Dustin Y. answered • 02/29/16

Tutor

New to Wyzant
I would love to help

Hi!

That's one hefty problem!

First lets pull that 3 out of there since it is a factor of 3, 24, and 28.

Now we have 3(16a

^{5}b^{2}- 8a^{4}b^{3}^{ + }a^{3}b^{4})Now lets look at A. a

^{3}is the smallest A we have so lets pull that out too!Now we have 3a

^{3}(16a^{2}b^{2}- 8ab^{3}+ b^{4})Were getting close! There's only one more variable left that looks like it repeats and that's B! Looks like b

^{2}can be pulled out as well so lets do that!Now we have 3a

^{3}b^{2}(16a^{2}-8ab + b^{2})It almost looks like were done considering we've factored out a 3, an a

^{3}, and a b^{2}but there's one more thing we can do. notice that what's inside the parenthesis is a quadratic. That means there's a chance that we can factor that too!Lets do it.

(4a - b)(4a - b)

Lets double check to make sure it works. To do that we foil it.

4a * 4a= 16a

^{2}4a * -b= -4ab

-b * 4a= -4ab

-b * -b= b

^{2}adding those up we get 16a

^{2}-8ab+b^{2}Looks like it worked so were finally done!

Our answer is

**3a**

^{3}b^{2}(4a-b)^{2}Best of Luck!