Question

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

Answer 1

`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)`

Answer 2

`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)))`