Remainder and Factor Theorems

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  Mathematics NQF4 SB

  TVET FIRST

  • M Van Rensburg, I Mapaling M Trollope(Authors)
  • 2017(Publication Date)
  • Macmillan

    (Publisher)

  53 Module 3 Work with algebraic expressions using the Remainder and Factor Theorems Module 3 Overview By the end of this module you should be able to: • Unit 3.1: Use and apply the remainder theorem and the factor theorem to: – Find the remainder. – Prove that an expression is a factor. – Find an unknown variable to make an expression a factor or to leave a remainder. • Unit 3.2: Factorise third-degree polynomials including examples that require the factor theorem. Unit 3.1: Use and apply the remainder theorem and the factor theorem 3.1.1 Polynomials Topic 2: Functions and algebra Polynomials are mostly used to approximate unknown functions. One method of doing this, is to divide a polynomial into factors. In this module, we will discuss the division of polynomials, first by using long division, and then by using the remainder theorem. We also explain the factor theorem, which is used to find factors of a polynomial. A polynomial is an algebraic expression that contains two or more terms, with each term consisting of a number (called the coefficient ) and a variable raised to a positive power (counting number). The math symbols (operators) + and – separate the terms. The term with the highest exponent is written first in a polynomial. For example, 2 a 3 + 4 a 2 – a + 9 is a polynomial with four terms. Here, the first term, 2 a 3 , is a ‘third-degree’ term. The second term, 4 a 2 , is a ‘second-degree’ term and the third term, a , is a ‘first-degree’ term. (This is because a = a 1 ). The fourth term, 9, contains no variable such as x , y or a , and is the constant. The degree of the polynomial is given by the exponent of the highest power. Therefore, 2 a 3 + 4 a 2 – a + 9 is a third-degree polynomial . Later in this module you will learn how to factorise third-degree polynomials. Here are different types of polynomials: • In the form ax + b , for example 2 x + 3: this is a linear expression, or polynomial of degree 1.

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  Precalculus, Enhanced Edition

  • David Cohen, Theodore Lee, David Sklar, , David Cohen, Theodore Lee, David Sklar(Authors)
  • 2016(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  According to the fundamental theorem of algebra, the equation f ( x ) 0 has a root; let’s call this root r 1 . By the factor theorem, x r 1 is a factor of f ( x ), and we can write for some polynomial Q 1 ( x ) that has degree n 1 and leading coefficient a n . If the degree of Q 1 ( x ) happens to be zero, we’re done. On the other hand, if the degree of Q 1 ( x ) is at least 1, another application of the fundamental theorem of algebra fol-lowed by the factor theorem gives us Q 1 ( x ) ( x r 2 ) # Q 2 ( x ) f ( x ) ( x r 1 ) # Q 1 ( x ) f ( x ) a n ( x r 1 )( x r 2 ) p ( x r n ) a n x n a n 1 x n 1 p Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. where the degree of Q 2 ( x ) is n 2 and the leading coefficient of Q 2 ( x ) is a n . We now have We continue this process until the quotient is Q n ( x ) a n . As a result, we obtain as we wished to show. The linear factors theorem tells us that any polynomial can be expressed as a product of linear factors. The theorem gives us no information, however, as to how those factors can actually be obtained. The next example demonstrates a case in which the factors are readily obtainable; this is always the case with quadratic polynomials. a n ( x r 1 )( x r 2 ) p ( x r n ) f ( x ) ( x r 1 )( x r 2 ) p ( x r n ) a n f ( x ) ( x r 1 )( x r 2 ) # Q 2 ( x ) 13.3 The Fundamental Theorem of Algebra 937 EXAMPLE 2 Writing Quadratics as the Products of Linear Factors Express the following second-degree polynomials in the form a n ( x r 1 )( x r 2 ): (a) 3 x 2 5 x 2; (b) x 2 4 x 5.

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  Abstract Algebra

  An Introduction

  • Thomas Hungerford(Author)
  • 2020(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Theorem 4.15 The Remainder Theorem Let F be a field, f ( x )  F [ x ], and a  F . The remainder when f ( x ) is divided by the polynomial x 2 a is f ( a ). Theorem 4.16 The Factor Theorem Let F be a field, f ( x )  F [ x ], and a  F . Then a is a root of the polynomial f ( x ) if and only if x 2 a is a factor of f ( x ) in F [ x ]. Proof • First assume that a is a root of f ( x ). Then we have f ( x ) 5 ( x 2 a ) q ( x ) 1 r ( x ) [ Division Algorithm ] f ( x ) 5 ( x 2 a ) q ( x ) 1 f ( a ) [ Remainder Theorem ] f ( x ) 5 ( x 2 a ) q ( x ) [ a is a root of f ( x ) , so f ( a ) 5 0 F .] Therefore, x 2 a is a factor of f ( x ). Conversely, assume that x 2 a is a factor of f ( x ), say f ( x ) 5 ( x 2 a ) g ( x ). Then a is a root of f ( x ) because f ( a ) 5 ( a 2 a ) g ( a ) 5 0 F g ( a ) 5 0 F . Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 108 Chapter 4 Arithmetic in F [ x ] Corollary 4.17 Let F be a field and f ( x ) a nonzero polynomial of degree n in F [ x ]. Then f ( x ) has at most n roots in F . Proof* • If f ( x ) has a root a 1 in F , then by the Factor Theorem, f ( x ) 5 ( x 2 a 1 ) h 1 ( x ) for some h 1 ( x )  F [ x ]. If h 1 ( x ) has a root a 2 in F , then by the Factor Theorem f ( x ) 5 ( x 2 a 1 )( x 2 a 2 ) h 2 ( x ) for some h 2 ( x )  F [ x ]. If h 2 ( x ) has a root a 3 in F , repeat this procedure and continue doing so until you reach one of these situations: (1) f ( x ) 5 ( x 2 a 1 )( x 2 a 2 )    ( x 2 a n ) h n ( x ) (2) f ( x ) 5 ( x 2 a 1 )( x 2 a 2 )    ( x 2 a k ) h k ( x ) and h k ( x ) has no root in F .

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  Number Theory Revealed: A Masterclass

  • Andrew Granville(Author)
  • 2019(Publication Date)
  • American Mathematical Society

    (Publisher)

  Deduce that f ( x ) divides h ( x ). The Fundamental Theorem of Algebra is the well-known result that a polyno-mial f ( x ) = ∑ d j =0 f j x j with complex coefficients of degree d has exactly d roots (where the roots are counted according to their multiplicity; for example, x ( x − 1) 2 has the three roots 0, 1, and 1, the root 1 counted twice). This theorem, used without proof while a student learns basic mathematics, is rather more subtle to prove than one might guess. Four of the great mathematicians, Euler (1749), La-grange (1772), Laplace (1795), and Gauss (1799) published purported proofs which 117 118 Appendix 3F. Fundamental theorems and factoring polynomials had subtle errors, before the first correct proof appeared in 1814 due to Argand. The difficulty in proving the Fundamental Theorem of Algebra is the following first step, which appears to be innocuous but isn’t. Lemma 3.22.1. If f ( x ) ∈ C [ x ] has degree d ≥ 1 , then f ( x ) has at least one root in C . We will not prove this but we will deduce the Fundamental Theorem from it: Theorem 3.10 (The Fundamental Theorem of Algebra) . If f ( x ) ∈ C [ x ] has degree d ≥ 1 , then f ( x ) has exactly d roots in C , counted with multiplicity. Proof. By induction. For d = 1 we note that ax + b has the unique root − b/a . For higher degree, f has a root α by Lemma 3.22.1. Let g ( x ) = x − α in Proposition 2.10.1 to obtain a polynomial q ( x ) and a constant r for which f ( x ) = ( x − α ) q ( x )+ r . Substituting in x = α , we deduce that r = f ( α ) = 0; that is, f ( x ) = ( x − α ) q ( x ). Now q ( x ) has degree d − 1 so has d − 1 roots, counted with multiplicity, by the induction hypothesis. Therefore f ( x ) = ( x − α ) q ( x ) has 1 + ( d − 1) = d roots, counted with multiplicity. We note that if f ( x ) = ∑ d j =0 f j x j , then q ( x ) := f ( x ) − f ( α ) x − α = d j =0 f j x j − α j x − α = d j =0 f j ( x j − 1 + αx j − 2 + · · · + α j − 1 ) .

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  College Algebra

  • R. Gustafson, Jeff Hughes(Authors)
  • 2016(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Section 4.3 The Remainder and Factor Theorems; Synthetic Division 435 Getting Ready You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 1. The variables in a polynomial have -number exponents. 2. A zero of Psxd is any number c for which . 3. The Remainder Theorem holds when c is number. 4. If Psxd is a polynomial function and Psxd is divided by , the remainder will be Pscd. 5. If Psxd is a polynomial function, then Pscd 5 0 if and only if x 2 c is a of Psxd. 6. A shortcut method for dividing a polynomial by a binomial of the form x 2 c is called division. Practice Use long division to perform each division. 7. 4x 3 2 2x 2 2 x 1 1 x 2 1 8. 2x 3 1 3x 2 2 5x 1 1 x 1 3 9. 2x 4 1 x 3 1 2x 2 1 15x 2 5 x 1 2 10. x 4 1 6x 3 2 2x 2 1 x 2 1 x 2 1 Find each value by substituting the given value of x into the polynomial and simplifying. Then find the value by performing long division and finding the remainder. 11. Psxd 5 3x 3 2 2x 2 2 5x 2 7; Ps2d 12. Psxd 5 5x 3 1 4x 2 1 x 2 1; Ps22d 13. Psxd 5 7x 4 1 2x 3 1 5x 2 2 1; Ps21d 14. Psxd 5 2x 4 2 2x 3 1 5x 2 2 1; Ps2d 15. Psxd 5 2x 5 1 x 4 2 x 3 2 2x 1 3; Ps1d 16. Psxd 5 3x 5 1 x 4 2 3x 2 1 5x 1 7; Ps22d Use the Remainder Theorem to find the remainder that occurs when Psxd 5 3x 4 1 5x 3 2 4x 2 2 2x 1 1 is divided by each binomial. 17. x 1 2 18. x 2 1 19. x 2 2 20. x 1 1 21. x 1 3 22. x 2 3 23. x 2 4 24. x 1 4 Use the Factor Theorem to determine whether each state- ment is true. If the statement is not true, so indicate. 25. x 2 1 is a factor of Psxd 5 x 7 2 1.

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