Stitz-Zeager_College_Algebra_e-book

We 5 25 5 nd that a2 a1 9 1 2 and a3 a2 27 9 10

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Unformatted text preview: x) = 2x2 3 x3 − 2x2 + x 4. R(x) = x3 3 − x2 + x 4x3 x2 − 2 5. R(x) = x3 + 5 x − 1 x4 + 6x2 + 9 6. R(x) = 8x2 x4 + 16 Solution. 1. We begin by factoring the denominator to find 2x2 − x − 1 = (2x + 1)(x − 1). We get x = − 1 2 and x = 1 are both zeros of multiplicity one and thus we know x+5 x+5 A B = = + −x−1 (2x + 1)(x − 1) 2x + 1 x − 1 2x2 Clearing denominators, we get x +5 = A(x − 1)+ B (2x +1) so that x +5 = (A +2B )x + B − A. Equating coefficients, we get the system A + 2B = 1 −A + B = 5 This system is readily handled using the Addition Method from Section 8.1, and after adding both equations, we get 3B = 6 so B = 2. Using back substitution, we find A = −3. Our answer is easily checked by getting a common denominator and adding the fractions. x+5 2 3 = − −x−1 x − 1 2x + 1 2x2 526 Systems of Equations and Matrices 2. Factoring the denominator gives x3 − 2x2 + x = x x2 − 2x + 1 = x(x − 1)2 which gives x = 0 as a zero of multiplicity one and x = 1 as a zero of multip...
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