# Thread: More Factor By Grouping

1. ## More Factor By Grouping

Factor By grouping.

4a^2b^2 - 9 (ab + c)^2

Must I FOIL (ab + c)^2 and distribute 9 before factoring?

2. ## Re: More Factor By Grouping

Factor By grouping.

4a^2b^2 - 9 (ab + c)^2

Must I FOIL (ab + c)^2 and distribute 9 before factoring?
No, $\ \ \ 4a^2b^2 - 9(ab + c)^2$ is the difference of two squares.

$A^2 - B^2 \ = \ (A - B)(A + B)$

Determine what the square root of $\ \ 4a^2b^2 \ \ is \ \ and \ \ also \ \ what \ \ the \ \ square \ \ root \ \ of \ \ 9(ab + c)^2 \ \ is.$

3. ## Re: More Factor By Grouping

4a^2b^2 - 9 (ab + c)^2

(2ab - 3 (ab + c))(2ab + 3 (ab + c))

Correct?

4. ## Re: More Factor By Grouping

4a^2b^2 - 9 (ab + c)^2

(2ab - 3 (ab + c))(2ab + 3 (ab + c))

Correct?
there are clearer ways to post this

$4a^2b^2 - 9(a b+ c)^2 = x^2 - y^2$

$x = 2ab$

$y= 3(ab+c)$

$x^2 - y^2 = (x+y)(x-y) = (2ab + 3(ab+c))(2ab-3(ab+c))$

so yes, it looks like you are correct

5. ## Re: More Factor By Grouping

Of course, it would make sense to then "clean up" each factor:
(2ab - 3 (ab + c))(2ab + 3 (ab + c))= (2ab- 3ab- 3c)(2ab+ 3ab+ 3c)= -(ab+ 3c)(5ab+ 3c)

6. ## Re: More Factor By Grouping

Originally Posted by romsek
there are clearer ways to post this

$4a^2b^2 - 9(a b+ c)^2 = x^2 - y^2$

$x = 2ab$

$y= 3(ab+c)$

$x^2 - y^2 = (x+y)(x-y) = (2ab + 3(ab+c))(2ab-3(ab+c))$

so yes, it looks like you are correct
Good to know I got it right.