Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 3 - Section 3.5 - Complex Zeros and the Fundamental Theorem of Algebra - 3.5 Exercises - Page 293: 8

Answer

$(a)\quad $Zeros: $0$ and $\pm 3i$ $(b)\quad P(x)=x^{3}(x-3i)(x+3i)$

Work Step by Step

Complete Factorization Theorem (p.287) $P$ factors into $n$ linear factors : $P(x)=a(x-c_{1})(x-c_{2})\cdots(x-c_{n})$ where $a$ is the leading coefficient of $P$ and $c_{1}, c_{1}, \ldots, c_{n}$ are the zeros of $P$. ----- $(a)\quad$ $x^{5}+9x^{3}=0$ $x^{3}(x^{2}+9)=0$ So, either $x^{3}=0\Rightarrow x=0$ is a zero, (multiplicity 3), or $x^{2}+9=0\Rightarrow x^{2}=-3\Rightarrow x=\pm 3i$ Zeros: $0$ and $\pm 3i$ $(b)$ The leading coefficient of P is $a=1$, so, by the above theorem, $P(x)=x^{3}(x-3i)(x+3i)$
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.