Factor the expression by grouping. First, the expression needs to be rewritten as $15m^{2}+am+bm-12$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-8$$ $$ab=15\left(-12\right)=-180$$
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 $-180$.
Rewrite $15m^{2}-8m-12$ as $\left(15m^{2}-18m\right)+\left(10m-12\right)$.
$$\left(15m^{2}-18m\right)+\left(10m-12\right)$$
Factor out $3m$ in the first and $2$ in the second group.
$$3m\left(5m-6\right)+2\left(5m-6\right)$$
Factor out common term $5m-6$ by using distributive property.
$$\left(5m-6\right)\left(3m+2\right)$$
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 $ax^{2}+bx+c=0$.
$$15m^{2}-8m-12=0$$
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 $m=\frac{8±28}{30}$ when $±$ is plus. Add $8$ to $28$.
$$m=\frac{36}{30}$$
Reduce the fraction $\frac{36}{30}$ to lowest terms by extracting and canceling out $6$.
$$m=\frac{6}{5}$$
Now solve the equation $m=\frac{8±28}{30}$ when $±$ is minus. Subtract $28$ from $8$.
$$m=-\frac{20}{30}$$
Reduce the fraction $\frac{-20}{30}$ to lowest terms by extracting and canceling out $10$.
$$m=-\frac{2}{3}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{6}{5}$ for $x_{1}$ and $-\frac{2}{3}$ for $x_{2}$.
Multiply $\frac{5m-6}{5}$ times $\frac{3m+2}{3}$ by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
