allhvasstH

2021-02-25

Write the final factorization for each problem.
$12{a}^{3}+20{a}^{2}b-9a{b}^{2}-15{b}^{3}$

Demi-Leigh Barrera

Step 1
GCF of $12{a}^{3}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}20{a}^{2}bis=4{a}^{2}$
GCF of $-9a{b}^{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}-15{b}^{3}is=-3{b}^{2}$
Factor out $4{a}^{2}$ from the first two terms and then factor out $-3{b}^{2}$ from the last two terms.
$12{a}^{3}+20{a}^{2}b-9a{b}^{2}-15{b}^{3}$
$=4{a}^{2}\left(3a+5b\right)-3{b}^{2}\left(3a+5b\right)$
Step 2
Then we can factor out (3a+5b) from both terms.
$4{a}^{2}\left(3a+5b\right)-3{b}^{2}\left(3a+5b\right)$
$=\left(3a+5b\right)\left(4{a}^{2}-3{b}^{2}\right)$
Result:$\left(3a+5b\right)\left(4{a}^{2}-3{b}^{2}\right)$

Jeffrey Jordon