Factoring $x^2-y^2-z^2+2yz+x+y-z$

Question

This is a factorization problem on polynomials. I can't find a way to solve it, neither can a math program called MathWay. Help me, please.

$$x^2-y^2-z^2+2yz+x+y-z$$

My book says that the answer is $(x+y-z)(x-y+z+1)$.

Answer 1

$$\begin{align}x^2-y^2-z^2+2yz+x+y-z&=x^2-(y-z)^2+x+y-z\\&=(x+(y-z))(x-(y-z))+x+y-z\\&=(x+y-z)(x-y+z+1)\end{align}$$

Answer 2

You can consider $\,x^2+x-y^2-z^2+2yz+y-z\,$ as a quadratic in $\,x\,$, and find its roots using the quadratic formula, the discriminant being:

$$ \begin{align} \Delta_x = 1 - 4(-y^2-z^2+2yz+y-z) &= 4y^2 - 4(2z+1)y +4z^2+4z+1 \\ &= \left(2y\right)^2- 2 \cdot 2y(2z+1)+\left(2z+1\right)^2 \\ &= \left(2y-2z-1\right)^2 \end{align} $$

Therefore the roots are $\displaystyle x_{1,2} = \frac{-1 \pm \left(2y-2z-1\right)}{2}=\begin{cases}y-z-1 \\ -y+z\end{cases}\;$ and the original polynomial factors as $\,(x-x_1)(x-x_2)\,$.