How do you factor completely #−12x^3y^5 − 9x^2y^2 + 12xy^3#?
1 Answer
But note that:
#color(red)(9x^4y^4)-12x^3y^5-9x^2y^2+12xy^3 = 3xy^2(xy-1)(xy+1)(3x-4y)#
Explanation:
Given:
#-12x^3y^5-9x^2y^2+12xy^3#
First note that all of the terms are divisible by
#-12x^3y^5-9x^2y^2+12xy^3 = -3xy^2(4x^2y^3+3x-4y)#
It is not possible to factor this further.
I suspect that a leading term
#9x^4y^4-12x^3y^5-9x^2y^2+12xy^3 = 3xy^2(3x^3y^2-4x^2y^3-3x+4y)#
#color(white)(9x^4y^4-12x^3y^5-9x^2y^2+12xy^3) = 3xy^2((3x^3y^2-4x^2y^3)-(3x-4y))#
#color(white)(9x^4y^4-12x^3y^5-9x^2y^2+12xy^3) = 3xy^2(x^2y^2(3x-4y)-1(3x-4y))#
#color(white)(9x^4y^4-12x^3y^5-9x^2y^2+12xy^3) = 3xy^2(x^2y^2-1)(3x-4y)#
#color(white)(9x^4y^4-12x^3y^5-9x^2y^2+12xy^3) = 3xy^2((xy)^2-1^2)(3x-4y)#
#color(white)(9x^4y^4-12x^3y^5-9x^2y^2+12xy^3) = 3xy^2(xy-1)(xy+1)(3x-4y)#