How do you factor completely #−12x^3y^5 − 9x^2y^2 + 12xy^3#?

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Apr 13, 2017

Answer:

#-12x^3y^5-9x^2y^2+12xy^3 = -3xy^2(4x^2y^3+3x-4y)#

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 #3#, #x# and #y^2#, and hence by #3xy^2#. So we can separate that out as a factor. I will use #-3xy^2# because I prefer positive leading coefficients:

#-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# is missing from the question, since we find:

#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)#

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