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: 34

Answer

$-i, i, -5i,$ and $5i.$

Work Step by Step

First, because all coefficients of f(x) are real numbers, the Conjugate Zeros Theorem applies. If $i$ is a zero, so is its conjugate, $-i.$ 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 now $k=i$,, note the missing $x^{3}$ and $x$ terms $\left\{\begin{array}{lllllll} i & | & 1 & 0 & 26 & 0 & 25\\ & & & i & -1 & 25i & -25\\ & & -- & -- & -- & -- & --\\ & & 1 & i & 25 & 25i & \fbox{$0$} \end{array}\right.$ $f(x)=(x-i)(x^{3}+ix^{2}+25x+25i)$ Now, we use the fact that $-i$ is also a zero. Apply synthetic division on the quotient. $\left\{\begin{array}{llllll} -i & | & 1 & i & 25 & 25i\\ & & & -i & 0 & -25i\\ & & -- & -- & -- & --\\ & & 1 & 0 & 25 & \fbox{$0$} \end{array}\right.$ $f(x)=(x+i)(x-i)(x^{2}+25)$ The zeros of $x^{2}+25:$ $x^{2}+25=0$ $x^{2}=-25$ $x=\pm 5i$ The other zeros (other than the given) are $-i,\quad -5i$ and $5i.$
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