By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $-120$ and $q$ divides the leading coefficient $1$. One such root is $12$. Factor the polynomial by dividing it by $x-12$.
$$\left(x-12\right)\left(x^{2}-11x+10\right)$$
Consider $x^{2}-11x+10$. Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx+10$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-11$$ $$ab=1\times 10=10$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $10$.
$$-1,-10$$ $$-2,-5$$
Calculate the sum for each pair.
$$-1-10=-11$$ $$-2-5=-7$$
The solution is the pair that gives sum $-11$.
$$a=-10$$ $$b=-1$$
Rewrite $x^{2}-11x+10$ as $\left(x^{2}-10x\right)+\left(-x+10\right)$.
$$\left(x^{2}-10x\right)+\left(-x+10\right)$$
Factor out $x$ in the first and $-1$ in the second group.
$$x\left(x-10\right)-\left(x-10\right)$$
Factor out common term $x-10$ by using distributive property.
