## College Algebra (10th Edition)

$\dfrac{13}{25x^2-20x+4}$
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $\dfrac{(4x+1)\cdot5-(5x-2)\cdot4}{(5x-2)^2} ,$ use the Distributive Property first. Then remove the grouping symbols and combine like terms. Finally, use special products to simplify the denominator. $\bf{\text{Solution Details:}}$ Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{(4x\cdot5+1\cdot5)-(5x\cdot4-2\cdot4)}{(5x-2)^2} \\\\= \dfrac{(20x+5)-(20x-8)}{(5x-2)^2} .\end{array} Removing the grouping symbols and then combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{20x+5-20x+8}{(5x-2)^2} \\\\= \dfrac{13}{(5x-2)^2} \\\\= \dfrac{13}{(5x-2)^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} \dfrac{13}{(5x)^2-2(5x)(2)+(2)^2} \\\\= \dfrac{13}{25x^2-20x+4} .\end{array}