## College Algebra (10th Edition)

$\dfrac{x^2+4}{x^4-8x^2+16}$
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $\dfrac{x\cdot2x-(x^2-4)\cdot1}{(x^2-4)^2} ,$ multiply the factors first. Then remove the grouping symbols and combine like terms. Finally, use special products to simplify the denominator. $\bf{\text{Solution Details:}}$ Multiplying the factors, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{2x^2-(x^2-4)}{(x^2-4)^2} .\end{array} Removing the grouping symbols and then combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{2x^2-x^2+4}{(x^2-4)^2} \\\\= \dfrac{x^2+4}{(x^2-4)^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{x^2+4}{(x^2)^2-2(x^2)(4)+(4)^2} \\\\= \dfrac{x^2+4}{x^4-8x^2+16} .\end{array}