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: 33

Answer

$-i, i,-2i$ and $2i.$

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 & 5 & 0 & 4\\ & & & -i & -1 & -4i & -4\\ & & -- & -- & -- & -- & --\\ & & 1 & -i & 4 & -4i & \fbox{$0$} \end{array}\right.$ $f(x)=(x+i)(x^{3}+ix^{2}+4x+4i)$ Now, we use the fact that $i$ is also a zero. Apply synthetic division on the quotient. $\left\{\begin{array}{llllll} i & | & 1 & -i & 4 & -4i\\ & & & i & 0 & 4i\\ & & -- & -- & -- & --\\ & & 1 & 0 & 4 & \fbox{$0$} \end{array}\right.$ $f(x)=(x+i)(x-i)(x^{2}+4)$ The zeros of $x^{2}+4:$ $x^{2}+4=0$ $x^{2}=-4$ $x=\pm 2i$ The other zeros (other than the given) are $i,\quad -2i$ and $2i.$
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