Answer
$\dfrac{-3x^2-8x+3}{x^4+2x^2+1}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
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}
