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

Skilled2021-02-26Added 97 answers

Step 1

GCF of$12{a}^{3}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}20{a}^{2}bis=4{a}^{2}$

GCF of$-9a{b}^{2}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}-15{b}^{3}is=-3{b}^{2}$

Factor out$4{a}^{2}$ from the first two terms and then factor out $-3{b}^{2}$ from the last two terms.

$12{a}^{3}+20{a}^{2}b-9a{b}^{2}-15{b}^{3}$

$=4{a}^{2}(3a+5b)-3{b}^{2}(3a+5b)$

Step 2

Then we can factor out (3a+5b) from both terms.

$4{a}^{2}(3a+5b)-3{b}^{2}(3a+5b)$

$=(3a+5b)(4{a}^{2}-3{b}^{2})$

Result:$(3a+5b)(4{a}^{2}-3{b}^{2})$

GCF of

GCF of

Factor out

Step 2

Then we can factor out (3a+5b) from both terms.

Result:

Jeffrey Jordon

Expert2021-11-14Added 2575 answers

Answer is given below (on video)