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: 71

Answer

$(m+n-5)(m-n+1)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Group the first $3$ terms and the last $3$ terms of the given expression, $ m^2-4m+4-n^2+6n-9 ,$ and use the factoring of perfect square trinomials. Then use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ Grouping the first $3$ terms and last $3$ terms of the given expression results to \begin{array}{l}\require{cancel} (m^2-4m+4)-(n^2-6n+9) .\end{array} Using the factoring of perfect square trinomials which is given by $a^2-2ab+b^2=(a-b)^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} (m-2)^2-(n-3)^2 .\end{array} The expressions $ (m-2)^2 $ and $ (n-3)^2 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ (m-2)^2-(n-3)^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-2)+(n-3)][(m-2)-(n-3)] \\\\= [m-2+n-3][(m-2-n+3] \\\\ [m+n-5][m-n+1] \\\\= (m+n-5)(m-n+1) .\end{array}
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