Factor the expression by grouping. First, the expression needs to be rewritten as $3P^{2}+aP+bP-5$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-2$$ $$ab=3\left(-5\right)=-15$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-15$.
$$1,-15$$ $$3,-5$$
Calculate the sum for each pair.
$$1-15=-14$$ $$3-5=-2$$
The solution is the pair that gives sum $-2$.
$$a=-5$$ $$b=3$$
Rewrite $3P^{2}-2P-5$ as $\left(3P^{2}-5P\right)+\left(3P-5\right)$.
$$\left(3P^{2}-5P\right)+\left(3P-5\right)$$
Factor out $P$ in $3P^{2}-5P$.
$$P\left(3P-5\right)+3P-5$$
Factor out common term $3P-5$ by using distributive property.
$$\left(3P-5\right)\left(P+1\right)$$
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 $ax^{2}+bx+c=0$.
$$3P^{2}-2P-5=0$$
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.
Now solve the equation $P=\frac{2±8}{6}$ when $±$ is plus. Add $2$ to $8$.
$$P=\frac{10}{6}$$
Reduce the fraction $\frac{10}{6}$ to lowest terms by extracting and canceling out $2$.
$$P=\frac{5}{3}$$
Now solve the equation $P=\frac{2±8}{6}$ when $±$ is minus. Subtract $8$ from $2$.
$$P=-\frac{6}{6}$$
Divide $-6$ by $6$.
$$P=-1$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{5}{3}$ for $x_{1}$ and $-1$ for $x_{2}$.
