Question

Factor by using the difference of squares method: 36x^4 - 4x^2

A. (6x^2 – 2x)(6x^2 - 2x) = 4x^2(3x - 1)(3x - 1)
B. (6x^2 + 2x)(6x^2 + 2x) = 2x^2(3x + 1)(2x + 1)
C. (-6x^2 – 2x)(-6x^2 - 2x) = 4x^2(-3x - 1)(-3x - 1)
D. (6x^2 – 2x)(6x^2 + 2x) = 4x^2(3x - 1)(3x + 1)
E. (-6x^2 + 2x)(6x^2 - 2x) = 2x^2(-3x + 1)(3x - 1)
C. This polynomial cannot be factored by using the difference of squares method.

Thanks!

Answer 1

The difference of squares formula is a^2-b^2=(a+b)(a-b). a^2, in this case, is 36x^4, so we can say the square root(positive in this case) is 6x^2. This means that a is 6x^2. We can do this also to the b^2 part and get 4x^2=b^2 so b=2x. The difference of squares formula states that the answer is (6x^2+2x)(6x^2-2x), so the answer is D.