Factor the expression by grouping. First, the expression needs to be rewritten as $a^{2}+pa+qa-60$. To find $p$ and $q$, set up a system to be solved.
$$p+q=-7$$ $$pq=1\left(-60\right)=-60$$
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 $-60$.
Rewrite $a^{2}-7a-60$ as $\left(a^{2}-12a\right)+\left(5a-60\right)$.
$$\left(a^{2}-12a\right)+\left(5a-60\right)$$
Factor out $a$ in the first and $5$ in the second group.
$$a\left(a-12\right)+5\left(a-12\right)$$
Factor out common term $a-12$ by using distributive property.
$$\left(a-12\right)\left(a+5\right)$$
Steps Using the Quadratic Formula
Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$a^{2}-7a-60=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
