How do you factor 16x^4 - 81y^416x4−81y4?
1 Answer
Feb 11, 2017
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:
16x^4-81y^4 = (4x^2)^2-(9y^2)^216x4−81y4=(4x2)2−(9y2)2
color(white)(16x^4-81y^4) = (4x^2-9y^2)(4x^2+9y^2)16x4−81y4=(4x2−9y2)(4x2+9y2)
color(white)(16x^4-81y^4) = ((2x)^2-(3y)^2)(4x^2+9y^2)16x4−81y4=((2x)2−(3y)2)(4x2+9y2)
color(white)(16x^4-81y^4) = (2x-3y)(2x+3y)(4x^2+9y^2)16x4−81y4=(2x−3y)(2x+3y)(4x2+9y2)
The remaining quadratic factor has no linear factors with Real coefficients.
