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: 67

Answer

(a) $P(x)=(x+1)(x-1)(x^2+9)$ (b)$P(x)=(x+1)(x-1)(x+3i)(x-3i)$

Work Step by Step

(a) Factor P into linear and irreducible quadratic factors with real coefficients. Assume $y=x^2$, we have $P(x)=y^2+8y-9=(y+9)(y-1)$, hence $P(x)=(x^2+9)(x^2-1)=(x+1)(x-1)(x^2+9)$ (b) Factor P completely into linear factors with complex coefficients. Let $x^2+9=0$, we can find the complex roots as $x=\pm 3i$, hence $P(x)=(x+1)(x-1)(x+3i)(x-3i)$
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