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 294: 44

Answer

$x^4+13x^2+36$

Work Step by Step

The Complex Conjugate zeroes Theorem states that the conjugate of $i$ is also a zero of the polynomial $P(x)$. Here, we have three zeroes of the polynomial $P(x)$ of degree $4$. The factorization of $P(x)$ is given as follows: $P(x)=(x-2i)(x+2i)(x-3i)(x+3i)$ Apply the difference square formula. $P(x)=(x^2-(2i)^2(x^2-(3i)^2)=(x^2+4)(x^2+9)=x^4+9x^2+4x^2+36$ Hence, $P(x)=x^4+13x^2+36$
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.