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Module in solving polynomial
 

Module in solving polynomial

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Module in solving polynomial Module in solving polynomial Presentation Transcript

  • Module in Solving Polynomials for Second Year High School Cabrera, Ronalyn M. Sandagon, Alexandria M. contents next
  • 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
  • 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
  • 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
    • Produce graduates who can demonstrate and practice the professional and ethical requirement for the Bachelor of Secondary Education such as;
    • to serve as positive and role model in the pursuit of learning thereby maintaining high regard to professional growth;
    • focus on the significance of the providing wholesome and desirable learning environment;
    • facilitate learning process in diverse type of learners;
    • use varied approaches and activities, instructional materials, and learning resources;
    • use assessment date to plan and revise the teaching learning plans;
    • direct and strengthen the links between schools and community activities;
    • conduct research and development in teacher education and other related activities.
    contents back next
  • 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
  • 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
  • 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
  • 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
    • At the end of this module, the students are expected to:
    • classify the different operations used in solving polynomial;
    • identify the factors and the zeros of polynomial;
    • apply different laws in solving polynomial problem;
    • evaluate the degree of polynomial;
    • relate other equation like quadratic equation in polynomial equation;
    • follow the laws of exponent regarding polynomial problem;
    • compare and differentiate the graph of polynomial to other graph;
    • avoid confusion with the sign;
    • define what kind of graphs are form in different degree of polynomial; and
    • develop the skills of students in solving mathematical problems.
    contents back next
  • 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
  • POLYNOMIALS
    • Objectives:
    • At the end of this chapter, student must be able to:
        • name the parts of polynomial functions;
        • compare exponent from variables;
        • name an algebraic expression; and
        • simplify polynomial function.
    Chapter I contents back next
    • Objectives;
    • At the end of this lesson the students will be able to;
        • . define Polynomial Functions;
        • . write a polynomials in ascending and descending order; and
        • . formulate a Polynomial functions.
    Be familiar with variables and exponents, and you may have dealt with expressions like 3 x 4 or 6 x . Polynomials are sums of these "variables and exponents" expressions. Each piece of the polynomial, each part that is being added, is called a "term". Polynomial terms have variables which are raised to whole-number exponents (or else the terms are just plain numbers); there are no square roots of variables, no fractional powers, and no variables in the denominator of any fractions. Here are some examples: LESSON 1 Polynomial Functions next back contents
  • 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.
    • The exponent on a term tells you the "degree" of the term. For instance, the leading term in the above polynomial is a "second-degree term" or "a term of degree two". The second term is a "first degree" term. The degree of the leading term tells you the degree of the whole polynomial; the polynomial above is a "second-degree polynomial". Here are a couple more examples:
    • Give the degree of the following polynomial:  2 x 5 – 5 x 3 – 10 x + 9
    • This polynomial has four terms, including a fifth-degree term, a third-degree term, a first-degree term, and a constant term.
    • This is a fifth-degree polynomial.
    • Give the degree of the following polynomial:  7 x 4 + 6 x 2 + x
    • This polynomial has three terms, including a fourth-degree term, a second-degree term, and a first-degree term. There is no constant term.
    • This is a fourth-degree polynomial.
    • When a term contains both a number and a variable part, the number part is called the "coefficient". The coefficient on the leading term is called the "leading" coefficient.
    • In the above example, the coefficient of the leading term is 4; the coefficient of the second term is 3; the constant term doesn't have a coefficient.   
    • The "poly" in "polynomial" means "many". I suppose, technically, the term "polynomial" should only refer to sums of many terms, but the term is used to refer to anything from one term to the sum of a zillion terms. However, the shorter polynomials do have their own names:
    • a one-term polynomial, such as 2 x or 4 x 2 , may also be called a "monomial" ("mono" meaning "one")
    • a two-term polynomial, such as 2 x + y or x 2 – 4, may also be called a "binomial" ("bi" meaning "two")
    • a three-term polynomial, such as 2 x + y + z or x 4 + 4 x 2 – 4, may also be called a "trinomial" ("tri" meaning "three")
    contents back next
    • Polynomials are also sometimes named for their degree:
      • a second-degree polynomial, such as 4 x 2 , x 2 – 9, or ax 2 + bx + c , is also called a "quadratic"
      • a third-degree polynomial, such as –6 x 3 or x 3 – 27, is also called a "cubic"
      • a fourth-degree polynomial, such as x 4 or 2 x 4 – 3 x 2 + 9, is sometimes called a "quartic"
      • a fifth-degree polynomial, such as 2 x 5 or x 5 – 4 x 3 – x + 7, is sometimes called a "quintic“
    • Therefore, "quad" generally refers to "four", as when an ATV is referred to as a "quad bike". For polynomials, however, the "quad" from "quadratic" is derived from the Latin for "making square". As in, if you multiply length by width (of, say, a room) to find the area in "square" units, the units will be raised to the second power. The area of a room that is 6 meters by 8 meters is 48 m 2 . So the "quad" refers to the four corners of a square, from the geometrical origins of parabolas and early polynomial
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  • Activity 1
    • Name; ___________________________ Date; ___________________
    • Year & Section; _____________________ Score; __________________
    • A. Identify which of the following defines polynomial functions.
    • 3/ x 4 + 2/x 3 – 1/x 2 + 2x = 0
    • x 4 + 3x -1 + 1
    • P(x) = 0
    • x 3 – 2x + 1
    • x 5 – 3x – 1
    contents back next
    • Write each polynomial functions in descending power of x.
    • P(x) = 4 – 3x 3 + 5x + 2x 5
    • 2x + 23 – x 4 + 6x 3 + x 2 = 0
    • P(x) = x 6 + 16 – x 5 + 11x
    • 2x 3 + 3 – x 4 + 5xy
    • 5 – x 3 + 7x + x 2 = 0
    next back contents
  • LESSON 2
    • Objectives
    • At the end of this lesson, the students are expected to;
        • classify a polynomial functions according to degrees;
        • name a polynomial functions according to its terms; and
        • state the terms of polynomial functions.
    Classification of Polynomials Polynomials are classified according to many different properties. One classification of polynomials is based on the number of distinct variables. A polynomial in one variable is called a univariate polynomial; a polynomial in more than one variable is called a multivariate polynomial. These notions refer more to the kind of polynomials one is generally working with than to individual polynomials; for instance when working with univariate polynomials one does not exclude constant polynomials (which may result for instance from the subtraction of non-constant polynomials), although strictly speaking constant polynomials do not contain any variables at all. It is possible to further classify multivariate polynomials as bivariate, trivariate etc., according to the maximum number of variables used. It is common, for instance, to say simply "polynomials in x , y , and z ". A polynomial is called homogeneous of degree n if all its terms have degree n . Univariate polynomials have many properties not shared by multivariate polynomials. For instance, the terms of a univariate polynomial are completely ordered (in a natural way) by their degree. A univariate polynomial in x of degree n then takes the general form back contents next
  • 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
  • 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
  • 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
  • Activity 2 Name; ___________________________ Date; ___________________ Year & Section; _____________________ Score; __________________ A. Classify each polynomial according to its degree and the number of its non-zero term. contents back next Name Degree Number of non zero term
    • X 3 ­ + 3x
    2. X 2 - 2 3. X 5 + 2x – 2 4. X 3 + x + 28 5. X 2 + 4
  • LESSON 3 Evaluation of polynomial functions
    • Objectives;
    • At the end of this lesson, the students will be able to
        • . evaluate polynomial functions;
        • . solve for the value of a variable; and
        • . simplify the value of a variable.
    • "Evaluating" a polynomial is the same as evaluating anything else: you plug in the given value of x , and figure out what y is supposed to be. For instance:
    • Evaluate 2 x 3 – x 2 – 4 x + 2 at x = – 3
    • I need to plug in "–3" for the " x ", remembering to be careful with my parentheses and the negatives:
    • 2(–3) 3 – (–3) 2 – 4(–3) + 2   =  2(–27) – (9) + 12 + 2   =  –54 – 9 + 14   =  –63 + 14   =  –49
    • Always remember to be careful with the minus signs!
    • Polynomials: Combining "Like Terms"
    • Probably the most common thing you will be doing with polynomials is "combining like terms". This is the process of adding together whatever terms you can, but not overdoing it by trying to add together terms that can't actually be combined. Terms can be combined ONLY IF they have the exact same variable part. Here is a rundown of what's what:
    next back contents
    • Once you have determined that two terms are indeed "like" terms and can indeed therefore be combined, you can then deal with them in a manner similar to what you did in grammar school. When you were first learning to add, you would do "five apples and six apples is eleven apples". You have since learned that, as they say, "you can't add apples and oranges". That is, "five apples and six oranges" is just a big pile of fruit; it isn't something like "eleven applanges". Combining like terms works much the same way.
    • Simplify 3 x + 4 x
    • These are like terms since they have the same variable part, so I can combine the terms: three x 's and four x 's makes seven x 's:   
    • 3 x + 4 x = 7 x
    • Simplify 2 x 2 + 3 x – 4 – x 2 + x + 9
    • It is often best to group like terms together first, and then simplify:
    • 2 x 2 + 3 x – 4 – x 2 + x + 9   =  (2 x 2 – x 2 ) + (3 x + x ) + (–4 + 9)   =  x 2 + 4 x + 5
    • In the second line, write the coefficient of 1 in front of variable expressions with no written coefficient, as is shown in red below:
    • (2 x 2 – x 2 ) + (3 x + x ) + (–4 + 9)   = (2 x 2 – 1 x 2 ) + (3 x + 1 x ) + (–4 + 9)   = 1 x 2 + 4 x + 5   =  x 2 + 4 x + 5
    next back contents 4 x and 3 NOT like terms The second term has no variable 4 x and 3 y NOT like terms The second term now has a variable, but it doesn't match the variable of the first term 4 x and 3 x 2 NOT like terms The second term now has the same variable, but the degree is different 4 x and 3 x LIKE TERMS Now the variables match and the degrees match
    • Simplify 10 x 3 – 14 x 2 + 3 x – 4 x 3 + 4 x – 6
    • 10 x 3 – 14 x 2 + 3 x – 4 x 3 + 4 x – 6   =  (10 x 3 – 4 x 3 ) + (–14 x 2 ) + (3 x + 4 x ) – 6   =  6 x 3 – 14 x 2 + 7 x – 6
    • Note: When moving the terms around, remember that the terms' signs move with them. Don't mess yourself up by leaving orphaned "plus" and "minus" signs behind.
    • Simplify 25 – ( x + 3 – x 2 )
    • The first thing to do is take the negative through the parentheses:
    • 25 – ( x + 3 – x 2 )   =  25 – x – 3 + x 2   =  x 2 – x + 25 – 3   = x 2 – x + 22 0
    • Keep track of the negative sign, put the understood 1 in front of the parentheses:
    • 25 – ( x + 3 – x 2 )   =  25 – 1 ( x + 3 – x 2 )   =  25 – 1 x – 3 + 1 x 2   =  1 x 2 – 1 x + 25 – 3   =  1 x 2 – 1 x + 22   = x 2 – 1 x + 22
    • While the first format (without the 1's being written in) is the more "standard" format, either format should be acceptable. Use the format that works most successfully for you.
    • Simplify x + 2( x – [3 x – 8] + 3)
    • This is just an order of operations problem with a variable in it. If it work carefully from the inside out, paying careful attention to "minus" signs, then ;
    • x + 2( x – [3 x – 8] + 3)   =  x + 2( x – 1[3 x – 8] + 3)   =  x + 2( x – 3 x + 8 + 3)   =  x + 2(–2 x + 11)   =  x – 4 x + 22   =  –3 x + 22
    next back contents
    • Simplify [(6 x – 8) – 2 x ] – [(12 x – 7) – (4 x – 5)]
    • Work from the inside out:
    • [(6 x – 8) – 2 x ] – [(12 x – 7) – (4 x – 5)]   =  [6 x – 8 – 2 x ] – [12 x – 7 – 4 x + 5]   =  [4 x – 8] – [8 x – 2]   =  4 x – 8 – 8 x + 2   =  –4 x – 6
    • Simplify – 4 y – [3 x + (3 y – 2 x + {2 y – 7} ) – 4 x + 5]
    • – 4 y – [3 x + (3 y – 2 x + {2 y – 7} ) - 4 x + 5]   =  –4 y – [3 x + (3 y – 2 x + 2 y – 7) - 4 x + 5]   =  –4 y – [3 x + (–2 x + 5 y – 7) – 4 x + 5]   =  –4 y – [3 x – 2 x + 5 y – 7 – 4 x + 5]   =  –4 y – [3 x – 2 x – 4 x + 5 y – 7 + 5]   =  –4 y – [–3 x + 5 y – 2]   =  –4 y + 3 x – 5 y + 2   =  3 x – 4 y – 5 y + 2   =  3 x – 9 y + 2
    • Note: Don't get careless and confuse multiplication and addition. This may sound like a silly thing to say, but it is the most commonly-made mistake (after messing up the order of operations):
    • ( x )( x ) = x 2     (multiplication)
    • x + x = 2 x     (addition)
    • " x 2 "  DOES NOT EQUAL   " 2 x "
    next back contents
  • CHAPTER TEST
      • . Identify which of the following expressions defines polynomials.
    • x 3 + 3x – 5
    • P(g)= 5 -x + 4x – 2/3x
    • 0 = 3x – 1 -10 + 6x
    • P(x) = x 12 + 6x 11 – 3x
    • ax 2 + 7 + 5x =
    next back contents
  • 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
  • C. Evaluate the following polynomials.
    • 2x 4 – 3 x 12 + 6x 11 – 3x , x= 2
    • x 2 + 3 , x = -2
    • x 6 + 16 – x 5 + 11x, x = 1
    • 5 – x 3 + 7x + x 2 , x = 0
    • 2x + 23 – x 4 + 6x 3 + x 2 , x = 3
    next back contents
  • CHAPTER II LAW OF EXPONENTS
    • Objectives;
    • At the end of this chapter, student must be able to:
    • reduce expression to its simplest form;
    • derive negative exponent to positive exponent; and
    • realize the important relationship between exponent and radicals.
    back contents next
  • LESSON 4 Exponents; Basic Rules
    • Objectives;
    • At the end of this lesson, the students will be able to;
        • define exponents;
        • appreciate the importance of knowing the laws of exponents; and
        • apply the law of exponents.
    Exponents are shorthand for repeated multiplication of the same thing by itself. For instance, the shorthand for multiplying three copies of the number 5 three is shown on the right-hand side of of the "equals" sign in (5)(5)(5) = 5 3 . The "exponent", being 3 in this example, stands for however many times the value is being multiplied. The thing that's being multiplied, being 5 in this example, is called the "base". This process of using exponents is called "raising to a power", where the exponent is the "power". The expression "5 3 " is pronounced as "five, raised to the third power" or "five to the third". There are two specially-named powers: "to the second power" is generally pronounced as "squared", and "to the third power" is generally pronounced as "cubed". So "5 3 " is commonly pronounced as "five cubed". When we deal with numbers, we usually just simplify; rather deal with "27" than with "3 3 ". But with variables, we need the exponents, because we'd rather deal with " x 6 " than with " xxxxxx ". Exponents have a few rules that we can use for simplifying expressions contents back next
    • Simplify ( x 3 )( x 4 )    
    • To simplify this, think in terms of what those exponents mean. "To the third" means "multiplying three copies" and "to the fourth" means "multiplying four copies". Using this fact, I can "expand" the two factors, and then work backwards to the simplified form:
    • ( x 3 )( x 4 ) = ( xxx ) ( xxxx )            =  xxxxxxx             =  x 7
    • Note that  x 7  also equals  x (3+4) . This demonstrates the first basic exponent rule: Whenever you multiply two terms with the same base, you can add the exponents:
    • (  x  m  ) (  x  n  ) =  x ( m + n )
    • However, we can NOT simplify ( x 4 )( y 3 ), because the bases are different: ( x 4 )( y 3 ) =  xxxxyyy  = ( x 4 )( y 3 ). Nothing combines.
    • Simplify ( x 2 ) 4
    • Just as with the previous exercise, think in terms of what the exponents mean. The "to the fourth" means that multiplying four copies of  x 2 :
    • ( x 2 ) 4  = ( x 2 )( x 2 )( x 2 )( x 2 )         = ( xx )( xx )( xx )( xx )         =  xxxxxxxx          =  x 8
    • Note that  x 8  also equals  x ( 2×4 ) . This demonstrates the second exponent rule: Whenever exponent expression that is raised to a power, multiply the exponent and power:
    • (  x m  )  n  =  x  m n
    next back contents
  • If a product inside parentheses and a power on the parentheses, then the power goes on each element inside. For instance, ( xy 2 ) 3  = ( xy 2 )( xy 2 )( xy 2 ) = ( xxx )( y 2 y 2 y 2 ) = ( xxx )( yyyyyy ) =  x 3 y 6  = ( x ) 3 ( y 2 ) 3 . Another example would be: Warning: This rule does NOT work if a sum or difference within the parentheses. Exponents, unlike multiplications, do NOT "" over addition. For instance, given (3 + 4) 2 , do NOT succumb to the temptation to say "This equals 3 2  + 4 2  = 9 + 16 = 25", because this is wrong. Actually, (3 + 4) 2  = (7) 2  = 49, not 25. When in doubt, write out the expression according to the definition of the power. Given ( x  – 2) 2 , don't try to do this in your head. Instead, write it out: "squared" means "times itself", so ( x  – 2) 2  = ( x  – 2)( x  – 2) =  xx  – 2 x  – 2 x  + 4 =  x 2  – 4 x  + 4. Anything to the power zero is just "1".
    • This rule means that some exercises may be a lot easier than they may at first appear:
    • Simplify [(3 x 4 y 7 z 12 ) 5  (–5 x 9 y 3 z 4 ) 2 ] 0
    • The zero power on the outside means that the value of the entire thing is just  1 .
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  • Activity 4
    • Name; _________________________ Date; ____________________
    • Year & section; __________________ Score: ___________________
    • A. Simplify the following expressions and apply the Laws of Exponents.
    • (a-b) 3 / a-b
    • 25 100 / 125 50
    • (8x 2 - x) 3
    • (12x 3 + 2x 2 – 2x + 3) 4
    • 25 2 + 32 2
    next back contents
  • LESSON 5 Negative Exponents
    • Objectives;
    • At the end of this lesson, the student will be able to;
      • simplify negative exponents;
      • show the rule of simplifying negative exponent.
    A negative exponent just means that the base is on the wrong side of the fraction line, it needs to flip the base to the other side. For instance, " x –2 " (ecks to the minus two) just means " x 2 , but underneath, as in 1/( x 2 )". Write  x –4  using only positive exponents.
    • Write   x 2   /   x – 3   using only positive exponents.
    • Write   2 x – 1   using only positive exponents.
    Note that the "2" above does not move with the variable; the exponent is only on the " x ". next back contents
    • Write   (3 x ) – 2   using only positive exponents.
    Unlike the previous exercise, the parentheses meant that the negative power did indeed apply to the three as well as the variable.
    • Write   ( x – 2   /   y – 3 ) – 2   using only positive exponents.
    This one can also be done as ; Since exponents indicate multiplication, and since order doesn't matter in multiplication, there will often be more than one sequence of steps that will lead to a valid simplification of a given exercise. As long as steps were correct, it will end up with the exact answer . Anything to the power zero is just "1". It is stated that "because that's how the rules work out." Another would be to trace through a progression like the following : 3 5  = 3 6  ÷ 3 = 3 6  ÷ 3 1  = 3 6–1  = 3 5 = 243  3 4  = 3 5  ÷ 3 = 3 5  ÷ 3 1  = 3 5–1  = 3 4 = 81  3 3  = 3 4  ÷ 3 = 3 4  ÷ 3 1  = 3 4–1  = 3 3 = 27  3 2  = 3 3  ÷ 3 = 3 3  ÷ 3 1  = 3 3–1  = 3 2 = 9  3 1  = 3 2  ÷ 3 = 3 2  ÷ 3 1  = 3 2–1  = 3 1 = 3 next back contents
  • Then logically  3 0  = 3 1  ÷ 3 1  = 3 1–1  = 3 0  = 1.
    • A negative-exponents explanation of the "anything to the zero power is just 1" might be as follows:
    • m 0  =  m ( n  –  n )  =  m n  ×  m – n  =  m n  ÷  m n  = 1
    • ...since anything divided by itself is just "1".
    • Anything to the zero power is "1", so 0 0  = 1.
    • Zero to any power is zero, so 0 0  = 0.
    • The "math gods" have not yet settled on a "definition" of 0 0 . In fact, in calculus, "0 0 " will be called an "indeterminate form".
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  • Activity 5
    • Name; ___________________________ Date; ___________________
    • Year & Section; _____________________ Score; __________________
    • A. Simplify the polynomial expression following the law of exponent.
    • ( m -4 / n -2 ) -3
    • s -9 / s 2
    • x 6 / x -5
    • ( 4y) -2
    • 100 -9
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  • LESSON 6 Fractional (Rational) Exponents
    • Objectives;
    • At the end of this topic, the students will be able to;
          • know the law of simplifying Fractional exponents; and
          • simplify factional exponents.
    You already know of one relationship between exponents and radicals: the appropriate radical will "undo" an exponent, and the right power will "undo" a root. For example: But there is another relationship (which can make computations like those above much simpler): For the square (or "second") root, we can write it as the one-half power, like this: The cube (or "third") root is the one-third power: next back contents
  • 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
  • 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
  • Activity 6
    • Name; ___________________________ Date; ___________________
    • Year & Section; _____________________ Score; __________________
    • Change the following in fractional form.
    • 1.
    • 2.
    • 3.
    • 4.
    • 5.
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  • CHAPTER TEST
      • . Simplify the following expression and apply the law of exponent.
    • 64 n ) 2 + (x-2) 3
    • (1 / 81) 3 + (125 + x) 2
    • (Y - 3x - 2x + 2) 2
    • 6 6
    • ( x + 2 )
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  • B. Simplify the following exponent applying the law of negative exponent.
    • ( 2r -6 ) -3
    • ( 5a -7 ) -4
    • ( 10g ) -3
    • ( 3ab -2 / cd -3 ) -4
    • ( h -4 / k 5 ) -8
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  • C. Change the following into Fractional form. 1. 2. 3. 4. 5. next back contents
  • OPERATIONS OF POLYNOMIALS CHAPTER III
    • Objectives;
    • At the end of this chapter, student must be able to:
      • understand the simplest format of solving polynomials;
      • recognize the use of different operation and signs; and
      • apply the different theorem and process in simplifying polynomial functions.
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  • LESSON 7 Addition of polynomials
    • Objectives
        • At the end of this lesson, the students will be able to;
          • add polynomial functions;
          • combine similar terms; and
          • apply the properties of addition.
    • Adding polynomials is just a matter of combining like terms, with some order of operations considerations thrown in. As long as you're careful with the minus signs, and don't confuse addition and multiplication, you should do fine.
    • There are a couple formats for adding and subtracting, and they hearken back to earlier times, when you were adding and subtracting just plain old numbers. First, you learned addition "horizontally", add polynomials in the same way, grouping like terms and then simplify.
    • Simplify (2 x + 5 y ) + (3 x – 2 y )
    • I'll clear the parentheses, group like terms, and then simplify:
    • (2 x + 5 y ) + (3 x – 2 y )   =  2 x + 5 y + 3 x – 2 y   =  2 x + 3 x + 5 y – 2 y   =  5 x + 3 y
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    • Horizontal addition works fine for simple examples. But when adding plain old numbers, generally try not to add 432 and 246 horizontally; instead, "stack" them vertically, one on top of the other, and then add down the columns:
    • Do the same thing with polynomials. This is how the above simplification exercise looks when it is done "vertically":
    • Simplify (2 x + 5 y ) + (3 x – 2 y )
      • Put each variable in its own column; in this case, the first column will be the x -column, and the second column will be the y -column:
    We can get the same solution vertically as I got horizontally: 5 x + 3 y . The format use, horizontal or vertical, is a matter of taste (unless the instructions explicitly tell you otherwise). Given a choice, use whichever format that you're more comfortable and successful with. Note that, for simple additions, horizontal addition (so you don't have to rewrite the problem) is probably simplest, but, once the polynomials get complicated, vertical is probably safest bet (so you don't "drop", or lose, terms and minus signs).
      • Simplify (3 x 3 + 3 x 2 – 4 x + 5) + ( x 3 – 2 x 2 + x – 4)
    • add horizontally:
      • (3 x 3 + 3 x 2 – 4 x + 5) + ( x 3 – 2 x 2 + x – 4)   =  3 x 3 + 3 x 2 – 4 x + 5 + x 3 – 2 x 2 + x – 4   =  3 x 3 + x 3 + 3 x 2 – 2 x 2 – 4 x + x + 5 – 4   =  4 x 3 + 1 x 2 – 3 x + 1
    contents next back
    • Simplify (7 x 2 – x – 4) + ( x 2 – 2 x – 3) + (–2 x 2 + 3 x + 5)
    • It's perfectly okay to have to add three or more polynomials at once. I'll just go slowly and do each step thoroughly, and it should work out right.
    • Adding horizontally:
        • (7 x 2 – x – 4) + ( x 2 – 2 x – 3) + (–2 x 2 + 3 x + 5)   =  7 x 2 – x – 4 + x 2 – 2 x – 3 + –2 x 2 + 3 x + 5   =  7 x 2 + 1 x 2 – 2 x 2 – 1 x – 2 x + 3 x – 4 – 3 + 5   =  8 x 2 – 2 x 2 – 3 x + 3 x – 7 + 5   =  6 x 2 – 2
    Note the 1's in the third line. Any time you have a variable without a coefficient, there is an "understood" 1 as the coefficient. If you find it helpful to write that 1 in, then do so. Adding vertically:
    • Either way, I get the same answer: 6 x 2 – 2
    • Simplify ( x 3 + 5 x 2 – 2 x ) + ( x 3 + 3 x – 6) + ( – 2 x 2 + x – 2)
    • Horizontally:
        • ( x 3 + 5 x 2 – 2 x ) + ( x 3 + 3 x – 6) + (–2 x 2 + x – 2)   =  x 3 + 5 x 2 – 2 x + x 3 + 3 x – 6 + –2 x 2 + x – 2   =  x 3 + x 3 + 5 x 2 – 2 x 2 – 2 x + 3 x + x – 6 – 2   =  2x 3 + 3 x 2 + 2 x – 8
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  • Activity 7
    • Name; ___________________________ Date; ___________________
    • Year & Section; _____________________ Score; __________________
    • Give the sum of each pair of polynomials.
    • 3x 2 - 4x + 5 , 2x 3 – 7x – 2
    • 10 ( 2a – 3b c) , 4a + 2b – 5c
    • ( xy- 2xy 2 ) , 4xy 2
    • 2x 2 – 5x + 5 , 3x 2 + x – 7
    • 3p 3 + 2p + 6 , 4p 3 + p + 2
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  • LESSON 8 Subtracting Polynomials
    • Objectives;
    • At the end of this topic, the students will be able to;
          • subtract polynomial functions;
          • find the difference of the two similar terms; and
          • apply the properties of subtraction.
    • Subtracting polynomials is quite similar to adding polynomials, but you have that pesky minus sign to deal with. Here are some examples, done both horizontally and vertically:
    • Simplify ( x 3 + 3 x 2 + 5 x – 4) – (3 x 3 – 8 x 2 – 5 x + 6)
    • The first thing to do is take that negative through the parentheses.
    • Horizontally:
        • ( x 3 + 3 x 2 + 5 x – 4) – (3 x 3 – 8 x 2 – 5 x + 6)   =  ( x 3 + 3 x 2 + 5 x – 4) – 1(3 x 3 – 8 x 2 – 5 x + 6)   =  ( x 3 + 3 x 2 + 5 x – 4) – 1(3 x 3 ) – 1 (–8 x 2 ) – 1(–5 x ) – 1(6)   =  x 3 + 3 x 2 + 5 x – 4 – 3 x 3 + 8 x 2 + 5 x – 6   =  x 3 – 3 x 3 + 3 x 2 + 8 x 2 + 5 x + 5 x – 4 – 6   =  –2 x 3 + 11 x 2 + 10 x –10
    next back contents
  • Vertically:    In the horizontal case, noticed that running the negative through the parentheses changed the sign on each term inside the parentheses. The shortcut here is to not bother writing in the subtraction sign or the parentheses; instead, just change all the signs in the second row. change all the signs in the second row, and add down:
    • Simplify (6 x 3 – 2 x 2 + 8 x ) – (4 x 3 – 11 x + 10)
    • Horizontally:
          • (6 x 3 – 2 x 2 + 8 x ) – (4 x 3 – 11 x + 10)   =  (6 x 3 – 2 x 2 + 8 x ) – 1(4 x 3 – 11 x + 10)   =  (6 x 3 – 2 x 2 + 8 x ) – 1(4 x 3 ) – 1(–11 x ) – 1(10)   =  6 x 3 – 2 x 2 + 8 x – 4 x 3 + 11 x – 10   =  6 x 3 – 4 x 3 – 2 x 2 + 8 x + 11 x – 10   =  2 x 3 – 2 x 2 + 19 x – 10
    Vertically: write out the polynomials, leaving gaps as necessary: change the signs in the second line, and add: next back contents
  • Activity 8
    • Name; ___________________________ Date; ___________________
    • Year & Section; _____________________ Score; __________________
    • A. Give the difference of the following polynomials and arrange your answer in descending order.
    • 4xy – 7xy
    • a 2 + b 2 + ab , a 2 – b 2 + 2ab
    • 3x 2 – 4x + 5 , 2x 3 – 7x – 2
    • 4x 2 + 3x – 10 , 3x 2 – 8x 12
    • 2p ( q- 4r ) , 3p-5r
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  • LESSON 9 Multiplying Polynomials
    • Objectives;
    • At the end of this lesson, the students are expected to;
          • multiply two different polynomials;
          • apply the law of exponents; and
          • develop the skills of solving polynomials and multiplying functions.
    • Simple Polynomial Multiplication
    • There were two formats for adding and subtracting polynomials: "horizontal" and "vertical". You can use those same two formats for multiplying polynomials. The very simplest case for polynomial multiplication is the product of two one-term polynomials. For instance:
    • Simplify (5 x 2 )( – 2 x 3 )
    • Applying the rules of exponent and the distributive property.
    • (5 x 2 )(–2 x 3 ) = –10 x 5
    The next step up in complexity is a one-term polynomial times a multi-term polynomial. For example: Simplify –3 x (4 x 2 – x + 10) To do this, distribute the –3 x through the parentheses ; –3 x (4 x 2 – x + 10)      =  –3 x (4 x 2 ) – 3 x (– x ) – 3 x (10)      =  –12 x 3 + 3 x 2 – 30 x contents back next
    • The next step up is a two-term polynomial times a two-term polynomial. This is the simplest of the "multi-term times multi-term" cases. There are actually three ways to do this. Since this is one of the most common polynomial multiplications that you will be doing,
      • Simplify ( x + 3)( x + 2)
      • The first way I can do this is "horizontally"; in this case, however, I'll have to distribute twice, taking each of the terms in the first parentheses "through" each of the terms in the second parentheses:   
        • ( x + 3)( x + 2)      =  ( x + 3)( x ) + ( x + 3)(2)      =  x ( x ) + 3( x ) + x (2) + 3(2)      =  x 2 + 3 x + 2 x + 6      =  x 2 + 5 x + 6
    • This is probably the most difficult and error-prone way to do this multiplication. The "vertical" method is much simpler. First, think back to when you were first learning about multiplication. When you did small numbers, it was simplest to work horizontally,
            • 3 × 4 = 12
    • But to a larger numbers, stacked the numbers vertically and, working from right to left, took one digit at a time from the lower number and multiplied it, right to left, across the top number. For each digit in the lower number, you formed a row underneath, stepping the rows off to the left as worked from digit to digit in the lower number. Then added down.
    • But it's easy when it was solve vertically:
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    • Multiply polynomials in this same manner, done "vertically" this time:
    • Simplify ( x + 3)( x + 2)
    • be sure to do very neatly.
    The answer will be x 2 + 5 x + 6 ...and then multiply: FOIL METHOD: A Special (and Misleading) Case There is also a special method, useful ONLY for a two-term polynomial times another two-term polynomial. The method is called "FOIL". The letters F-O-I-L come from the words "first", "outer", "inner", "last", and are a memory device for helping you remember how to multiply horizontally, without having to write out the distribution and without dropping any terms. Here is what FOIL stands for: next back contents set up the multiplication :
  • That is, FOIL tells you to multiply the first terms in each of the parentheses, then multiply the two terms that are on the "outside" (furthest from each other), then the two terms that are on the "inside" (closest to each other), and then the last terms in each of the parentheses. In other words, using the previous example :
    • Use FOIL to simplify ( x + 3)( x + 2)
    • "first": ( x )( x ) = x 2 "outer":  ( x )(2) = 2 x "inner":  (3)( x ) = 3 x "last":  (3)(2) = 6
    • ( x + 3)( x + 2) = x 2 + 2 x + 3 x + 6 = x 2 + 5 x + 6
    • FOIL tends to be taught as "the" way to multiply all polynomials, which is clearly not true. (As soon as either one of the polynomials has more than a "first" and "last" term in its parentheses, you're hosed if you try to use FOIL, because those terms won't "fit".) When multiplying larger polynomials, switch to vertical multiplication, because it's just much easier to use.
    • Simplify ( x – 4)( x – 3)
    So the answer is: x 2 – 7 x + 12 next back contents
    • Using FOIL would give:
      • "first": ( x )( x ) = x 2 "outer": ( x )(–3) = –3 x "inner": (–4)( x ) = –4 x "last": (–4)(–3) = +12
      • product: ( x 2 ) + (–3 x ) + (–4 x ) + (+12) = x 2 – 7 x + 1
    • Simplify ( x – 3 y )( x + y )
    So the answer is: x 2 – 2 xy – 3 y 2
    • Using FOIL would give:
      • "first": ( x )( x ) = x 2 "outer": ( x )( y ) = xy "inner": (–3 y )( x ) = –3 xy "last": (–3 y )( y ) = –3 y 2
      • product: ( x 2 ) + ( xy ) + (–3 xy ) + (–3 y 2 ) = x 2 – 2 xy – 3 y 2
    • "FOIL" works ONLY for the specific and special case of a two-term expression times another two-term expression. It does NOT apply in ANY other case.
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    • General Polynomial Multiplication
    • Sometimes you will have to multiply one multi-term polynomial by another multi-term polynomial. Do this horizontally if you want, but there is so much room for error that usually switch over to vertical multiplication once the polynomials get big. For bigger multiplications, vertical is usually faster.
    • Simplify (4 x 2 – 4 x – 7)( x + 3)
    • horizontally:
        • (4 x 2 – 4 x – 7)( x + 3)   =  (4 x 2 – 4 x – 7)( x ) + (4 x 2 – 4 x – 7)(3)   =  4 x 2 ( x ) – 4 x ( x ) – 7( x ) + 4 x 2 (3) – 4 x (3) – 7(3)   =  4 x 3 – 4 x 2 – 7 x + 12 x 2 – 12 x – 21
      =  4 x 3 – 4 x 2 + 12 x 2 – 7 x – 12 x – 21   =  4 x 3 + 8 x 2 – 19 x – 21 vertically: the answer is: 4 x 3 + 8 x 2 – 19 x – 21
    • Simplify ( x + 2)( x 3 + 3 x 2 + 4 x – 17)
    • Note that, since order doesn't matter for multiplication, still put the " x + 2" polynomial on the bottom for vertical multiplication, just as always put the smaller number on the bottom when you were doing regular vertical multiplication with just plain numbers.
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    • the answer is: x 4 + 5 x 3 + 10 x 2 – 9 x – 34
    • Simplify (3 x 2 – 9 x + 5)(2 x 2 + 4 x – 7)
    So the answer is:  6 x 4 – 6 x 3 – 47 x 2 + 83 x – 35
    • Simplify ( x 3 + 2 x 2 + 4)(2 x 3 + x + 1)
    • Notice that these polynomials have "gaps" in their terms. The first polynomial has an x 3 term, an x 2 term, and a constant term, but no x term; and the second polynomial has an x 3  term, an x term, and a constant term, but no x 2 term. Do the vertical multiplication, leave spaces in set-up, corresponding to the "gaps" in the degrees of the polynomials' terms
    • .
    • (This is similar to using zeroes as "place holders" in regular numbers. You have can a thousand digit of 3, a hundreds digit of 2, and a units digit of 5, so you'd put a 0 in for the tens digits, creating the number 3,205.)
    • Solution;
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  • Solution; The answer is:  2 x 6 + 4 x 5 + x 4 + 11 x 3 + 2 x 2 + 4 x + next back contents
  • Activity 9
    • Name; ___________________________ Date; ___________________
    • Year & Section; _____________________ Score; _________________
    • A. Find the product of the following polynomials.
    • X-3 , 2x 2 – 3x + 4
    • 8c 2 d ( 2cd )
    • 2x 3 – 7x – 2 ( x- 2 )
    • 4p 3 + 5p + 3 , x + 3
    • 3x 2 - 4x + 5 , x+ 2
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  • LESSON 10 Division of polynomials
    • Objectives;
    • At the end of this lesson, the students will be able to;
          • divide polynomial functions;
          • realize the significance of different methods in dividing polynomials; and
          • apply the law of exponents.
    There are two cases for dividing polynomials: either the "division" is just a simplification and reducting a fraction, long polynomial division. Polynomial Long Division Dividing a polynomial by something more complicated than just a simple monomial, then you'll need to use a different method for the simplification. That method is called "long (polynomial) division", and it works just like the long (numerical) division you did back in elementary school, except that now you're dividing with variables
    • Divide   x 2 – 9x – 10 by x + 1
    • Long division for polynomials works in much the same way:
    contents back next First, set up the division: For the moment, ignore the other terms and look just at the leading x of the divisor and the leading x 2 of the dividend.
  • 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:    
  • 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:  
  • Then the solution to this division is: x – 10 Since the remainder on this division was zero (that is, since there wasn't anything left over), the division came out "even". When you do regular division with numbers and the division comes out even, it means that the number you divided by is a factor of the number you're dividing. For instance, divide 50 by 10, the answer will be a nice neat "5" with a zero remainder, because 10 is a factor of 50. In the case of the above polynomial division, the zero remainder tells us that x + 1 is a factor of x 2 – 9x – 10, which you can confirm by factoring the original quadratic dividend, x 2 – 9x – 10.
    • Examples;
      • Simplify  
      • This can be done in either of two ways: factor the quadratic and then cancel the common factor, like this:
    use long division : The answer to the division is quotient, the polynomial across the top:  x + 2 next back contents
    • Divide 3x 3 – 5x 2 + 10x – 3   by   3x + 1
    This division did not come out even. Think back to when you did long division with plain numbers. Sometimes there would be a remainder; for instance, if you divide 132 by 5: ...there is a remainder of 2. You made a fraction, put the remainder on top of the divisor, and wrote the answer as "twenty-six and two-fifths": next back contents
  • The first form, without the "plus" in the middle, is how "mixed numbers" are written, but the meaning of the mixed number is actually the addition. Do the same thing with polynomial division. Since the remainder is –7 and since the divisor is 3x + 1, then turn the remainder into a fraction (the remainder divided by the original divisor), and add this fraction to the polynomial across the top of the division symbol. Since the division looks like this: ...then the answer is this:    Note; Do not write the polynomial "mixed number" in the same format as numerical mixed numbers! Just append the fractional part to the polynomial part, this will be interpreted as polynomial multiplication, since it is not.
    • Divide 2x 3 – 9x 2 + 15   by   2x – 5
    • First off, note that there is a gap in the degrees of the terms of the dividend: the polynomial 2x 3 – 9x 2 + 15 has no x term. Leave space for a x-term column, just in case. Create this space by turning the dividend into 2x 3 – 9x 2 + 0x + 15. Do the division
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  • Remember to add the remainder to the polynomial part of the answer:
    • Divide 4x 4 + 3x 3 + 2x + 1   by x 2 + x + 2
    • Add a 0x 2 term to the dividend (inside the division symbol) to make space for my work, and then do the division in the usual manner:
    The answer is: next back contents
  • The Factor Theorem is a result of the Remainder Theorem, and is based on the same reasoning. As the Remainder Theorem points out, in dividing polynomials p ( x ) by a factor x – a of that polynomial, then you will get a zero remainder. Look again at that Division Algorithm expression of the polynomial: p ( x ) = ( x – a ) q ( x ) + r ( x ) If x – a is indeed a factor of p ( x ), then the remainder after division by x – a will be zero. That is: p ( x ) = ( x – a ) q ( x ) The Factor Theorem In terms of the Remainder Theorem, this means that, if x – a is a factor of p ( x ), then the remainder, when we do synthetic division by x = a , will be zero. The point of the Factor Theorem is the reverse of the Remainder Theorem: If synthetic-division of a polynomial by x = a and get a zero remainder, then, not only is x = a a zero of the polynomial (courtesy of the Remainder Theorem), but x  –  a is also a factor of the polynomial (courtesy of the Factor Theorem). Just as with the Remainder Theorem, the point here is not to do the long division of a given polynomial by a given factor. This Theorem isn't repeating what you already know, but is instead trying to make your life simpler. When faced with a Factor Theorem exercise, apply synthetic division and then check for a zero remainder.
    • Use the Factor Theorem to determine whether x – 1 is a factor of        f ( x ) = 2 x 4 + 3 x 2 – 5 x + 7.
    For x – 1 to be a factor of  f ( x ) = 2 x 4 + 3 x 2 – 5 x + 7, the Factor Theorem says that x = 1 must be a zero of  f ( x ). To test whether x – 1 is a factor, first set x – 1 equal to zero and solve to find the proposed zero, x = 1. next back contents
  • Then use synthetic division to divide f ( x ) by x = 1. Since there is no cubed term., be careful to remember to insert a "0" into the first line of the synthetic division to represent the omitted power of x in 2 x 4 + 3 x 2 – 5 x + 7:
    • Since the remainder is not zero, then the Factor Theorem says that:
    • x – 1 is not a factor of f ( x ).
    • Using the Factor Theorem, verify that x + 4 is a factor of        f ( x ) = 5 x 4 + 16 x 3 – 15 x 2 + 8 x + 16.
    If x + 4 is a factor, then (setting this factor equal to zero and solving) x = –4 is a root. To do the required verification, check that, when using synthetic division on  f ( x ), with x = –4, gets a zero remainder: The remainder is zero, so the Factor Theorem says that: x + 4 is a factor of 5 x 4 + 16 x 3 – 15 x 2 + 8 x + 16. Factor Theorem is used when factoring polynomials "completely". Rather than trying various factors by using long division, you will use synthetic division and the Factor Theorem. Any time divide by a number (being a potential root of the polynomial) and get a zero remainder in the synthetic division, next back contents
  • this means that the number is indeed a root, and thus " x minus the number" is a factor. Then continue the division with the resulting smaller polynomial, continuing until you arrives at a linear factor (so you've found all the factors) or a quadratic .
    • Using the fact that –2 and  1 / 3 are zeroes of  f ( x ) = 3 x 4 + 5 x 3 + x 2 + 5 x – 2, factor the polynomial completely.   
    • If x = –2 is a zero, then x + 2 = 0, so x + 2 is a factor. Similarly, if x = 1 / 3 is a zero, then x  –  1 / 3 = 0, so x – 1 / 3 is a factor. By two of the zeroes, the two factors: x + 2 and x – 1 / 3 .
    • The Factor Theorem says that need not to do the long division with the known factors of x + 2 and x – 1 / 3 . Instead, use synthetic division with the associated zeroes –2 and  1 / 3 .
    The remainder is zero, which is expected because at the start that –2 was a known zero of the polynomial. Rather than starting over again with the original polynomial, now work on the remaining polynomial factor of 3 x 3 – x 2 + 3 x – 1 (from the bottom line of the synthetic division), divide this by the other given zero, x = 1 / 3 : next back contents
    • This leaves with the quadratic 3 x 2 + 3, which can be solve:
      • 3 x 2 + 3 = 0 3( x 2 + 1) = 0 x 2 + 1 = 0 x 2 = – 1 x = ± i
    If the zeroes are x = – i and x = i , then the factors are x – (– i ) and x – ( i ), or x + i and x – i . remember that divided off a "3" when solving the quadratic; it is still part of the polynomial, and needs to be included as a factor. Then the fully-factored form is: 3 x 4 + 5 x 3 + x 2 + 5 x – 2 = 3( x + 2)( x – 1 / 3 )( x + i )( x – i ) Synthetic Division: The   Process   Synthetic division is a shorthand, or shortcut, method of polynomial division in the special case of dividing by a linear factor -- and it only works in this case. Synthetic division is generally used, however, not for dividing out factors but for finding zeroes (or roots) of polynomials. If the polynomial equation y = x 2 + 5x + 6, factor the polynomial as y = (x + 3)(x + 2). Then find the zeroes of y by setting each factor equal to zero and solving. You will find that x = –2 and x = –3 are the two zeroes of y. However, also work backwards from the zeroes to find the originating polynomial. For instance, if x = –2 and x = –3 are the zeroes of a quadratic, then x + 2 = 0, so x + 2 is a factor, and x + 3 = 0, so x + 3 is a factor. Therefore, the quadratic must be of the form y = a(x + 3)(x + 2). next back contents
  • (The extra number "a" in that last sentence is in there because, when you are working backwards from the zeroes, you don't know toward which quadratic you're working. For any non-zero value of "a", your quadratic will still have the same zeroes. But the issue of the value of "a" is just a technical consideration; as long as you see the relationship between the zeroes and the factors, that's all you really need to know for this lesson.) Anyway, the above is a long-winded way of saying that, if x – n is a factor, then x = n is a zero, and if x = n is a zero, then x – n is a factor. And this is the fact you use when you do synthetic division. Let's look again at the quadratic from above: y = x 2 + 5x + 6. From the Rational Roots Test, you know that ± 1, 2, 3, and 6 are possible zeroes of the quadratic. (And, from the factoring above, you know that the zeroes are, in fact, –3 and –2.) How would you use synthetic division to check the potential zeroes? Well, think about how long polynomial division works. If we guess that x = 1 is a zero, then this means that x – 1 is a factor of the quadratic. And if it's a factor, then it will divide out evenly; that is, if we divide x 2 + 5x + 6 by x – 1, we would get a zero remainder. Let's check:
    • In Synthetic Division;
    next back contents First, write the coefficients ONLY inside an upside-down division symbol:       
  • 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.
  • 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):
  • Return to this relationship between factors and zeroes throughout what follows; the two topics are inextricably intertwined.
    • Synthetic Division Examples;
    • Complete the indicated division.
    For this first exercise, display the entire synthetic-division process step-by-step. next back contents First, carry down the "2" that indicates the leading coefficient: Multiply by the number on the left, and carry the result into the next column:   Add down the column: Multiply by the number on the left, and carry the result into the next column:   Add down the column:
  • This exercise never said anything about polynomials, factors, or zeroes, but this division says that, if you divide 2x 4 – 3x 3 – 5x 2 + 3x + 8 by x – 2, then the remainder will be 2, and therefore x – 2 is not a factor of 2x 4 – 3x 3 – 5x 2 + 3x + 8, and x = 2 is not a zero (that is, a root or x-intercept) of the initial polynomial.
    • Divide 3x 3 – 2x 2 + 3x – 4 by x – 3 using synthetic division.
    • Write the answer in the form " q(x) + r(x) / d(x) ".
    next back contents Multiply by the number on the left, and carry the result into the next column:   Add down the column: Multiply by the number on the left, and carry the result into the next column:   Add down the column for the remainder:   The completed division is:
  • 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:   
  • 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
  • From long division of regular numbers, the remainder (if there is one) has to be smaller than whatever you divided by. In polynomial terms, since we're dividing by a linear factor (that is, a factor in which the degree on x is just an understood "1"), then the remainder must be a constant value. That is, when you divide by " x – a ", remainder will just be some number. The Remainder Theorem then points out the connection between division and multiplication. For instance, since 12 ÷ 3 = 4, then 4 × 3 = 12. If the remainder, do the multiplication and then add the remainder back in. For instance, since 13 ÷ 5 = 2 R 3, then 13 = 5 × 2 + 3. This process works the same way with polynomials. That is: If   p ( x ) / ( x – a )  =  q ( x )  with remainder   r ( x ), then  p ( x )  =  ( x – a ) q ( x )  +   r ( x ). (Technically, this "if - then" statement is the "Division Algorithm for Polynomials". But the Algorithm is the basis for the Remainder Theorem.)
    • In terms of our concrete example;
    • Since  ( x ^3 – 7 x – 6) / ( x – 4)   =   x 2 + 4 x + 9   with remainder 30,
    • then   x 3 – 7 x – 6  =   ( x – 4) ( x 2 + 4 x + 9)  +  30.
    • The Remainder Theorem says that we can restate the polynomial in terms of the divisor, and then evaluate the polynomial at x = a . But when x = a , the factor " x – a " is just zero! Then evaluating the polynomial at x = a gives us:
      • p ( a ) = ( a – a ) q ( a ) + r ( a )        = (0) q ( a ) + r ( a )        = 0 + r ( a )        = r ( a )
    next back contents
    • But remember that the remainder term r ( a ) is just a number! So the value of the polynomial p ( x ) at x = a is the same as the remainder you get when you divide that polynomial p ( x ) by x – a . In terms of our concrete example:
      • p (4) = (4 – 4)((4) 2 + 4(4) + 9) + 30        = (0)(16 + 16 + 9) + 30        = 0 + 30        = 30
    When dividing by a linear factor, don't "have" to use long polynomial division; instead, use synthetic division, which is much quicker. In example, Note that the last entry in the bottom row is 30, the remainder from the long division (as expected) and also the value of   p ( x ) = x 3 – 7 x – 6 at x = 4. And that is the point of the Remainder Theorem: There is a simpler, quicker way to evaluate a polynomial p ( x ) at a given value of x , and this simpler way is not to evaluate p ( x ) at all, but to instead do the synthetic division at that same value of x . Here are some examples:
    • Use the Remainder Theorem to evaluate  f ( x ) = 6 x 3 – 5 x 2 + 4 x – 17 at x = 3.
    • First off, even though the Remainder Theorem refers to the polynomial and to long division and to restating the polynomial in terms of a quotient, a divisor, and a remainder, that's not actually what meant to be done. Instead, do the synthetic division, using "3" as the divisor:
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    • Since the remainder (the last entry in the bottom row) is 112, then the Remainder Theorem says that:
    • f (3) = 112.
    • Using the Remainder Theorem, find the value of f (–5), for   f ( x ) = 3 x 4 + 2 x 3 + 4 x .
    • Do the synthetic division, remembering to put zeroes in for the powers of  x that are not included in the polynomial:
    • Since the remainder is 1605, then, thanks to the Remainder Theorem, I know that:
    • f (–5) = 1605.
    • Use the Remainder Theorem to determine whether x = 2 is a zero of       f ( x ) = 3 x 7 – x 4 + 2 x 3 – 5 x 2 – 4
    • For x = 2 to be a zero of  f ( x ), then  f (2) must evaluate to zero. In the context of the Remainder Theorem, this means that my remainder, when dividing by x = 2, must be zero:
    The remainder is not zero. Then   x = 2 is not a zero of f ( x ). next back contents
    • Use the Remainder Theorem to determine whether x = – 4 is a solution of        x 6 + 5 x 5 + 5 x 4 + 5 x 3 + 2 x 2 – 10 x – 8 = 0
    For x = –4 to be a solution of f ( x ) = x 6 + 5 x 5 + 5 x 4 + 5 x 3 + 2 x 2 – 10 x – 8 = 0, it must be that  f (–4) = 0. In the context of the Remainder Theorem, this means that the remainder, when dividing by x = –4, must be zero: The remainder is zero. Then x = –4 is a solution of the given equation. contents back next
  • Activity 10
    • Name; ___________________________ Date; ___________________
    • Year & Section; _____________________ Score; __________________
    • Find the quotient of each polynomials using long division.
    • ( x 4 – x + 1 ) / (x 3 + x 2 – 1)
    • ( x 8 + 1 ) / x 2 + 1
    • (2x 3 + 5x 2 – 13) / x + 2
    next back contents
    • B. Use synthetic Division to find the quotient and the remainder of the following polynomials.
    • ( 3x 3 ­ – 4x 2 + x + 2 ) / ( x- 1 )
    • ( 3x 4 – 3x 2 + 4x + 3 ) / ( x + 2 )
    • ( 2x 3 – 3x 2 + 2x – 6 ) / ( x + 3 )
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    • C. Use the remainder theorem to solve for the quotient and for the remainder of the following polynomials.
    • ( 3x 3 – x 4 – x + 4) / ( x + 4 )
    • ( 6x 3 + 7x 2 – 7x + 6 ) / ( x + 2)
    • D. Use the factor theorem to prove if the first polynomial is the factor of the second.
    • x – 1 ; x 4 – 3x 2 – 2x – 4
    • X + 3 ; 5x 4 + 7x + 6x 2 + 9x + 27
    back next contents
  • CHAPTER TEST
      • . Find the sum of each Polynomials.
    • X 3 + 3 , x-3
    • 6+ 8p 3 + 5p 2 , 4p 3 + 5p + 3
    • 12x 2 y 2 – 8x 2 y + 15x , x 2 y 3
    • a 2 - 5ab – 3b 2 , 2a 2 – 9b + 2b 2
    • 2a + 2b , 3a+ b
    contents next back
        • . Give the difference of the following polynomial expression.
    • 5x( 2a – 3b ) , 3b ( a- 3x)
    • 2x 3 y 3 , x 3 y 3
    • 3x 2 - 4x + 5 , 2x 3 – 7x – 2
    • a 2 - 5ab – 3b 2 , 2a 2 – 9b + 2b 2
    • 6+ 8p 3 + 5p 2 , 4p 3 + 5p
    • 5x( 2a – 3b ) , 3b ( a- 3x)
    • 2x 3 y 3 , x 3 y 3
    back next contents
  • 8. 3x 2 - 4x + 5 , 2x 3 – 7x – 2 9. a 2 - 5ab – 3b 2 , 2a 2 – 9b + 2b 2 10. 6+ 8p 3 + 5p 2 , 4p 3 + 5p + 3 C. Multiply each Polynomial expression
    • , 2x 3 – 7x – 2 , x 3
    • 2a + 2b , 3a+ b
    • X 3 + 3 , x-3
    • 3x 2 + x – 7 , x- 3
    • 3p 3 + 2p + 6 , x 2
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  • D. Perform the indicated operations. First use synthetic division and then check your work by long division:
    • (a 3 - 3a 2 + a) ÷ a.
    • (10x 3 - 7x 2 y - 16xy 2 + 12y 3 ) ÷ ( 2x 2 + xy - 2y2)
    • ( x 2 + 11x + 30) ÷ ( x + 6 )
    • (12 + x 2 - 7x) ÷ (x - 3)
    • (a 2 - 11a + 30) ÷ (a - 5)
    next back contents
  • ZEROS OF POLYNOMIALS CHAPTER IV
    • Objectives:
    • At the end of this chapter, student must be able to:
      • evaluate the zeros of polynomials;
      • reduce the zeros of polynomial functions to its original form of equation; and
      • apply the quadratic formula and other methods in finding the zeros of polynomials.
    next back contents
  • LESSON 11 Finding zeros of polynomial functions
    • Objectives;
    • At the end of this lesson the students will be able to;
          • know the principle of zero products.
          • find the zeros of polynomials
          • give the multiplicities of the zeros of polynomials
    • Once you know how to do synthetic division, use the technique as a shortcut to finding factors and zeroes of polynomials. Here are some examples:
    • Use synthetic division to determine whether x = 1 is a zero of x 3  – 1.
    • Set up the synthetic division, and check to see if the remainder is zero. If the remainder is zero, then x = 1 is a zero of x 3  – 1.
    To do the initial set-up, note that I needed to leave "gaps" for the powers of x that are not included in the polynomial. That is, followed the practice used with long division, and wrote the polynomial as x 3  + 0x 2  + 0x – 1 for the purposes of doing the division contents back next
    • Since the remainder is zero, then x = 1 is a zero of x 3  – 1.
    • Since x = 1 is a zero of x 3  – 1, then x – 1 is a factor, so the polynomial x 3  – 1 factors as (x – 1)(x 2  + x + 1). 
    • Use synthetic division to find all the zeroes of x 4  + x 3  – 11x 2  – 5x + 30.
    • Comparing the results of the Rational Roots Test to a quick graph, I decide to test x = 2 as a possible zero. Set up the divison:
    and here is the result next back contents
  • 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
  • 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
  • Activity 11
    • Name; ___________________________ Date; ___________________
    • Year & Section; _____________________ Score; __________________
    • Give all the zeros of the polynomial functions.
    • P(x) = x 2 – 5x + 6
    • P(x) = x 3 – 9x
    • P(x) = x 3 + 2x 2 – 5x – 6
    • P(x) = 6x 2 + 8x – 8
    • P(x) = x 3 – 1
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  • LESSON 12 Rational Zero Theorem
    • Objectives;
    • At the end of this lesson, the student is expected to;
          • define the Rational Root Theorem;
          • find the rational zeros of a polynomials; and
          • identify the multiplicities of the rational zeros of polynomials.
    A root or zero of a function is a number that, when plugged in for the variable, makes the function equal to zero. Thus, the roots of a polynomial P(x) are values of x such that P(x) = 0 . If P(x) is a polynomial with integer coefficients and if is a zero of P(x) ( P then p is a factor of the constant term of P(x) and q is a factor of the leading coefficient of P(x) . Use the Rational Zeros Theorem to find all the rational zeros of a polynomial. Here are the steps: = 0)
    • Arrange the polynomial in descending order
    • Write down all the factors of the constant term. These are all the possible values of p .
    • Write down all the factors of the leading coefficient. These are all the possible values of q.
    4.Write down all the possible values of Remember that since factors can be negative and -
    • must both be included. Simplify each value and cross out any duplicates.
    next back contents
    • Example: Find all the rational zeros of P(x) = x 3 -9x + 9 + 2x 4 -19x 2 .
    • P(x) = 2x 4 + x 3 -19x 2 - 9x + 9
    • Factors of constant term: ±1 , ±3 , ±9 .
    • Factors of leading coefficient: ±1 , ±2 .
    Use the rational zeros theorem to factor a polynomial. Using synthetic division, we can find one real root a and we can find the quotient when P(x) is divided by x - a . Next, use synthetic division to find one factor of the quotient. Then continue this process until the polynomial has been completely factored. Example (as above): Factor P(x) = 2x 4 + x 3 -19x 2 - 9x + 9 . As seen from the second synthetic division above, 2x 4 + x 3 -19x 2 -9x + 9÷x + 1 = 2x 3 - x 2 - 18x + 9 . Thus, P(x) = (x + 1)(2x 3 - x 2 - 18x + 9) . The second term can be divided synthetically by x + 3 to yield 2x 2 - 7x + 3 . Thus, P(x) = (x + 1)(x + 3)(2x 2 - 7x + 3) . The trinomial can then be factored into (x - 3)(2x - 1) . Thus, P(x) = (x + 1)(x + 3)(x - 3)(2x - 1) . We can see that this solution is correct because the four rational roots found above are zeros of our result. contents back next
  • Activity 12
    • Name; ___________________________ Date; ___________________
    • Year & Section; _____________________ Score; __________________
    • Find only the rational zero of the polynomials.
    • P(x) = x 3 – 27
    • P(x) = 2x 3 – 7x 2 + 4x + 4
    • P(x) = 6x 3 – 7x 2 – 7x + 3
    • P(x) = 2x 3 – 3x 2 – 11x + 6
    • P(x) = x 3 – 3x 2 + 3x – 1
    next back contents
  • CHAPTER TEST
        • . Find the zeros of the polynomial Functions.
    • P(x) = x 4 + x 2 – 1
    • P(x) = x 2 + 1
    • P(x) = x 3 – 3x – 2
    • P(x) = x 3 – 5x
    • P(x) = 2x 3 – 3x 2 – 72x + 76
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        • B. Give only the rational zero of each polynomial functions.
    • P(x) = 2x 2 – x + 2
    • P(x) = x 2 + 3x + 10
    • P(x) = x 2 -64
    • P(x) = x 4 – 1
    • P(x) = x 4 + x 3 + 4x 2 + 6x – 12
    next back contents
  • GRAPHS OF POLYNOMIALS CHAPTER V
    • Objectives
    • At the end of this chapter, the students must be able to:
      • find the x -intercepts and y -intercept of a polynomial function;
      • write the equation of a polynomial function given the zeros and a point on the function;
      • determine the minimal degree of a polynomial given its graph;
      • use a graphing utility to find the absolute maximum or absolute minimum of a polynomial function;
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  • LESSON 13 Polynomial Graphs
    • Objectives
    • At the end of this topic, the students will be able to;
          • sketch a graph of polynomial functions;
          • plot the zeros of polynomials;
          • determine the turning point of a functions;
          • identify the intersection o the graph; and
          • classify the different types of graphs.
    When you're graphing (or looking at a graph of) polynomials, it can help to already have an idea of what basic polynomial shapes look like. One of the aspects of this is "end behavior", and it's pretty easy. Look at these graphs: next back contents with a positive leading coefficient with a negative leading coefficient
  • 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
  • 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
    • Describe the end behavior of   f ( x ) = 3 x 7 + 5 x + 1004
    • This polynomial is much too large for me to view in the standard screen on my graphing calculator, so either I can waste a lot of time fiddling with WINDOW options, or I can quickly use my knowledge of end behavior.
    • This function is an odd-degree polynomial, so the ends go off in opposite directions, just like every cubic I've ever graphed. A positive cubic enters the graph at the bottom, down on the left, and exits the graph at the top, up on the right. Since the leading coefficient of this odd-degree polynomial is positive, then its end-behavior is going to mimic a positive cubic.
    • " Down" on the left and "up" on the right
    Zeroes and Their Multiplicities: "Flexing" at the Axis There's an extra detail I'd like to mention regarding the multiplicity of a zero and the graph of the polynomial: You can tell from the graph whether an odd-multiplicity zero occurs only once or if it occurs more than once.
    • What is   the multiplicity of x = 5, given that the graph shows a fifth-degree polynomial with all real-number roots, and the root x = – 5 has a multiplicity of   2?
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  • 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
  • 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
  • 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 the polynomial .
    • What is the minimum possible degree of the polynomial graphed below?
    Since there are four bumps on the graph, and since the end-behavior says that this is an odd-degree polynomial, then the degree of the polynomial is 5, or 7, or 9, or... But: The minimum possible degree is 5. next back contents
  • Activity 13
    • Name; ___________________________ Date; ___________________
    • Year & Section; _____________________ Score; __________________
    • Draw the graph of the following polynomials. Then, identify the turning points of the graph.
    • P(x) = ( x- 1 ) ( x + 2 ) ( x + 5 )
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  • 2. P(x) = ( x + 3 ) ( x- 3 ) ( x – 1 ) 3. P(x) = x 3 – x 2 – 4x + 4 next back contents
  • 4. P(x) = x 3 – 3x 5. P(x) = 4x + 3 contents back next
  • CHAPTER TEST
        • . Graph each Polynomials.
    • A box without a lid is constructed from a 36 inch by 36 inch piece of cardboard by cutting x inch squares from each corner and folding up the sides.
      • Determine the volume of the box as a function of the variable x .
      • Use a graphing utility to approximate the values of x that produce a volume of 3280.5 cubic inches.
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    • 2. For the function  x 3 + 3x – 4
      • find the x -intercepts
      • find the y -intercept
      • describe the end behaviors
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    • 3. For the function x 2 + y 2 + 4x + 16y = 5
      • find the x -intercepts
      • find the y -intercept
      • describe the end behaviors
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  • References Bernabe, Julieta G., ADVANCE ALGEBRA TRIGONOMETRY AND STATISTICS JTW.CORPORATION, 2003 Conception, Benjamin, COLLEGE ALGEBRA WITH RECREATIONAL MATHEMATICS, pp,25-43 Coronel,Antonio C., MATHEMATICS IV: AN INTEGRATED APPROACH, pp.69-107 Frisk, Peter D., BEGINNING ALGEBRA WORDSWORTH INC.1985 Mc Connel, John W. Algebra Scott., FORESMAN AND COMPANY 1996 Pascual, Leonarda B. MATHEMATICS IV BOOK MEDIA PRESS. 1997 Leithold. Louis et. Al. ,COLLEGE A;GEBRA, pp. 164-217 Brown, Dolciani & Sorgenfery Kaye, ALGEBRA AND TRIGONOMETRY ,pp. 98-126 Ferido, Junita, MATHEMATICS IV,( ADVANCE ALGEBRA ) ,pp. 32-49 next back contents
  • URL http://www.singaporemath.com/Mathematics_s/1.htm 11-09-09 http://www.sleeplessinkl.com/2009/04/24/relearning-math 11-09-09 http://www.singaporemath.com/Mathematics_s/1.htm 11-28-09 http://www.singaporemath.com/Learning_Mathematics_p/smmlm.htm 11-26-09 http://search.yahoo.com/search?p=&toggle=1&cop=mss&ei=UTF-8&fr=yfp-t-701 10-25-09 http://www.google.com.ph/#hl=tl&source=hp&q=purplemath=&aq=0&oq=singapore+model+me&fp=ffec8c89bd59b524 09-18-09 http://home.sandiego.edu/~learningmathematics.pdf 11-15-09 http://en.wikipedia.org/wiki/polynomialfunction.ph 10-23-09 http://hubpages.com/hub/learning-Maths--A-review 10-23-09 http://www.angelfire.com/kstutoriali/what.html 10-23-09 contents back next
  • Image 001. http://images.search.yahoo.com/search/images?p=mathematics+logo&ni=20&ei=utf-8&fr=yfp-701&fr2=sq-gac&xargs=0&pstart=1&b6 01-05-10 002. http://google.com/images?q=tbn:qwRJDhznTK6zQM:http://www.kumonbb.com/wp-content/uploads/2008/05/home_ed.jpg 01-28-10 003. http://www.goshen.edu/.cWtools/download.php/mnF=pen%20and%20paper.JPG,mnOD=Pictures,mnOD=My%20Documents,dc=forms,dc=www,dc=goshen,dc=edu 01-28-10 back contents next
  • About the Authors Alexandria M. Sandagon graduated from Laguna State Polytechnic University at Siniloan Campus with the degrre of Bachelor of Secondary Education Major in Mathematics. She was taught during her High School level at Ungos National High School Extension Class at Brgy. Llavac Real, Quezon. She finished her elementary level at Llavac Elementary School at Llavac Real, Quezon. Ronalyn M. Cabrera graduated at Laguna State Polytechnic University with the degree of Bachelor of Secondary Education and chose Mathematics as her field of specialization. She finish her elementary grade at Nanguma Elementary School, Mabitac Laguna. And High School grade at Mabitac National High School, Mabitac Laguna. back contents