How do you factor 2d^4-32f^42d4−32f4?
1 Answer
Dec 28, 2016
Explanation:
The difference of squares identity can be written:
a^2-b^2 = (a-b)(a+b)a2−b2=(a−b)(a+b)
Hence we find:
2d^4-32f^4 = 2(d^4-16f^4)2d4−32f4=2(d4−16f4)
color(white)(2d^4-32f^4) = 2((d^2)^2-(4f^2)^2)2d4−32f4=2((d2)2−(4f2)2)
color(white)(2d^4-32f^4) = 2(d^2-4f^2)(d^2+4f^2)2d4−32f4=2(d2−4f2)(d2+4f2)
color(white)(2d^4-32f^4) = 2(d^2-(2f)^2)(d^2+4f^2)2d4−32f4=2(d2−(2f)2)(d2+4f2)
color(white)(2d^4-32f^4) = 2(d-2f)(d+2f)(d^2+4f^2)2d4−32f4=2(d−2f)(d+2f)(d2+4f2)
The remaining sum of squares can only be factored further with Complex coefficients.
