Perform the multiplication or division and simplify. x+34x2−9÷x2+7x+122x2+7x−15

Step 1
Simplify: ${\displaystyle \frac{x+3}{4x^2-9}÷\frac{x^2+7x+12}{2x^2+7x-15}}$
Step 2
Inverting and changing to multiplication:
${\displaystyle \frac{x+3}{4x^2-9}÷\frac{x^2+7x+12}{2x^2+7x-15}}$
${\displaystyle =\frac{x+3}{4x^2-9}\times \frac{2x^2+7x-15}{x^2+7x+12}}$
factorizing the quadratics:
${\displaystyle =\frac{x+3}{{\left(2x\right)}^2-3^2}\times \frac{2x^2+10x-3x-15}{x^2+4x+3x+12}}$
${\displaystyle =\frac{x+3}{(2x-3)(2x+3)}\times \frac{2x(x+5)-3(x+5)}{x(x+4)+3(x+4)}}$
${\displaystyle =\frac{x+3}{(2x-3)(2x+3)}\times \frac{(2x-3)(x+5)}{(x+3)(x+4)}}$
Step 3
cancelling the same terms, we get:
${\displaystyle =\frac{(x+3)}{(2x-3)(2x+3)}\times \frac{(2x-3)(x+5)}{(x+3)(x+4)}}$
${\displaystyle =\frac{x+5}{(2x+3)(x+4)}}$

Step 1
Given: ${\displaystyle \frac{x+3}{4x^2-9}÷\frac{x^2+7x+12}{2x^2+7x-15}}$
Divide ${\displaystyle \frac{x+3}{4x^2-9}}$ by ${\displaystyle \frac{x^2+7x+12}{2x^2+7x-15}}$ by multiplying ${\displaystyle \frac{x+3}{4x^2-9}}$ by the reciprocal of ${\displaystyle \frac{x^2+7x+12}{2x^2+7x-15}}$
${\displaystyle \frac{(x+3)(2x^2+7x-15)}{(4x^2-9)(x^2+7x+12)}}$
Factor the expressions that are not already factored.
${\displaystyle \frac{(2x-3)(x+3)(x+5)}{(2x-3)(x+3)(x+4)(2x+3)}}$
Cancel out ${\displaystyle (2x-3)(x+3)}$ in both numerator and denominator.
${\displaystyle \frac{x+5}{(x+4)(2x+3)}}$
Expand the expression.
${\displaystyle \frac{x+5}{2x^2+11x+12}}$