Answer
$\dfrac{4n^{2}+4n-3}{6n^{2}-n-15}\cdot\dfrac{8n^{2}+32n+30}{4n^{2}+16n+15}=\dfrac{2(2n-1)}{3n-5}$
Work Step by Step
$\dfrac{4n^{2}+4n-3}{6n^{2}-n-15}\cdot\dfrac{8n^{2}+32n+30}{4n^{2}+16n+15}$
Factor both rational expressions completely:
$\dfrac{4n^{2}+4n-3}{6n^{2}-n-15}\cdot\dfrac{8n^{2}+32n+30}{4n^{2}+16n+15}=...$
$...=\dfrac{(2n+3)(2n-1)}{(2n+3)(3n-5)}\cdot\dfrac{2(2n+5)(2n+3)}{(2n+5)(2n+3)}=...$
Evaluate the product of the two rational expressions:
$...=\dfrac{2(2n+3)^{2}(2n-1)(2n+5)}{(2n+3)^{2}(3n-5)(2n+5)}=...$
Simplify by removing the factors that appear both in the numerator and in the denominator of the resulting rational expression:
$...=\dfrac{2(2n-1)}{3n-5}$
