## College Algebra (10th Edition)

$\dfrac{-3x^2-8x+3}{x^4+2x^2+1}$
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $\dfrac{(x^2+1)\cdot3-(3x+4)\cdot2x}{(x^2+1)^2} ,$ 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{(x^2\cdot3+1\cdot3)-(3x\cdot2x+4\cdot2x)}{(x^2+1)^2} \\\\= \dfrac{(3x^2+3)-(6x^2+8x)}{(x^2+1)^2} .\end{array} Removing the grouping symbols and then combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{3x^2+3-6x^2-8x}{(x^2+1)^2} \\\\= \dfrac{-3x^2-8x+3}{(x^2+1)^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{-3x^2-8x+3}{(x^2)^2+2(x^2)(1)+(1)^2} \\\\= \dfrac{-3x^2-8x+3}{x^4+2x^2+1} .\end{array}