Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$x^{2}+4x-2=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
$$x=\frac{-4±\sqrt{4^{2}-4\left(-2\right)}}{2}$$
Square $4$.
$$x=\frac{-4±\sqrt{16-4\left(-2\right)}}{2}$$
Multiply $-4$ times $-2$.
$$x=\frac{-4±\sqrt{16+8}}{2}$$
Add $16$ to $8$.
$$x=\frac{-4±\sqrt{24}}{2}$$
Take the square root of $24$.
$$x=\frac{-4±2\sqrt{6}}{2}$$
Now solve the equation $x=\frac{-4±2\sqrt{6}}{2}$ when $±$ is plus. Add $-4$ to $2\sqrt{6}$.
$$x=\frac{2\sqrt{6}-4}{2}$$
Divide $-4+2\sqrt{6}$ by $2$.
$$x=\sqrt{6}-2$$
Now solve the equation $x=\frac{-4±2\sqrt{6}}{2}$ when $±$ is minus. Subtract $2\sqrt{6}$ from $-4$.
$$x=\frac{-2\sqrt{6}-4}{2}$$
Divide $-4-2\sqrt{6}$ by $2$.
$$x=-\sqrt{6}-2$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $-2+\sqrt{6}$ for $x_{1}$ and $-2-\sqrt{6}$ for $x_{2}$.
