To factor this we shall take it as this as a polynomial of Let us write in decreasing power of x
$2x^2−y^2−3z^2−xy+4yz+5zx$
$=2x^2+x(-y+5z) -(y^2–4yz+3z^2)$
$= 2x^2+x(-y+5z) -(y-z)(y-3z)$ by factoring the the term independent of z.
We let $a = y-z$ and $b= y-3z$ which are factors of the independent term and they are co prime to each other
Now a and b are co-primes so 2 the coefficient of $x^2$ $must go with one of them and we have to chose the same so that the sum (or difference) is co-efficient of x.
We have $a - 2b = -y + 5z$
So we get
$2x^2 +(a-2b) x - ab$
$= 2x^2 + ax - 2bx -ab$+
$= x(2x+a)-b(2x+ a)$
$= (x-b)(2x+a) = (x-y+3z)(2x+y-z)$
