By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $20$ and $q$ divides the leading coefficient $1$. One such root is $-10$. Factor the polynomial by dividing it by $x+10$.
$$\left(x+10\right)\left(x^{2}+3x+2\right)$$
Consider $x^{2}+3x+2$. Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx+2$. To find $a$ and $b$, set up a system to be solved.
$$a+b=3$$ $$ab=1\times 2=2$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. The only such pair is the system solution.
$$a=1$$ $$b=2$$
Rewrite $x^{2}+3x+2$ as $\left(x^{2}+x\right)+\left(2x+2\right)$.
$$\left(x^{2}+x\right)+\left(2x+2\right)$$
Factor out $x$ in the first and $2$ in the second group.
$$x\left(x+1\right)+2\left(x+1\right)$$
Factor out common term $x+1$ by using distributive property.
