Answer
$\pm 2i, \pm 3i$
Step 2. We have $f(x)=(x+3i)(x-3i)(x+2i)(x-2i)$
Work Step by Step
Step 1. $f(x)=x^4+13x^2+36=(x^2+9)(x^2+4)$, thus the zeros are $x=\pm 3i, \pm 2i$
Step 2. We have $f(x)=(x+3i)(x-3i)(x+2i)(x-2i)$
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
Chapter 3 - Polynomial and Rational Functions - Section 3.3 Complex Zeros; Fundamental Theorem of Algebra - 3.3 Assess Your Understanding - Page 231: 36
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