Factor the expression by grouping. First, the expression needs to be rewritten as $a^{2}+pa+qa+15$. To find $p$ and $q$, set up a system to be solved.
$$p+q=-8$$ $$pq=1\times 15=15$$
Since $pq$ is positive, $p$ and $q$ have the same sign. Since $p+q$ is negative, $p$ and $q$ are both negative. List all such integer pairs that give product $15$.
$$-1,-15$$ $$-3,-5$$
Calculate the sum for each pair.
$$-1-15=-16$$ $$-3-5=-8$$
The solution is the pair that gives sum $-8$.
$$p=-5$$ $$q=-3$$
Rewrite $a^{2}-8a+15$ as $\left(a^{2}-5a\right)+\left(-3a+15\right)$.
$$\left(a^{2}-5a\right)+\left(-3a+15\right)$$
Factor out $a$ in the first and $-3$ in the second group.
$$a\left(a-5\right)-3\left(a-5\right)$$
Factor out common term $a-5$ by using distributive property.
$$\left(a-5\right)\left(a-3\right)$$
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 $ax^{2}+bx+c=0$.
$$a^{2}-8a+15=0$$
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.
