Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 3 - Review - Test - Page 323: 9

Answer

$x^4+2x^3+10x^2+18x+9$

Work Step by Step

A zero of $-1$ with multiplicity 2 give factors of $(x+1)^2$, a complex zero of $3i$ means another zero of $-3i$, so the function can be written as: $P(x)=(x+1)^2(x-3i)(x+3i)=(x^2+2x+1)(x^2+9)=x^4+2x^3+10x^2+18x+9$
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