Factor the expression by grouping. First, the expression needs to be rewritten as $3x^{2}+ax+bx-15$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-4$$ $$ab=3\left(-15\right)=-45$$
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 $-45$.
$$1,-45$$ $$3,-15$$ $$5,-9$$
Calculate the sum for each pair.
$$1-45=-44$$ $$3-15=-12$$ $$5-9=-4$$
The solution is the pair that gives sum $-4$.
$$a=-9$$ $$b=5$$
Rewrite $3x^{2}-4x-15$ as $\left(3x^{2}-9x\right)+\left(5x-15\right)$.
$$\left(3x^{2}-9x\right)+\left(5x-15\right)$$
Factor out $3x$ in the first and $5$ in the second group.
$$3x\left(x-3\right)+5\left(x-3\right)$$
Factor out common term $x-3$ by using distributive property.
$$\left(x-3\right)\left(3x+5\right)$$
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 $ax^{2}+bx+c=0$.
$$3x^{2}-4x-15=0$$
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 $x=\frac{4±14}{6}$ when $±$ is plus. Add $4$ to $14$.
$$x=\frac{18}{6}$$
Divide $18$ by $6$.
$$x=3$$
Now solve the equation $x=\frac{4±14}{6}$ when $±$ is minus. Subtract $14$ from $4$.
$$x=-\frac{10}{6}$$
Reduce the fraction $\frac{-10}{6}$ to lowest terms by extracting and canceling out $2$.
$$x=-\frac{5}{3}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $3$ for $x_{1}$ and $-\frac{5}{3}$ for $x_{2}$.
