College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 3 - Polynomial and Rational Functions - Exercise Set 3.4 - Page 387: 27



Work Step by Step

The Linear Factorization Theorem... in words: An nth-degree polynomial can be expressed as the product of a nonzero constant and $n$ linear factors, where each linear factor has a leading coefficient of 1. Imaginary roots, if they exist, occur in conjugate pairs. ------------------------- If $4+3i$ is a zero, so is $4-3i.$ $f(x)=a_{n}(x+5)(x-4-3i)(x-4+3i)$ $f(x)=a_{n}(x+5)(x^{2}-4x+3ix-4x+16-12i-3ix+12i-9i^{2})$ ...$ i^{2}=-1$ ... $f(x)=a_{n}(x+5)(x^{2}-8x+25)$ Using the given $f(2),$we find $a_{n}:$ $f(2)=a_{n}(2+5)(4-16+25)$ $91=a_{n}(7)(13)$ $91=a_{n}(91)$ $a_{n}=1$ so $f(x)=1\cdot(x+5)(x^{2}-8x+25)$ $f(x)=(x^{3}-8x^{2}+25x+5x^{2}-40x+125)$ $f(x)=x^{3}-3x^{2}-15x+125$
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