Factor the expression by grouping. First, the expression needs to be rewritten as $3x^{2}+ax+bx-14$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-1$$ $$ab=3\left(-14\right)=-42$$
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 $-42$.
$$1,-42$$ $$2,-21$$ $$3,-14$$ $$6,-7$$
Calculate the sum for each pair.
$$1-42=-41$$ $$2-21=-19$$ $$3-14=-11$$ $$6-7=-1$$
The solution is the pair that gives sum $-1$.
$$a=-7$$ $$b=6$$
Rewrite $3x^{2}-x-14$ as $\left(3x^{2}-7x\right)+\left(6x-14\right)$.
$$\left(3x^{2}-7x\right)+\left(6x-14\right)$$
Factor out $x$ in the first and $2$ in the second group.
$$x\left(3x-7\right)+2\left(3x-7\right)$$
Factor out common term $3x-7$ by using distributive property.
$$\left(3x-7\right)\left(x+2\right)$$
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 $ax^{2}+bx+c=0$.
$$3x^{2}-x-14=0$$
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 $x=\frac{1±13}{6}$ when $±$ is plus. Add $1$ to $13$.
$$x=\frac{14}{6}$$
Reduce the fraction $\frac{14}{6}$ to lowest terms by extracting and canceling out $2$.
$$x=\frac{7}{3}$$
Now solve the equation $x=\frac{1±13}{6}$ when $±$ is minus. Subtract $13$ from $1$.
$$x=-\frac{12}{6}$$
Divide $-12$ by $6$.
$$x=-2$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{7}{3}$ for $x_{1}$ and $-2$ for $x_{2}$.
