Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 3 - 3.4 - Zeroes of Polynomial Functions - 3.4 Exercises - Page 284: 65


$f(x)=(x+2)(x-2)(x+2i)(x-2i)$ The zeros: $x=-2$ $x=2$ $x=-2i$ $x=2i$

Work Step by Step

$f(x)=x^4-16$ $f(x)=(x^2)^2-4^2$ $f(x)=(x^2-4)(x^2+4)$ $f(x)=(x^2-2^2)[x^2-(2i)^2]$ $f(x)=(x+2)(x-2)(x+2i)(x-2i)$ The zeros: $x+2=0$ $x=-2$ $x-2=0$ $x=2$ $x+2i=0$ $x=-2i$ $x-2i=0$ $x=2i$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.