M1410205Part1

# M1410205Part1 - 2.46 SECTION 2.5 FINDING ZEROS OF...

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2.46 SECTION 2.5: FINDING ZEROS OF POLYNOMIAL FUNCTIONS Assume f x ( ) is a nonconstant polynomial with real coefficients written in standard form. PART A: TECHNIQUES WE HAVE ALREADY SEEN Refer to: Notes 1.31 to 1.35 Section A.5 in the book Notes 2.45 Refer to 1) Factoring ( Notes 1.33 ) 2) Methods for Dealing with Quadratic Functions ( Book Section A.5: pp.A49-51 ) a) Square Root Method ( Notes 1.31, 2.45 ) b) Factoring ( Notes 1.33 ) c) QF ( Notes 1.34, 2.45 ) d) CTS (Completing the Square) ( Book Section A.5: p.A49 ) 3) Bisection Method (for Approximating Zeros) ( Notes 2.20 to 2.21 ) 4) Synthetic Division and the Remainder Theorem (for Verifying Zeros) ( Notes 2.33 )

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2.47 PART B: RATIONAL ZERO TEST Rational Zero Test (or Rational Roots Theorem) Let f x ( ) be a polynomial with integer (i.e., only integer) coefficients written in standard form: a n x n + a n 1 x n 1 + ... + a 1 x + a 0 each constant a i Z ; a n 0; a 0 0; n Z + ( ) If f x ( ) has rational zeros, they must be in the list of ± p q candidates, where: p is a factor of a 0 , the constant term, and q is a factor of a n , the leading coefficient. Note: We require a 0 0 . If a 0 = 0 , try factoring out the GCF first. Example Factor f x ( ) = 4 x 3 5 x 2 7 x + 2 completely, and find all of its real zeros. Solution Since the GCF = 1 , and Factoring by Grouping does not seem to help, we resort to using the Rational Zero Test. We will now list the candidates for possible rational zeros of f x ( ) . p (factors of the constant term, 2): ± 1, ± 2 q (factors of the leading coefficient, 4): ± 1, ± 2, ± 4 Note: You may omit the ± symbols above if you use them below. List of ± p q candidates: ± 1 1 , ± 1 2 , ± 1 4 , ± 2 1 , ± 2 2 , ± 2 4 Simplified: ± 1, ± 1 2 , ± 1 4 , ± 2, ± 1, ± 1 2 Redundant   
2.48 Use Synthetic Division to divide f x ( ) by x k ( ) , where k is one of our rational candidates. Remember that the following are equivalent for a nonzero polynomial f x ( ) and a real number k : x k ( ) is a factor of f x ( ) k is a zero of f x ( ) i.e., f k ( ) = 0 ( ) We get a 0 remainder in the Synthetic Division process. The first

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M1410205Part1 - 2.46 SECTION 2.5 FINDING ZEROS OF...

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