Precalculus (6th Edition) Blitzer

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

Chapter P - Section P.5 - Factoring Polynomials - Exercise Set: 71

Answer

$(x+2)(x-3)(x+3)$

Work Step by Step

Regroup to obtain: $=(x^3+2x^2) + (-9x-18)$ Factor out $x^2$ in the first group and $-9$ in the second to obtain: $=x^2(x+2) + (-9)(x+2) \\=x^2(x+2) - 9(x+2)$ Factor out the GCF of $x+2$ to obtain: $=(x+2)(x^2-9) \\=(x+2)(x^2-3^2)$ Factor the difference of two squares using the formula $a^2-b^2=(a-b)(a+b)$ to obtain: $=(x+2)(x-3)(x+3)$
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