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: 57

Answer

$(x+3)(x^2-3x+9)$

Work Step by Step

The given expression can be written as $x^3+3^3.$ RECALL: $a^3+b^3=(a+b)(a^2-ab+b^2).$ Factor the given sum of two cubes using the formula above with $a=x$ and $b=3$ to obtain $(x+3)(x^2-x(3)+3^2) \\=(x+3)(x^2-3x+9).$
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