Answer
$\dfrac{k^{2}+4k-12}{k^{2}+10k+24}\cdot\dfrac{k^{2}+k-12}{k^{2}-9}=\dfrac{k-2}{k+3}$
Work Step by Step
$\dfrac{k^{2}+4k-12}{k^{2}+10k+24}\cdot\dfrac{k^{2}+k-12}{k^{2}-9}$
Factor both rational expressions completely:
$\dfrac{k^{2}+4k-12}{k^{2}+10k+24}\cdot\dfrac{k^{2}+k-12}{k^{2}-9}=...$
$...=\dfrac{(k+6)(k-2)}{(k+6)(k+4)}\cdot\dfrac{(k+4)(k-3)}{(k-3)(k+3)}=...$
Evaluate the product of the two fractions:
$...=\dfrac{(k+6)(k-2)(k+4)(k-3)}{(k+6)(k+4)(k-3)(k+3)}=...$
Simplify the resulting expression by removing the factors that appear both in the numerator and in the denominator:
$...=\dfrac{k-2}{k+3}$
