How do you factor completely #12x^3y + 6x^2y^2 - 9xy^3#?

2 Answers
Feb 19, 2017

Answer:

#12x^3y+6x^2y^2-9xy^3 = 3/4xy(4x+(1-sqrt(13))y)(4x+(1+sqrt(13))y)#

Explanation:

Given:

#12x^3y+6x^2y^2-9xy^3#

Note that all of the terms are of degree #4# and all are divisible by #3xy#:

#12x^3y+6x^2y^2-9xy^3 = 3xy(4x^2+2xy-3y^2)#

We can factor the remaining quadratic by completing the square as follows:

#4x^2+2xy-3y^2 = 1/4(16x^2+8xy-12y^2)#

#color(white)(4x^2+2xy-3y^2) = 1/4((4x)^2+2(4x)(y)+y^2-13y^2)#

#color(white)(4x^2+2xy-3y^2) = 1/4((4x+y)^2-(sqrt(13)y)^2)#

#color(white)(4x^2+2xy-3y^2) = 1/4((4x+y)-sqrt(13)y)((4x+y)+sqrt(13)y)#

#color(white)(4x^2+2xy-3y^2) = 1/4(4x+(1-sqrt(13))y)(4x+(1+sqrt(13))y)#

Putting it all together:

#12x^3y+6x^2y^2-9xy^3 = 3/4xy(4x+(1-sqrt(13))y)(4x+(1+sqrt(13))y)#

Feb 19, 2017

Answer:

#3^2x y(y-x/3(1-sqrt(13)))(y-x/3(1+sqrt(13)))#

Explanation:

#f(x,y)=12 x^3 y + 6 x^2 y^2 - 9 x y^3# is a homogeneous function

Making #y=lambda x# and substituting

#f(x,lambda)=3 x^4 lambda(4 + (2 - 3 lambda)lambda)#

Solving # lambda(4 + (2 - 3 lambda)lambda)=0# we have

#lambda = {0,1/3(1-sqrt(13)),1/3(1+sqrt(13))}# or

#f(x,lambda)=3^2x^4lambda(lambda-1/3(1-sqrt(13)))(lambda-1/3(1+sqrt(13)))# so

#f(x,y)=3^2x y(y-x/3(1-sqrt(13)))(y-x/3(1+sqrt(13)))#