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: 40

Answer

$Q(x)=x^{3}+x$

Work Step by Step

The conjugate of i is also a zero (Conjugate Roots Theorem) so we have three zeros, 0, $i,$ and $-i$ for a polynomial of degree 3, as there should be. Factorization of Q(x): $Q(x)=a(x-0)(x+i)(x-i)$ ... taking a=1, apply difference of squares... $=x(x^{2}-(-1))=x(x^{2}+1)$ $=x^{3}+x$
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