Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx-20$. To find $a$ and $b$, set up a system to be solved.
$$a+b=1$$ $$ab=1\left(-20\right)=-20$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-20$.
$$-1,20$$ $$-2,10$$ $$-4,5$$
Calculate the sum for each pair.
$$-1+20=19$$ $$-2+10=8$$ $$-4+5=1$$
The solution is the pair that gives sum $1$.
$$a=-4$$ $$b=5$$
Rewrite $x^{2}+x-20$ as $\left(x^{2}-4x\right)+\left(5x-20\right)$.
$$\left(x^{2}-4x\right)+\left(5x-20\right)$$
Factor out $x$ in the first and $5$ in the second group.
$$x\left(x-4\right)+5\left(x-4\right)$$
Factor out common term $x-4$ by using distributive property.
