# Math Help - Complicated factorisation technique

1. ## Complicated factorisation technique

Hi,

Does anyone know a reasonably standard technique to factorise something like,

$x^4 + y^4 + 9x^2 + y^2 - 6x^3 + 2x^2y^2 - 6xy^2 - 4x = 0$

into:

$(x^2-4x+y^2) (x^2-2x+y^2+1) = 0$

i.e. when not given any other information (like one of the factors), how to recognise it can be factored, and do it etc

2. notice that the degree of y is 2 or 4, we can let $t=y^2,\text{ then }t^2=y^4$.
consider the left hand side as a quadratical polynomial of t , we can rewrite it as
$t^2+(2x^2-6x+1)t+(x^4-6x^3+9x^2-4x)=0$.
then factorise $(x^4-6x^3+9x^2-4x)$:
since the root of $(x^4-6x^3+9x^2-4x)$ is 0 , 1(multiple root) , 4, we have $(x^4-6x^3+9x^2-4x)=x(x-4)(x-1)^2$.
then we find the fact that $(2x^2-6x+1)$ is exactly the sum of $x(x-4)\text{ and }(x-1)^2$.
therefore the factorization follows immediately.

I hope I've made myself clear enough, and can help you out. Good luck!