first, notice that the first three terms, x^2-12x+36, is a perfect square, and can be factored into (x-6)(x-6),, now rewriting the expression, it becomes:
(x-6)^2 - 49y^2
(x-6)^2 - (7y)^2
this looks like difference of two squares,, let a = x-6 and b=7y, thus:
a^2 - b^2
(a-b)(a+b)
(x-6-7y)(x-6+7y)
so there,, :)
having trouble factoring this particular problem
x^2-12x+36-49y2
factored by grouping
x(x-12) (7y-6) (7y+6)
the answer in the book has
(x-6+7y)(x-6-7y)
confused.
2 answers
Gotta remember those damn steps. I immediately assumed it was four terms so I factor by grouping. I forgot that you first look for common factor, then squares etc.
Thanks.
Thanks.