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: 72

Answer

$(x+3)(x-5)(x+5)$

Work Step by Step

Regroup to obtain: $=(x^3+3x^2) + (-25x-75)$ Factor out $x^2$ in the first group and $-25$ in the second to obtain: $=x^2(x+3) + (-25)(x+3) \\=x^2(x+3) - 25(x+3)$ Factor out the GCF of $x+3$ to obtain: $=(x+3)(x^2-25) \\=(x+3)(x^2-5^2)$ Factor the difference of two squares using the formula $a^2-b^2=(a-b)(a+b)$ to obtain: $=(x+3)(x-5)(x+5)$
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