How do you factor #12x^4-75y^4#?

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Answer 1 · 2015-06-26T21:26:55

#12x^4-75y^4#

#=3(2x^2-5y^2)(2x^2+5y^2)#

#=3(sqrt(2)x-sqrt(5)y)(sqrt(2)x+sqrt(5)y)(2x^2+5y^2)#

The difference of squares identity is #a^2-b^2 = (a-b)(a+b)#

First separate out the common factor #3# then use the difference of squares identity to get:

#12x^4-75y^4#

#=3((2x^2)^2-(5y^2)^2)#

#=3(2x^2-5y^2)(2x^2+5y^2)#

The first quadratic factor can be treated as a difference of squares with irrational coefficients, so...

#=3((sqrt(2)x)^2-(sqrt(5)y)^2)(2x^2+5y^2)#

#=3(sqrt(2)x-sqrt(5)y)(sqrt(2)x+sqrt(5)y)(2x^2+5y^2)#

The second quadratic factor is irreducible unless you use complex coefficients.