Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Section 5.4 - A General Approach to Factoring - 5.4 Exercises: 22

Answer

$7(2k-5)(4k^2+10k+25)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 56k^3-875 ,$ factor first the $GCF.$ Then use the factoring of the sum or difference of $2$ cubes. $\bf{\text{Solution Details:}}$ The $GCF$ of the terms in the given expression is $ 7 ,$ since it is the greatest expression that can divide all the terms evenly (no remainder.) Factoring the $GCF$ results to \begin{array}{l}\require{cancel} 7(8k^3-125) .\end{array} The expressions $ 8k^3 $ and $ 125 $ are both perfect cubes (the cube root is exact). Hence, $ 8k^3-125 $ is a $\text{ difference }$ of $2$ cubes. Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent to \begin{array}{l}\require{cancel} 7[(2k)^3-(5)^3] \\\\= 7(2k-5)[(2k)^2+2k(5)+(5)^2] \\\\= 7(2k-5)(4k^2+10k+25) .\end{array}
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