Factor the expression by grouping. First, the expression needs to be rewritten as $-3p^{2}+ap+bp-4$. To find $a$ and $b$, set up a system to be solved.
$$a+b=8$$ $$ab=-3\left(-4\right)=12$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $12$.
$$1,12$$ $$2,6$$ $$3,4$$
Calculate the sum for each pair.
$$1+12=13$$ $$2+6=8$$ $$3+4=7$$
The solution is the pair that gives sum $8$.
$$a=6$$ $$b=2$$
Rewrite $-3p^{2}+8p-4$ as $\left(-3p^{2}+6p\right)+\left(2p-4\right)$.
$$\left(-3p^{2}+6p\right)+\left(2p-4\right)$$
Factor out $3p$ in the first and $-2$ in the second group.
$$3p\left(-p+2\right)-2\left(-p+2\right)$$
Factor out common term $-p+2$ by using distributive property.
$$\left(-p+2\right)\left(3p-2\right)$$
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 $ax^{2}+bx+c=0$.
$$-3p^{2}+8p-4=0$$
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 $p=\frac{-8±4}{-6}$ when $±$ is plus. Add $-8$ to $4$.
$$p=-\frac{4}{-6}$$
Reduce the fraction $\frac{-4}{-6}$ to lowest terms by extracting and canceling out $2$.
$$p=\frac{2}{3}$$
Now solve the equation $p=\frac{-8±4}{-6}$ when $±$ is minus. Subtract $4$ from $-8$.
$$p=-\frac{12}{-6}$$
Divide $-12$ by $-6$.
$$p=2$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{2}{3}$ for $x_{1}$ and $2$ for $x_{2}$.
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form $x^2+Bx+C=0$.
$$x ^ 2 -\frac{8}{3}x +\frac{4}{3} = 0$$
Let $r$ and $s$ be the factors for the quadratic equation such that $x^2+Bx+C=(x−r)(x−s)$ where sum of factors $(r+s)=−B$ and the product of factors $rs = C$
$$r + s = \frac{8}{3} $$ $$ rs = \frac{4}{3}$$
Two numbers $r$ and $s$ sum up to $\frac{8}{3}$ exactly when the average of the two numbers is $\frac{1}{2}*\frac{8}{3} = \frac{4}{3}$. You can also see that the midpoint of $r$ and $s$ corresponds to the axis of symmetry of the parabola represented by the quadratic equation $y=x^2+Bx+C$. The values of $r$ and $s$ are equidistant from the center by an unknown quantity $u$. Express $r$ and $s$ with respect to variable $u$.
$$r = \frac{4}{3} - u$$ $$s = \frac{4}{3} + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = \frac{4}{3}$
