How do you factor #16x^4 - 81y^4#?
1 Answer
Feb 11, 2017
Explanation:
The difference of squares identity can be written:
#a^2-b^2 = (a-b)(a+b)#
Hence we find:
#16x^4-81y^4 = (4x^2)^2-(9y^2)^2#
#color(white)(16x^4-81y^4) = (4x^2-9y^2)(4x^2+9y^2)#
#color(white)(16x^4-81y^4) = ((2x)^2-(3y)^2)(4x^2+9y^2)#
#color(white)(16x^4-81y^4) = (2x-3y)(2x+3y)(4x^2+9y^2)#
The remaining quadratic factor has no linear factors with Real coefficients.