Answer
$\pm i, \pm 2i$
$f(x)=(x+i)(x-i)(x+2i)(x-2i)$
Work Step by Step
Step 1. $f(x)=x^4+5x^2+4=(x^2+1)(x^2+4)$, thus the zeros are $x=\pm i, \pm 2i$
Step 2. We have $f(x)=(x+i)(x-i)(x+2i)(x-2i)$
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: 35
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