Factor the expression by grouping. First, the expression needs to be rewritten as $2a^{2}+pa+qa-15$. To find $p$ and $q$, set up a system to be solved.
$$p+q=-1$$ $$pq=2\left(-15\right)=-30$$
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 $-30$.
$$1,-30$$ $$2,-15$$ $$3,-10$$ $$5,-6$$
Calculate the sum for each pair.
$$1-30=-29$$ $$2-15=-13$$ $$3-10=-7$$ $$5-6=-1$$
The solution is the pair that gives sum $-1$.
$$p=-6$$ $$q=5$$
Rewrite $2a^{2}-a-15$ as $\left(2a^{2}-6a\right)+\left(5a-15\right)$.
$$\left(2a^{2}-6a\right)+\left(5a-15\right)$$
Factor out $2a$ in the first and $5$ in the second group.
$$2a\left(a-3\right)+5\left(a-3\right)$$
Factor out common term $a-3$ by using distributive property.
