Answer
$\dfrac{3x^2+2x}{9x^2+6x+1}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To simplify the given expression, $
\dfrac{(3x+1)\cdot2x-x^2\cdot3}{(3x+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{(3x\cdot2x+1\cdot2x)-x^2\cdot3}{(3x+1)^2}
\\\\=
\dfrac{(6x^2+2x)-3x^2}{(3x+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{6x^2+2x-3x^2}{(3x+1)^2}
\\\\=
\dfrac{3x^2+2x}{(3x+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+2x}{(3x)^2+2(3x)(1)+(1)^2}
\\\\=
\dfrac{3x^2+2x}{9x^2+6x+1}
.\end{array}
