1. Module in Solving Polynomials for Second Year High School Cabrera, Ronalyn M. Sandagon, Alexandria M. contents next

2. A premier university at CALABARZON offering academic programs and related services designed to respond to the requirements of the Philippines Economy particularly Asian countries. next back contents

3. The university shall primarily provide advance educational professions, technical and vocational instructions in agriculture, fisheries, forestry, science engineering, industrial technology, teachers education, medicine, law, arts and sciences, information technology, and other related fields. It shall undertake research and extension services provide progressive leadership in its area of specialization. contents back next

4. In pursuit of college vision/mission, the college of education is committed to develop the full potential of individual and equip them with knowledge, skills and attitudes in teacher education allied fields to the increasing demands, challenges and opportunities of changing time for global competitiveness. next back contents

6. This Teacher’s “ Module in solving Polynomials ” is part of the requirements in Educational Technology 2 under the revised curriculum for Bachelor in Elementary Education based on CHED Memorandum Order (CMO)-30, Series of 2004. Educational Technology 2 is a three (3)-unit course designed to introduce both traditional and innovative technologies to facilitate and foster meaningful and effective learning where students are expected to demonstrate a sound understanding of the nature, application and production of the various types of educational technologies. The students are provided with guidance and assistance of selected faculty members of the College on the selection, production and utilization of appropriate technology tools in developing technology-based teacher support materials. Through the role and functions of computers especially the Internet, the student researchers and the advisers are able to design and develop various types of alternative delivery systems. These kinds of activities offer a remarkable learning experience for the education students as future mentors especially in the preparation and utilization of instructional materials. The output of the group’s effort on this enterprises may serve as a contribution to the existing body instructional materials that the institution may utilize in order to provide effective and quality education. The lessons and evaluations presented in this module may also function as a supplementary reference for secondary teachers and students. CABRERA, RONALYN M. Module Developer SANDAGON, ALEXANDRIA M. Module Developer next back contents

7. This Teacher’s “ Module in solving Polynomials ” is part of the requirements in Educational Technology 2 under the revised curriculum for Bachelor in Elementary Education based on CHED Memorandum Order (CMO)-30, Series of 2004. Educational Technology 2 is a three (3)-unit course designed to introduce both traditional and innovative technologies to facilitate and foster meaningful and effective learning where students are expected to demonstrate a sound understanding of the nature, application and production of the various types of educational technologies. The students are provided with guidance and assistance of selected faculty members of the College on the selection, production and utilization of appropriate technology tools in developing technology-based teacher support materials. Through the role and functions of computers especially the Internet, the student researchers and the advisers are able to design and develop various types of alternative delivery systems. These kinds of activities offer a remarkable learning experience for the education students as future mentors especially in the preparation and utilization of instructional materials. The output of the group’s effort on this enterprise may serve as a contribution to the existing body instructional materials that the institution may utilize in order to provide effective and quality education. The lessons and evaluations presented in this module may also function as a supplementary reference for secondary teachers and students. FOR-IAN V. SANDOVAL Computer Instructor/Adviser Educational Technology 2 DELIA F. MERCADO Module consultant/Instructor 3 LYDIA R. CHAVEZ Dean College of Education contents back next

8. The authors’ wishes to acknowledge with profound gratitude the many individuals by which in a one way or another gave their in valuable supports and assistant for the completion of this module; Dr. Corazon San Agustin , professor in Educational Technology I, for her guidance and encouragement in completing this requirement. Mr. For-Ian Sandoval , professor in Educational Technology II, for giving us opportunity to finish this module. Prof. Delia Mercado , Director of Laboratory High School, her valuable suggestion. Prof. Lydia R. Chavez , Dean of College Education , for her moral support. To Our loving and understanding parents for giving us moral and financial support and especially for their inspirations. To our beloved friends and dearest classmates who are always there to give support and inspiration. Thank You. And most of all, to our Almighty God , who is source of knowledge and wisdom. THE AUTHORS next back contents

9. This module is prepared for mathematics’ students in secondary levels with the aim enhancing the students’ skill in solving mathematical equations and functions. The main objective of this module for the students to learn by the different ways and steps in solving polynomials such as adding, subtracting, multiplying and dividing polynomial functions. This module is also prepared to enable the students to be able to find the degree and different terms of the polynomial functions. Different theorems and principles are included in this module in order to show how to find the zeros of the polynomial functions. Further contained are the ways to evaluate and simplify Polynomial Functions with real coefficients. The presentation develops a systematic procedure for determining the exact and approximate value of all real zeros of a function. Graphing and sketching the different kinds of graph of polynomials are likewise included. contents back next

11. Acknowledgement Introduction General Objectives CHAPTER 1 POLYNOMIAL FUNCTIONS Lesson 1; Polynomial functions Lesson 2; Classification of Polynomials Lesson 3; Evaluation of Polynomials CHAPTER 2 LAW OF EXPONENTS Lesson 4; Exponents; basic rules Lesson 5; Negative Exponents Lesson 6; Fractional Exponents CHAPTER 3 OPERATIONS OF POLYNOMIALS Lesson 7; Addition of Polynomials Lesson 8; Subtraction of Polynomials Lesson 9; Multiplying Polynomials Lesson 10; Dividing Polynomials Foreword CHAPTER 4 ZEROS OF POLYNOMIALS Lesson 11; Finding Zeros of Polynomials Lesson 12; Rational Zero Theorem CHAPTER 5 GRAPHS OF POLYNOMIAL FUNCTIONS Lesson 13; Polynomial Graphs REFERENCES VMGOs contents back next DEMO SLIDESHARE

14. Here is a typical polynomial: Notice the exponents on the terms. The first term has an exponent of 2; the second term has an "understood" exponent of 1; and the last term doesn't have any variable at all. Polynomials are usually written this way, with the terms written in "decreasing" order; that is, with the largest exponent first, the next highest next, and so forth, until you get down to the plain old number. Any term that doesn't have a variable in it is called a "constant" term because, no matter what value you may put in for the variable x , that constant term will never change. In the picture above, no matter what x might be, 7 will always be just 7. The first term in the polynomial, when it is written in decreasing order, is also the term with the biggest exponent, and is called the "leading term". contents back next 6 x –2 This is NOT a polynomial term... ...because the variable has a negative exponent. 1 / x 2 This is NOT a polynomial term... ...because the variable is in the denominator. sqrt ( x ) This is NOT a polynomial term... ...because the variable is inside a radical. 4 x 2 This IS a polynomial term... ...because it obeys all the rules.

20. Where; c n ≠ 0, c n -1 , ..., c 2 , c 1 and c 0 are constants, the coefficients of this polynomial. Here the term c n x n is called the leading term and its coefficient c n the leading coefficient; if the leading coefficient is 1, the univariate polynomial is called monic. Note that apart from the leading coefficient c n (which must be non-zero or else the polynomial would not be of degree n ) this general form allows for coefficients to be zero; when this happens the corresponding term is zero and may be removed from the sum without changing the polynomial. It is nevertheless common to refer to c i as the coefficient of x i , even when c i happens to be 0, so that x i does not really occur in any term; for instance one can speak of the constant term of the polynomial, meaning c 0 even if it should be zero. Polynomials can similarly be classified by the kind of constant values allowed as coefficients. One can work with polynomials with integral, rational, real or complex coefficients, and in abstract algebra polynomials with many other types of coefficients can be defined. Like for the previous classification, this is about the coefficients one is generally working with; for instance when working with polynomials with complex coefficients one includes polynomials whose coefficients happen to all be real, even though such polynomials can also be considered to be a polynomials with real coefficients. Polynomials can further be classified by their degree and/or the number of non-zero terms they contain. next back contents

21. Usually, a polynomial of degree 4 or higher is referred to as a polynomial of degree n , although the phrases quartic polynomial and quintic polynomial are also used. The names for degrees higher than 5 are even less common. The names for the degrees may be applied to the polynomial or to its terms. For example, a constant may refer to a zero degree polynomial or to a zero degree term. The polynomial 0, which may be considered to have no terms at all, is called the zero polynomial. Unlike other constant polynomials, its degree is not zero. Rather the degree of the zero polynomial is either left explicitly undefined, or defined to be negative (either –1 or –∞). [4] These conventions are important when defining Euclidean division of polynomials. contents back next Polynomials classified by degree Degree Name Example −∞ zero 0 0 (non-zero) constant 1 1 linear x + 1 2 quadratic x 2 + 1 3 cubic x 3 + 1 4 quartic (or biquadratic) x 4 + 1 5 quintic x 5 + 1 6 sextic (or hexic) x 6 + 1 7 septic (or heptic) x 7 + 1 8 octic x 8 + 1 9 nonic x 9 + 1 10 decic x 10 + 1

22. Further, polynomials may be classified by the number of terms (using the minimal number of terms, that is, not counting zero terms and combining like terms). The word monomial can be ambiguous, used either to refer to a polynomial with just a single term, as above, or to refer to the particular case of monic monomials, that is, having coefficient 1. next back contents Polynomials classified by degree Degree Name Example −∞ zero 0 0 (non-zero) constant 1 1 linear x + 1 2 quadratic x 2 + 1 3 cubic x 3 + 1 4 quartic (or biquadratic) x 4 + 1 5 quintic x 5 + 1 6 sextic (or hexic) x 6 + 1 7 septic (or heptic) x 7 + 1 8 octic x 8 + 1 9 nonic x 9 + 1 10 decic x 10 + 1

29. B. Classify each polynomials according to its degree and the number of non zero term. next back contents name degree Non zero term 1. x 2 + x + 10 2. x 8 – 3 3. x 9 + 6 4. x + 3 5. x 3 – 3

41. The fourth root is the one-fourth power: The fifth root is the one-fifth power; and so on. Looking at the first examples, we can re-write them like this: You can enter fractional exponents on your calculator for evaluation, but you must remember to use parentheses. If you are trying to evaluate, say, 15 (4/5) , you must put parentheses around the "4/5", because otherwise your calculator will think you mean "(15 4 ) ÷ 5". Fractional exponents allow greater flexibility (you'll see this lot in calculus), are often easier to write than the equivalent radical format, and permit you to do calculations that you couldn't before. For instance: Whenever you see a fractional exponent, remember that the top number is the power, and the lower number is the root (if you're converting back to the radical format). For instance: Some decimal powers can be written as fractional exponents, too. If "3 5.5 ", recall that 5.5 = 11 / 2 , so: 3 5.5 = 3 11/2 next back contents

42. Generally, though, when you get a decimal power (something other than a fraction or a whole number), you should just leave it as it is, or, if necessary, evaluates it in your calculator. For instance, 3 pi , where pi is the number approximately equal to 3.14159, cannot be simplified or rearranged as a radical. Note: When you are dealing with these exponents with variables, you might have to take account of the fact that you are sometimes taking even roots. Think about it: Suppose you start with the number –2. Then In other words, you put in a negative number, and got out a positive number! This is the official definition of absolute value: If x 3/6 , then x had better not be negative, because x 3 would still be negative, take the sixth root of a negative number. If x 4/6 , then a negative x becomes positive (because of the fourth power) and is then sixth-rooted, so it becomes | x | 2/3 (by reducing the fractional power). On the other hand, if x 4/5 , x is positive or negative, because a fifth root doesn't have any problem with negatives. next back contents

46. C. Change the following into Fractional form. 1. 2. 3. 4. 5. next back contents

62. Solution; The answer is: 2 x 6 + 4 x 5 + x 4 + 11 x 3 + 2 x 2 + 4 x + next back contents

65. next back contents Divide the leading x 2 inside by the leading x in front' Now take that x, and multiply it through the divisor, x + 1. First, multiply the x (on top) by the x (on the "side"), and carry the x 2 underneath: Then multiply the x (on top) by the 1 (on the "side"), and carry the 1x underneath: Then draw the "equals" bar, do the subtraction. To subtract the polynomials, change all the signs in the second line... ...and then add down. The first term (the x 2 ) will cancel out: remember to carry down that last term, the "subtract ten", from the dividend:

66. next back contents Now look at the x from the divisor and the new leading term, the –10x, in the bottom line of the division. divide the –10x by the x, It end up with a –10, so put that on top: Now multiply the –10 (on top) by the leading x (on the "side"), and carry the –10x to the bottom: ...and multiply the –10 (on top) by the 1 (on the "side"), and carry the –10 to the bottom: draw the equals bar, and change the signs on all the terms in the bottom row: Then add down:

76. next back contents Make sure you leave room inside, underneath the row of coefficients, to write another row of numbers later. Put the test zero, x = 1, at the left: Take the first number inside, representing the leading coefficient, and carry it down, unchanged, to below the division symbol: Multiply this carry-down value by the test zero, and carry the result up into the next column: Add down the column: Multiply the previous carry-down value by the test zero, and carry the new result up into the last column: Add down the column: This last carry-down value is the remainder.

77. Comparing, see that there is the same result from the synthetic division, the same quotient (namely, 1x + 6) and the same remainder at the end (namely, 12), as when we did the long division: The results are for result, being a quotient of x + 6, and a remainder of 12. matted differently, but you should recognize that each format provided us with the As you can see above, while the results are formatted differently, the results are otherwise the same: In the long division, divide the factor x + 3, and arrived at the result of x + 2 with a remainder of zero. This means that x + 3 is a factor, and that x + 2 is left after factoring out the x + 3. Setting the factors equal to zero, x = –3 and x = –2 are the zeroes of the quadratic. In the synthetic division, divide x = –3, and arrived at the same result of x + 2 with a remainder of zero. Because the remainder is zero, this means that x + 3 is a factor and x = –3 is a zero. Also, because of the zero remainder, x + 2 is the remaining factor after division. Setting this equal to zero, I get that x = –2 is the other zero of the quadratic. next back contents (from the factoring above) x + 3 is a factor of the polynomial, and therefore that x = –3 is a zero. Compare the results of long division and synthetic division when using the factor x + 3 (for the long division) and the zero x = –3 (for the synthetic division):

80. This question is asking, in effect, to convert an "improper" polynomial "fraction" into a polynomial "mixed number". That is, being asked to do something similar to converting the improper fraction 17 / 5 to the mixed number 3 2 / 5 , which is really the shorthand for the addition expression "3 + 2 / 5 ". To convert the polynomial division into the required "mixed number" format, do the division; As you can see, the remainder is 68. Started with a polynomial of degree 3 and then divided by x – 3 (that is, by a polynomial of degree 1), left with a polynomial of degree 2. Then the bottom line represents the polynomial 3x 2 + 7x + 24 with a remainder of 68. Putting this result into the required "mixed number" format, I get the answer as being: next back contents First, write down all the coefficients, and put the zero from x – 3 = 0 (so x = 3) at the left. Next, carry down the leading coefficient: Multiply by the potential zero, carry up to the next column, and add down: Repeat this process: Repeat this process again:

81. It is always true that, when you use synthetic division, your answer (in the bottom row) will be of degree one less than what you'd started with, because you have divided out a linear factor. The Remainder Theorem The Remainder Theorem is useful for evaluating polynomials at a given value of x , though it might not seem so, at least at first blush. This is because the tool is presented as a theorem with a proof, Fortunately, it need not to understand the proof of the Theorem; it needs to understand how to use the Theorem. The Remainder Theorem starts with an unnamed polynomial p ( x ), where " p ( x )" just means "some polynomial p whose variable is x ". Then the Theorem talks about dividing that polynomial by some linear factor x – a , where a is just some number. Then, as a result of the long polynomial division, end up with some polynomial answer q ( x ) (the " q " standing for "the quotient polynomial") and some polynomial remainder r ( x ). As a concrete example of p , a , q , and r , let's look at the polynomial p ( x ) = x 3 – 7 x – 6, and let's divide by the linear factor x – 4 (so a = 4): The answer is q ( x ) = x 2 + 4 x + 9 on top, with a remainder of r ( x ) = 30. next back contents

96. Since the remainder is zero, then x = 2 is indeed a zero of the original polynomial. To continue on and find the rest of the zeroes, should I start over again with x 4 + x 3 –11x 2 – 5x + 30? Well, think about when factoring something like 72. After dividing a 2 out and get a 36, do you go back to the 72 to try the next factor, or do you see what will go into the 36? Of course, try factors into the 36. Follow the same procedure here. Return to the original polynomial, but instead see what divides into my result. (Recall that synthetic-dividing out x = 2 is the same as long-dividing out x – 2, so the result has a degree that is one lower than what it started with. That is, to continue, deal not with the original fourth-degree polynomial x 4 + x 3 –11x 2 – 5x + 30, but with the third-degree result from the synthetic division: x 3 + 3x 2 – 5x – 15.) Continuing, and again comparing the Rational Roots Test with a quick graph, try x = –3. Set up the division: ...and here is the result: contents back next

97. Since the remainder is zero, then x = –3 is a zero of the original polynomial. At this point, the final result is a quadratic, (x 2 – 5), and I can apply the Quadratic Formula or other methods to get the remaining zeroes: Then all the zeroes are The above example shows how synthetic division is most-commonly used: given some polynomial, and told to find all of its zeroes. Create a list of possibilities, using the Rational Roots Test; plug various of these possible zeroes into the synthetic division until one of them "works" (divides out evenly, with a zero remainder); then try additional zeroes on the resulting (and lower-degree) polynomial until something else "works"; and keep going like this until getting down to a quadratic, at which point use the Quadratic Formula or other methods to get the last two of the original polynomial's zeroes. next back contents

106. As you can see, even-degree polynomials are either "up" on both ends (entering and then leaving the graphing "box" through the "top") or "down" on both ends (entering and then leaving through the "bottom"), depending on whether the polynomial has, respectively, a positive or negative leading coefficient. On the other hand, odd-degree polynomials have ends that head off in opposite directions. If they start "down" (entering the graphing "box" through the "bottom") and go "up" (leaving the graphing "box" through the "top"), they're positive polynomials; if they start "up" and go "down", they're negative polynomials. All even-degree polynomials behave, on their ends, like quadratics, and all odd-degree polynomials behave, on their ends, like cubics. contents back next with a negative leading coefficient with a positive leading coefficient with a positive leading coefficient

107. Which of the following could be the graph of a polynomial whose leading term is "–3 x 4 "? The important things to consider are the sign and the degree of the leading term. The exponent says that this is a degree-4 polynomial, so the graph will behave roughly like a quadratic: up on both ends or down on both ends. Since the sign on the leading coefficient is negative, the graph will be down on both ends. (The actual value of the negative coefficient, –3 in this case, is actually irrelevant for this problem. All I need is the "minus" part of the leading coefficient.) Clearly Graphs A and C represent odd-degree polynomials, since their two ends head off in opposite directions. Graph D shows both ends passing through the top of the graphing box, just like a positive quadratic would. The only graph with both ends down is: Graph B next back contents

109. The intercept at x = –5 is of multiplicity 2. The polynomial is of degree 5, so the zero at x = 5, the only other zero, must use up the rest of the multiplicities. Since 5 – 2 = 3, then x = 5 must be of multiplicity 3. The zero at x = 5 had to be of odd multiplicity, since the graph went through the x -axis. But the graph flexed a bit (the "flexing" being that bendy part of the graph) right in the area of x = 5. This flexing is what tells you that the multiplicity of x = 5 had to be more than just 1. In this particular case, the multiplicity couldn't have been 5 or 7 or more, because the degree of the whole polynomial was only 5,but the multiplicity certainly had to be more than just 1. Keep this in mind: Any odd-multiplicity zero that flexes at the crossing point, like this graph did at x = 5, is of multiplicity 3 or more. Note: If you get that odd flexing behavior at some location on the graph that is off the x -axis (above or below the axis), then you're probably looking at the effect of complex zeroes; namely, the zeroes that you'd find by using the Quadratic Formula , the zeroes that don't correspond to the graph crossing the x -axis. Degrees, Turnings, and "Bumps " Graphs don't always head in just one directly, like nice neat straight lines ; they can turn around and head back the other way. It isn't standard terminology, and you'll learn the proper terms when you get to calculus, but refer to the "turnings" of a polynomial graph as its "bumps". next back contents

110. Compare the numbers of bumps in the graphs below to the degrees of their polynomials: contents back next degree two degree 3 degree 3 degree 4 degree 4 one bump no bumps, but one flex point two bumps one (flattened) bump three bumps degree 5 degree 5 degree 5 degree 6 degree 6 degree 6 no bumps, but one flex point two bumps (one flattened) four bumps one (flat) bump three bumps (one flat) five bumps

111. From these graphs that, for degree n , the graph will have, at most, n – 1 bumps. The bumps represent the spots where the graph turns back on itself and heads back the way it came. This change of direction often happens because of the polynomial's zeroes or factors. But extra pairs of factors don't show up in the graph as much more than just a little extra flexing or flattening in the graph. Because pairs of factors have this habit of disappearing from the graph (or hiding as a little bit of extra flexure or flattening), the graph may have two fewer, or four fewer, or six fewer, etc, bumps than you might otherwise expect, or it may have flex points instead of some of the bumps. That is, the degree of the polynomial gives you the upper limit (the ceiling) on the number of bumps possible for the graph (this upper limit being one less than the degree of the polynomial), and the number of bumps gives you the lower limit (the floor) on degree of