Factor the expression by grouping. First, the expression needs to be rewritten as $-3P^{2}+aP+bP+7$. To find $a$ and $b$, set up a system to be solved.
$$a+b=4$$ $$ab=-3\times 7=-21$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-21$.
$$-1,21$$ $$-3,7$$
Calculate the sum for each pair.
$$-1+21=20$$ $$-3+7=4$$
The solution is the pair that gives sum $4$.
$$a=7$$ $$b=-3$$
Rewrite $-3P^{2}+4P+7$ as $\left(-3P^{2}+7P\right)+\left(-3P+7\right)$.
$$\left(-3P^{2}+7P\right)+\left(-3P+7\right)$$
Factor out $-P$ in the first and $-1$ in the second group.
$$-P\left(3P-7\right)-\left(3P-7\right)$$
Factor out common term $3P-7$ by using distributive property.
$$\left(3P-7\right)\left(-P-1\right)$$
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 $ax^{2}+bx+c=0$.
$$-3P^{2}+4P+7=0$$
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 $P=\frac{-4±10}{-6}$ when $±$ is plus. Add $-4$ to $10$.
$$P=\frac{6}{-6}$$
Divide $6$ by $-6$.
$$P=-1$$
Now solve the equation $P=\frac{-4±10}{-6}$ when $±$ is minus. Subtract $10$ from $-4$.
$$P=-\frac{14}{-6}$$
Reduce the fraction $\frac{-14}{-6}$ to lowest terms by extracting and canceling out $2$.
$$P=\frac{7}{3}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $-1$ for $x_{1}$ and $\frac{7}{3}$ for $x_{2}$.
