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 315: 23

Answer

$f(x)= (x-3i)(x+3)(x+4)$

Work Step by Step

If k is a zero of f, then, by the factor theorem, (x-k) is a factor of f(x). Synthetic division gives us a quotient q(x), with degree one less than f(x). and we know the remainder is zero, so $f(x)=(x-k)\cdot q(x).$ We use this first. $k=3i$ $\left\{\begin{array}{lllllll} 3i & | & 1 & 7-3i & 12-21i & -36i & \\ & & & 3i & 21i & 36i & \\ & & -- & -- & -- & -- & --\\ & & 1 & 7 & 12 & \fbox{$0$} & \end{array}\right.$ $f(x)= (x-3i)(x^{2}+7x+12)$ We now factor the trinomial $x^{2}+bx+c$ by searching for two factors of $c$ whose sum is $b.$ Factors of $12$ whose sum is $7$ ... are $4$ and $3$: $x^{2}+7x+12=(x+3)(x+4)$ So, $f(x)= (x-3i)(x+3)(x+4)$
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