# Factoring completely

• Aug 9th 2012, 09:46 AM
gashbell1
Factoring completely
how to factor this completely

6a^2+12ab+6b^2+7ac+7bc-20c^2?
thanks
• Aug 9th 2012, 10:21 AM
ebaines
Re: Factoring completely
(2a+2b+5c)(3a+3b-4c)

It's a matter of trial and error, but there are few clues you can use. First notice that the coefficients for all the a and b terms are the same - that tells you that the coefficients for a and b in the factored form are most likely the same as well. The fact that all coeefficients are positive except for the c^2 term tells you that all coefficients of the factored terms are positive except for one of the c's. The factors of 6 are either 1 and 6 or 2 and 3, so try both and see where it leads.
• Aug 9th 2012, 10:29 AM
HallsofIvy
Re: Factoring completely
What do you mean by "factor completely"? What I see is that \$\displaystyle 6^2+ 12ab+ 6b^2= 6(a^2+ 2ab+ b^2)= 6(a+ b)^2\$ and that \$\displaystyle 7ac+ 7bc= 7c(a+ b)\$. That gives \$\displaystyle 6a^2+ 12ab+ 6b^2= 6(a+ b)^2+ 7c(a+ b)- 20c^2\$. I would not consider that "factored completely" but that is the best I can do.
• Aug 9th 2012, 07:47 PM
gashbell1
Re: Factoring completely
Quote:

Originally Posted by ebaines
(2a+2b+5c)(3a+3b-4c)

It's a matter of trial and error, but there are few clues you can use. First notice that the coefficients for all the a and b terms are the same - that tells you that the coefficients for a and b in the factored form are most likely the same as well. The fact that all coeefficients are positive except for the c^2 term tells you that all coefficients of the factored terms are positive except for one of the c's. The factors of 6 are either 1 and 6 or 2 and 3, so try both and see where it leads.

Nice! Thanks. But can you show the solution? I didnt get the 7ac+7bc-20c^2 part there.

Quote:

Originally Posted by HallsofIvy
What do you mean by "factor completely"? What I see is that \$\displaystyle 6^2+ 12ab+ 6b^2= 6(a^2+ 2ab+ b^2)= 6(a+ b)^2\$ and that \$\displaystyle 7ac+ 7bc= 7c(a+ b)\$. That gives \$\displaystyle 6a^2+ 12ab+ 6b^2= 6(a+ b)^2+ 7c(a+ b)- 20c^2\$. I would not consider that "factored completely" but that is the best I can do.

Uhm. I think Factoring completely is, all answers must be form in a parentheses, no other operations. Thanks for help and solution but I need to know how to factor that 7ac+7bc-20c^2. :((
• Aug 10th 2012, 04:38 AM
ebaines
Re: Factoring completely
Quote:

Originally Posted by gashbell1
Nice! Thanks. But can you show the solution? I didnt get the 7ac+7bc-20c^2 part there.

\$\displaystyle (2a+2c+5c)(3a+3b-4c) = 2a(3a+3b-4c)+2b(3a+3b-4c)+5c(3a+3b-4c) =6a^2+6ab-8ac+6ab+6b^2-8bc+15ac+15bc-20c^2 = 6a^2+12ab+6b^2+7ac+7bc-20c^2\$