. Therefore, when wesay a quadratic can be factored, wemean that we can write the factorswith only integer coefficients.If a quadratic can be factored,it will be the product of two first-degree binomials, except for verysimple cases that just involvemonomials. For example x2byitself is a quadratic expressionwhere the coefficient a is equal to 1,and b and c are zero. We know thatx2factors into (x)(x).Another situation occurs whenonly the coefficient c is zero. Thenyou get something that looks like2x2+ 3xThis can be factored verysimply by factoring out(‘undistributing’) the commonfactor of x:2x2+ 3x = x(2x + 3)The most general case is whenall three terms are present, as inx2+ 5x + 6Mr. 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.
We look at two cases of this type.The easiest to factor are the oneswhere the coefficient of x2(which weare calling ‘a’) is equal to 1, as in theabove example. This is the simplestcase; let’s begin by looking at a = 1examples.Coefficient of x2is 1Since the trinomial comes frommultiplying two first-degreebinomials, let’s review what happenswhen we multiply binomials using theFOIL method. Remember that to dofactoring we will have to think aboutthis process in reverse (you could saywe want to ‘de-FOIL’ the trinomial).Suppose we are given(x + 2)(x + 3)Using the FOIL method, we get(x + 2)(x + 3) = x2 + 3x + 2x + 6Then, collecting like terms gives(x + 2)(x + 3) = x2 + 5x + 6Notice where the terms in thetrinomial came from. The x2camefrom x times x. The interesting part iswhat happens with the other parts, the‘+ 2’ and the ‘+ 3’.The last term in the trinomial, the6 in this case, came from multiplyingthe 2 and the 3. Where did the 5x inthe middle come from? We got the 5xby adding the 2x and the 3x when wecollected like terms. We can state thisas a rule:If the coefficient of x2is one, then tofactor the quadratic you need to findtwo numbers that:1. Multiply to give the constant term(which we call c)2. Add to give the coefficient of x(which we call b)This rule works even if there areminus signs in the quadraticexpression, if we remember how toadd/subtract and multiply/dividepositive and negative numbers.Special Case: Perfect SquareTrinomialRecall from special products ofbinomials :(a+b)2= a2+2ab+b2(a-b)2= a2– 2ab + b2The trinomials on the right are calledperfect squares because they are theMr. Breitsprecher’s Edition March 3, 2005 Algebra Connections, Page 2squares of a single binomial, ratherthan the product of two differentbinomials. A quadratic trinomial canalso have this form:(x + 3)2 = (x + 3)(x + 3) = x2+ 6x + 9Notice that the coefficient of x is thesum 3 + 3, and the constant term is theproduct 3 * 3. One can also say:The coefficient of x is twice thenumber 3The constant term is the numberthree squaredIn general, if a quadratic trinomial is aperfect square, thenThe coefficient of x is twice thesquare root of the constant termOr to restate it another way,The constant term is the square ofhalf the coefficient of xIn symbolic form, we can express thisas:(x+a)2= x2+2ax+a2It is important to be able to recognizeperfect square trinomials. This is thekey to solving quadratic equations.Coefficient of x2is not 1A quadratic is more difficult to factorwhen the coefficient of the squaredterm is not 1, because that coefficientis mixed in with the other productsfrom FOILing the two binomials.There are two methods for attackingthese: either you can use a systematicguess-and-check method, or a methodcalled factoring by grouping. We willfirst look at the guess-and-checkmethod.If you need to factor a trinomial suchas2x2+ x - 3Think about what combinationscould give the 2x2as well as the othertwo terms.In this example the 2x2mustcome from (x)(2x), and the constantterm might come from either (-1)(3)or (1)(-3). Now we must figure outwhich combination will give thecorrect middle term. By “trial anderror” we get:(x-1)(2x+3)Notice these patterns.The first term in the trinomial(the 2x2) is just the product of theleading terms in the binomials.The constant term in the trinomial(the -3) is the product of theconstant terms in the binomials(so far this is the same as in thecase where the coefficient of x2is1)The middle term in the trinomial(the x) is the sum of the outer andinner products. This involves allthe constants and coefficients andis not always obvious.Because 1 and 2 are relatively simpleand 3 is complicated, it makes sense tothink of the possible candidates thatwould satisfy conditions 1 and 2, andthen test them in every possiblecombination by multiplying theresulting binomials to see if you getthe correct middle term.Step-by-Step Process1. List all the possible ways to getthe coefficient of x2(which wecall “a”) by multiplying twonumbers2. List all the possible ways to getthe constant term (which we call“c”) by multiplying two numbers3. Try all possible combinations ofthese to see which ones give thecorrect middle termDon’t forget that the numberitself times 1 is a possibilityIf the number (a or c) isnegative, remember to try theplus and minus signs bothwaysIn our example: 2x2+ x – 3, we makea list of the possible factors of 2x2:The only choice is (2x)(x).Then we make a list of the possiblefactors of the constant term -3: it iseither (1)(-3) or (-1)(3).The possible factors of the trinomialare the binomials that we can make outof these possible factors, taken inevery possible order.(2x + 1)(x – 3)(x + 1)(2x – 3)(2x + 3)(x – 1)(x + 3)(2x – 1)Multiplying these and find the one thatworks (the third one). All you reallyneed to check is to see if the sum ofthe outer and inner multiplicationswill give you the correct middleterm, since we already know that wewill get the correct first and last terms.
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!
Mr. Breitsprecher’s Edition March 16, 2005 Algebra Connections, Page 21(6) + 12(1) = 18.(1, 12) and (2, 3): 1(2) + 12(3) =38. 1(3) + 12(2) = 27.(2, 6) and (1, 6): 2(1) + 6(6) = 38.2(6) + 6(1) = 18.(2, 6) and (2, 3): 2(2) + 6(3) = 22.2(3) + 6(2) = 18.(3, 4) and (1, 6): 3(1) + 4(6) = 27.3(6) + 4(1) = 22.(3, 4) and (2, 3): 3(2) + 4(3) = 18.3(3) + 4(2) = 17.(3x + 2)(4x + 3).Check: (3x + 2)(4x + 3) = 12x2+9x + 8x + 6 = 12x2+ 17x + 6.Example 3: Factor 4x2- 5x - 21Numbers that produce 4: (1, 4),(2, 2).Numbers that produce 21: (1, 21),(3, 7).Continuing:(1,4) and (1,21): 1(1)-4(21)= - 83.1(21) - 4(1) = 17.(1,4) and (3,7): 1(3) - 4(7) = - 25.1(7) - 4(3) = - 5.(x - 3)(4x + 7).Check: (x - 3)(4x + 7) = 4x2+7x -12x - 21 = 4x2- 5x - 21.Analysis: “Old School”Probably, most will find thisprocedure complex – it requires a adifferent rule depending on the signof “c.” Mr. B’s AlgebraConnections DOES NOTRECOMMEND STUDENTS USETHIS METHOD!It is actually the same processpresented in our text, but it relies ona different set of rules usingabsolute values) depending on thesign of “c.” Many will finddifferent set of rules (depending onthe sign of “c”) frustrating – itseems like more to memorize.For most of us, fewer rules arebetter. Our text uses the sameconcepts, but relies on anunderstanding of sign patterns whenfactoring.The key is to remember: anegative number MUST havefactors with different signs (onepositive, the other negative). Apositive number MUST have factorswith the same signs (either bothpoitive or both negative) Seeingpatterns in numbers is important inalgebra. It is often easier to learnthese if we allow someone to“guide” our practice.Factor by Grouping (Unit 4.4)1. Factor out the GCF, if it exists.This should always be your firststep in EVERY factoringproblem (no matter whatmethod you choose to use).2. Find 2 "mystery" numbers.These are similar to thenumbers you sought whenfactoring simpler trinomials -but there is a slight difference.You still want the sum to be thex-coefficient (b), but now youwant their product to equal ac,the product of the leadingcoefficient and the constant.3. Replace the x-coefficient.Rewrite the polynomial, butwhere b once stood, write thesum of the two "mystery"numbers in parentheses. This isthe key - we will now "blowthis trinomial up" into twobinomials.4. Distribute x through theparentheses. Theres still an xmultiplied times the quantity inparentheses from step 3multiply each of the mysterynumbers by x to eliminate thoseparentheses. AT THIS POINTWE HAVE A FACTOR BYGROUPING PROBLEM.5. Factor by Grouping.Example: 6x2-x-12Solution: This polynomial has noGCF. Skip right to calculating the"mystery" numbers; they should addto “b,” in this case, -1 and have aproduct of 6 * (-12) = -72. Thosenumbers, therefore, must be -9 and8. Replace the x-coefficient of -1with the sum of those "mystery"numbers in parentheses (groupingsymbols). This leaves us with:6x2+(-9+8)x-12Distribute the x lying outsidethe parenthesis: 6x2-9x+8x-12Now you can factor by grouping:= 3x(2x-3) + (3x+4)= (2x-3)(3x+4)Analysis: Factor by GroupingThis procedure is direct andappears simpler than other methods.It is based on re-writing trinomialsinto 4 term polynomials (theopposite of simplifying) and issometimes called “decomposition.”There is not as much "guess andby golly" or "trial and error" that isused in the textbook’s method or the“old school” approach. Be carefulabout deciding whether The Bombor our textbooks method is simpler."Trial and error" might seem likeextra work, but with practice, youwill see patterns and choose theright solution more and more often.Our textbook shows exampleswhere we work out every possiblesolution WRONG before we findthe right one. As we learn the“patterns,” we will tend to start withthe RIGHT solution withoutlooking at all of the wrong ones.THIS IS WHY GUIDEDPRACTICE, working with anotherperson (like your professor or theMath Center), is important.Please remember, text usescarefully selected problems toillustrate important patterns innumbers - patterns that reinforcealgebra skills and concepts. Ourtextbook is a beautiful presentationof a complex subject. Pleasepractice Units 4.3 and 4.4. Thiswill build your algebra skills andthen you can decide which to use.To factor ax2+bx+c, try various combinations of factors of ax2and c until a middle term of bx is obtained when checking. K.Elayn Martin-Gay has written our textbook to illustrate these“patterns.” Please practice Unit 4.3 with the Math Center or me.Working together, this presentation will work for you too!Example: 3x2+ 14x - 5Factors of 3x2: (3x, x); Factors of –5: (-1,5) & (1,-5)(3x-1)(x+5) checks – the product of the outer terms (15x)plus the product of the inner terms (-1x) equals the middleterm of our trinomial (14x). Our solution is (3x-1)(x+5)
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
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!
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
Mr. Breitsprecher’s Edition October 3, 2005 Algebra Connections, Page 2Area of a Triangle. Think ofthe area of a triangle as ½ that of animaginary rectangle that includesthe triangle. We need to know thebase (b) or bottom of the triangle.We also need the height (h), theperpendicular distance from the baseto the top angle of the triangle. Theformula is: A = 1/2 bhArea of a Square. Because oftheir square corners (90 degrees)and opposites sides, the area of asquare is the same formula as thearea of a rectangle. It is theproducts of 2 sides (remember, theyare equal). The formula is:A = l (length) * w (width)Area of a Rectangle. For anexplanation, please see above. Theformula is: A = l (length) * w(width).Area of a Circle. The area of acircle is based on the radius (r),which is ½ the diameter (D/2). Theformula is A=πr2VolumeArea is a 2 dimensional, or flatmeasurement. Volume is a 3-dimensional measurement. Think ofit as being based on themeasurement across, front-to-back,and up-and-down. Imaginecounting how many 1-inch squaresugar cubes would fit in a box – thatis what volume does. Volume isalways measure in cubed units(cubic inches, cubic feet, cubiccentimeters, and so forth).Volume of Pyramid (ThreeDimensional Triangle). Thevolume of a pyramid with a squarebase is one-third the area of the base(b) multiplied by the height (h). Theformula is: V= 1/3(s2)*hVolume of a Square/Cube.The 90-degree corners and equalsides make this one straightforward.Note that a 3-dimensional square iscalled a cube. The formula is: V = s3Volume of a Rectangle. Thisone is based on the same idea as thevolume of a square – it’s the productof the length, width, and height.The formula is:V=l(length) *w(width)*h(height)Volume of a Sphere. Thisexpression, like the area, is based onthe radius (r), which is ½ thediameter (D). We also need to usepi. The formula is: V = 4/3(πr3)Volume of a Cylinder. Thinkof this shape as a 3-dimensionalcross between a rectangular box anda sphere. The volume is based onthe radius (r), the height (h) and pi.The formula is: V = πr2hVolume of a Cone. This shapeis a 3-dimensional cross between atriangle and a circle and is based onthe radius (r), the height (h) and pi.The formula is: V = 1/3(πr2h)Temperature ConversionsBoth Fahrenheit and Celsius arebased on the freezing and boilingpoints of water. Fahrenheit uses 32for freezing and 212 for boiling.Celsius uses 0 for freezing and 100for boiling. It is probably easiest towork with Celsius, but we typicallyuse Fahrenheit.Fahrenheit to Celsius.Subtract 32 from the F temperature,multiply by 5 and divide by 9. Theformula is: C= (5/9) (F-32)Celsius to Fahrenheit.Multiple the Celsius temperaturesby 9, divide by 5, and add 32. Theformula is: F = (9/5)C+32More Useful FormulasDistance. The formula forcalculating distances is based on rate(speed expressed as a ratio ofdistance per unit of time) and time.Be sure that the distance isexpressed in the same unit as therate. Read these problems carefully– identify the variables that areknow, express all variables based onthe same units, and then plug theknown variables (with the sameunits) into the formula and solve forthe unknown. The formula is:d = r (rate)*t (time)Percent. Always representinga ratio of some percentage to 100,we calculate this based on the baseand rate. If we express ratios interms of 100, we make comparisonsmore meaningful. When 2 ratios areinvolved, solve percentage problemswith proportions using the crossproduct. The formula is:p (percentage) = b (base) * r(rate)Pythagorean Theorem. Aspecial case of a triangle is when 1angle equals 90 degrees – this is acalled right triangle. ThePythagorean Theorem states that ina right triangle, the sum of thesquares of the lengths of the legs isequal to the square of the length ofthe hypotenuse (the side directlyopposite of the 90 degree or rightangle). Many distance and heightproblems can be solved by applyingthis theorem. The formula isc2(hypotenuse) = a2(side adjacentright angle) + b2(other sideadjacent right angle)Simple Interest. We expect topay people for the privilege ofborrowing money – we pay backmore than we borrow. The “extra”amount is “interest.” There aredifferent ways to calculate this;some are complex. Simple interestis computed only on the originalamount of a loan. To calculate thisamount, we need to know how muchis borrowed (principal), the annualinterest rate (expressed as a decimal,NOT as a percent), and the amountof time the money will be borrowedin years (if given in months, divideby 12 to express as part of a year).The formula is:I (interest) = P (principal)*r(annual rate) * t (time in years)Perimeter• Triangle: P = s1+s2+s3• Square: P = 4s• Rectangle: P = 2l + 2w• Circle: C=πDArea• Triangle: A = 1/2 bhSquare: A = lw• Rectangle: A = lw• Circle: A=πr2Volume:• Pyramid (4--sided base) = 1/3(s2)*h• Square/Cube = s3• Rectangle = lwh• Sphere = 4/3(πr3)• Cylinder = πr2h• Cone = 1/3(πr2h)Temperature Conversions• Fahrenheit to Celsius, C= (5/9) (F-32)• Celsius to Fahrenheit, F = (9/5)C-32Distance• d = rtPercent• p = brSimple Interest• I = PrtNote: Many other geometric shapes can be considered combinationsof these. When appropriate, use these formulas for each part.
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
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.
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)
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.
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.