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 $ax^{2}+bx+c=0$.
$$n^{2}+n=0$$
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.
$$n=\frac{-1±\sqrt{1^{2}}}{2}$$
Take the square root of $1^{2}$.
$$n=\frac{-1±1}{2}$$
Now solve the equation $n=\frac{-1±1}{2}$ when $±$ is plus. Add $-1$ to $1$.
$$n=\frac{0}{2}$$
Divide $0$ by $2$.
$$n=0$$
Now solve the equation $n=\frac{-1±1}{2}$ when $±$ is minus. Subtract $1$ from $-1$.
$$n=-\frac{2}{2}$$
Divide $-2$ by $2$.
$$n=-1$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $0$ for $x_{1}$ and $-1$ for $x_{2}$.
$$n^{2}+n=n\left(n-\left(-1\right)\right)$$
Simplify all the expressions of the form $p-\left(-q\right)$ to $p+q$.
