How do you factor #7x^2y^2-63y^4# ?

1 Answer
Aug 23, 2017

Answer:

#7x^2y^2-63y^4 = 7y^2(x-3y)(x+3y)#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We will use this with #a=x^2# and #b=(3y)^2#, but first separate out the common factor #7y^2#...

#7x^2y^2-63y^4 = 7y^2(x^2-9y^2)#

#color(white)(7x^2y^2-63y^4) = 7y^2(x^2-(3y)^2)#

#color(white)(7x^2y^2-63y^4) = 7y^2(x-3y)(x+3y)#