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: 26

Answer

$Q(x)=(x-5)(x+5)(x-5i)(x+5i)$ Zeros: $5$ , $-5$ , $5i$ , $-5i$, each with multiplicity 1.

Work Step by Step

$Q(x)=(x^{2})^{2}-25^{2}= \qquad $...difference of squares... $=(x^{2}-25)(x^{2}+25)$ $x^{2}-25= x^{2}-5^{2}\quad $...difference of squares...$=(x-5)(x+5)$ $x^{2}+25=x^{2}-(-1)\cdot 5^{2}=x^{2}-(5i)^{2} \qquad $...difference of squares...$=(x-5i)(x+5i)$ $Q(x)=(x-5)(x+5)(x-5i)(x+5i)$ Zeros: $5$ , $-5$ , $5i$ , $-5i$, each with multiplicity 1.
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