Precalc0205to0207

Precalc0205to0207 - 2.46 SECTION 2.5: FINDING ZEROS OF...

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2.46 SECTION 2.5: FINDING ZEROS OF POLYNOMIAL FUNCTIONS Assume fx () 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 R e f e r t o 1) F a c t o r i n g ( 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) Q F ( 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 fx () 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 ² 0; a 0 ² n ± Z + If 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 = 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 . p (factors of the constant term, 2): ± 1, ± 2 q (factors of the leading coefficient, 4): ± ± 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 2 , ± 1 4 , ± 2, ± ± 1 2 Redundant ±² ³´ ³
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2.48 Use Synthetic Division to divide fx () by x ± k , where k is one of our rational candidates. Remember that the following are equivalent for a nonzero polynomial and a real number k : x ± k is a factor of ± k is a zero of i.e., fk = 0 ± We get a 0 remainder in the Synthetic Division process. The first ± is the Factor Theorem, and the second ± comes from the Remainder Theorem. See Notes on Section 2.3: 2.32-2.34 . We use trial-and-error and proceed through our list of candidates ( k ) for rational zeros: ± 1, ± 1 2 , ± 1 4 , ± 2 Let’s try k = 1 . Method 1 We may directly evaluate f 1 and see if it is 0. = 4 x 3 ± 5 x 2 ± 7 x + 2 f 1 = 41 3 ± 51 2 ± 71 + 2 = ± 6 ² 0 Therefore, 1 is not a zero of .
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2.49 Method 2 We may also use the Synthetic Division process and see if we get a 0 remainder. We do not get a 0 remainder, so 1 is not a zero of fx () . Method 3 Observe from both previous methods that we can compute f 1 by simply adding up the coefficients of in standard form. This does not work in general for other values of k , though. Let’s try k = 2 . Let’s use the Synthetic Division / Remainder Theorem method: We do get a 0 remainder, so 2 is a zero of .
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Precalc0205to0207 - 2.46 SECTION 2.5: FINDING ZEROS OF...

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