Factor the expression by grouping. First, the expression needs to be rewritten as $P^{2}+aP+bP+91$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-20$$ $$ab=1\times 91=91$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $91$.
$$-1,-91$$ $$-7,-13$$
Calculate the sum for each pair.
$$-1-91=-92$$ $$-7-13=-20$$
The solution is the pair that gives sum $-20$.
$$a=-13$$ $$b=-7$$
Rewrite $P^{2}-20P+91$ as $\left(P^{2}-13P\right)+\left(-7P+91\right)$.
$$\left(P^{2}-13P\right)+\left(-7P+91\right)$$
Factor out $P$ in the first and $-7$ in the second group.
$$P\left(P-13\right)-7\left(P-13\right)$$
Factor out common term $P-13$ by using distributive property.
$$\left(P-13\right)\left(P-7\right)$$
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 $ax^{2}+bx+c=0$.
$$P^{2}-20P+91=0$$
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.
