Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Review Exercises - Page 143: 104



Work Step by Step

$ x^{3}+5x^{2}-2x-10=\quad $ Factor in pairs; factor out $ x^{2}$ from the first pair, and $-2$ from the second $=x^{2}(x+5)-2(x+5)$ $=(x+5)(x^{2}-2)$ We could treat the second parentheses as a difference of squares, using $(\sqrt{2})^{2}=2$, but factoring implies integer coefficients, so we stop here.
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