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: 68

Answer

(a) $P(x)=(x^2+4)(x^2+4)$ (b) $P(x)=(x+2i)(x+2i)(x-2i)(x-2i)$

Work Step by Step

(a) Factor P into linear and irreducible quadratic factors with real coefficients. Assume $y=x^2$, we have $P(x)=y^2+8y+16=(y+4)^2$, hence $P(x)=(x^2+4)^2=(x^2+4)(x^2+4)$ (b) Factor P completely into linear factors with complex coefficients. Let $x^2+4=0$, we can find the complex roots as $x=\pm 2i$, hence $P(x)=(x+2i)^2(x-2i)^2=(x+2i)(x+2i)(x-2i)(x-2i)$
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