# Chapter 5 - Section 5.4 - A General Approach to Factoring - 5.4 Exercises - Page 348: 34

$9(5k+2r)(5k-2r)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $225k^2-36r^2 ,$ factor first the $GCF.$ Then use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ The $GCF$ of the terms in the given expression is $9 ,$ 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} 9(25k^2-4r^2) .\end{array} The expressions $25k^2$ and $4r^2$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $25k^2-4r^2 ,$ 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} 9[(5k)^2-(2r)^2] \\\\= 9[(5k+2r)(5k-2r)] \\\\= 9(5k+2r)(5k-2r) .\end{array}

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