Consider $a^{2}-3a-10$. Factor the expression by grouping. First, the expression needs to be rewritten as $a^{2}+pa+qa-10$. To find $p$ and $q$, set up a system to be solved.
$$p+q=-3$$ $$pq=1\left(-10\right)=-10$$
Since $pq$ is negative, $p$ and $q$ have the opposite signs. Since $p+q$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-10$.
$$1,-10$$ $$2,-5$$
Calculate the sum for each pair.
$$1-10=-9$$ $$2-5=-3$$
The solution is the pair that gives sum $-3$.
$$p=-5$$ $$q=2$$
Rewrite $a^{2}-3a-10$ as $\left(a^{2}-5a\right)+\left(2a-10\right)$.
$$\left(a^{2}-5a\right)+\left(2a-10\right)$$
Factor out $a$ in the first and $2$ in the second group.
$$a\left(a-5\right)+2\left(a-5\right)$$
Factor out common term $a-5$ by using distributive property.
