Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.7 - Factoring by Special Products - Exercise Set - Page 309: 37



Work Step by Step

Factoring the $GCF= 2 $ results to $ 2(125y^3-8x^3) $. Using $a^3+b^3=(a+b)(a^2-ab+b^2)$, or the factoring of the sum/difference of two cubes, then, \begin{array}{l} 2(125y^3-8x^3) \\= 2[(5y)+(-2x)][(5y)^2-(5y)(-2x)+(-2x)^2) \\= 2(5y-2x)(25y^2+10xy+4x^2) .\end{array}
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