how to factorize $x^2+10yz-2xz-2xy-3y^2-3z^2$?

Question

How to factorize $$x^2+10yz-2xz-2xy-3y^2-3z^2$$

It is expanded and we should make them into parts and factorize each part individually.

the last answer is $$(x+y-3z)(x-3y+z)$$

but how to get it ?

Answer 1

Write it as $x^2 - 2 (z+y)x + (10yz-3z^2-3y^2)$. It is known how to factor quadratic polynomials. Here, the roots are $$(z+y) \pm \sqrt{(z+y)^2 - (10yz-3z^2-3y^2)}=(z+y) \pm \sqrt{4y^2-8yz+4z^2}$$ $$=(z+y) \pm 2(y-z) \in \{3y-z,3z-y\}.$$ Therefore the polynomial factors as $(x-(3y-z)) \cdot (x-(3z-y))$.

Answer 2

One way to do it is \begin{align*} x^2+10yz-2xz-2xy-3y^2-3z^2&=(x^2+y^2+z^2-2xz-2xy+2yz)-(4y^2+4z^2-8yz)\\ &=(x-y-z)^2-(2y-2z)^2\\ &=(x+y-3z)(x-3y+z) \end{align*} where going from the second to the third line, we have used $a^2-b^2=(a-b)(a+b)$.