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: 67

Answer

$(2r+t)(r^2-rt+19t^2)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ (r+2t)^3+(r-3t)^3 ,$ use the factoring of the sum or difference of $2$ cubes. $\bf{\text{Solution Details:}}$ The expressions $ (r+2t)^3 $ and $ (r-3t)^3 $ are both perfect cubes (the cube root is exact). Hence, $ (r+2t)^3+(r-3t)^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} [(r+2t)+(r-3t)][(r+2t)^2-(r+2t)(r-3t)+(r-3t)^2] \\\\= [2r+t][(r+2t)^2-(r+2t)(r-3t)+(r-3t)^2] .\end{array} Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} [2r+t][(r^2+4rt+4t^2)-(r+2t)(r-3t)+(r^2-6rt+9t^2)] .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel} [2r+t][(r^2+4rt+4t^2)-(r^2-rt-6t^2)+(r^2-6rt+9t^2)] .\end{array} By combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} [2r+t][r^2+4rt+4t^2-r^2+rt+6t^2+r^2-6rt+9t^2] \\\\= [2r+t][(r^2-r^2+r^2)+(4rt+rt-6rt)+(4t^2+6t^2+9t^2)] \\\\= [2r+t][r^2-rt+19t^2] \\\\= (2r+t)(r^2-rt+19t^2) .\end{array}
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