College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 3 - Section 3.3 - Zeros of Polynomial Functions - 3.3 Exercises - Page 316: 57

Answer

$f(x)=x^{2}-10x+26$ (sample answer, taking $a=1$)

Work Step by Step

Complex zeros come in conjugate pairs (conjugate zeros theorem). The given zeros are a conjugate pair, so there is no need to search further. If $k$ is a zero, then $(x-k)$ is a factor of $f(x)$ ... (factor theorem) So, $(x-(5+i))$ is a factor of f, $(x-(5-i))$ is a factor of f. We assume multiplicity 1, so we have $f(x)=a(x-(5+i))(x-(5-i))$ We are asked for $a$ polynomial, so we choose the one for which $a=1.$ Use FOIL to distribute $f(x)=x^{2}-x(5-i)-x(5+i)+(5+i)(5-i)$ ... distribute, recognize a difference of squares $f(x)=x^{2}-5x+ix-5x-ix+25-i^{2}$ ... the terms with $ix$ cancel, and $i^{2}=-1$.... $f(x)=x^{2}-5x-5x+25-(-1)$ $f(x)=x^{2}-10x+26$
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