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: 45

Answer

$(m-n)(m^2+mn+n^2+m+n)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Group the terms of the given expression, $ m^3+m^2-n^3-n^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 third terms and the second and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (m^3-n^3)+(m^2-n^2) .\end{array} The expressions $ m^3 $ and $ n^3 $ are both perfect cubes (the cube root is exact). Hence, $ m^3-n^3 $ 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} [(m)^3-(n)^3]+(m^2-n^2) \\\\= (m-n)[(m)^2+m(n)+(n)^2]+(m^2-n^2) \\\\= (m-n)(m^2+mn+n^2)+(m^2-n^2) .\end{array} The expressions $ m^2 $ and $ n^2 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ m^2-n^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} (m-n)(m^2+mn+n^2)+[(m)^2-(n)^2] \\\\= (m-n)(m^2+mn+n^2)+(m-n)(m+n) .\end{array} Factoring the $GCF= (m-n) $ of the entire expression above results to \begin{array}{l}\require{cancel} (m-n)[(m^2+mn+n^2)+(m+n)] \\\\ (m-n)(m^2+mn+n^2+m+n) .\end{array}
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