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: 46

Answer

$(4x+y)(16x^2-4xy+y^2-4x+y)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Group the terms of the given expression, $ 64x^3+y^3-16x^2+y^2 ,$ such that the factored form of the groupings will result to a factor that is common to the entire expression. Then, use the factoring of $2$ cubes and the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (64x^3+y^3)-(16x^2-y^2) .\end{array} The expressions $ 64x^3 $ and $ y^3 $ are both perfect cubes (the cube root is exact). Hence, $ 64x^3+y^3 $ is a $\text{ sum }$ 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} [(4x)^3+(y)^3)]-(16x^2-y^2) \\\\= [(4x+y)(16x^2-4xy+y^2)]-(16x^2-y^2) \\\\= (4x+y)(16x^2-4xy+y^2)-(16x^2-y^2) .\end{array} The expressions $ 16x^2 $ and $ y^2 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ 16x^2-y^2 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} (4x+y)(16x^2-4xy+y^2)-[(4x)^2-(y)^2] \\\\= (4x+y)(16x^2-4xy+y^2)-[(4x+y)(4x-y)] \\\\= (4x+y)(16x^2-4xy+y^2)-(4x+y)(4x-y) .\end{array} Factoring the $GCF= (4x+y) $ of the entire expression above results to \begin{array}{l}\require{cancel} (4x+y)[(16x^2-4xy+y^2)-(4x-y)] \\\\ (4x+y)[16x^2-4xy+y^2-4x+y] \\\\ (4x+y)(16x^2-4xy+y^2-4x+y) .\end{array}
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