Factor By grouping. 4a^2b^2 - 9 (ab + c)^2 Must I FOIL (ab + c)^2 and distribute 9 before factoring?
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Originally Posted by mathdad1965 Factor By grouping. 4a^2b^2 - 9 (ab + c)^2 Must I FOIL (ab + c)^2 and distribute 9 before factoring? No, is the difference of two squares. Determine what the square root of Those will be your A and your B, respectively.
Last edited by greg1313; Jan 3rd 2017 at 01:30 PM.
4a^2b^2 - 9 (ab + c)^2 (2ab - 3 (ab + c))(2ab + 3 (ab + c)) Correct?
Originally Posted by mathdad1965 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
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)
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.