Factor the expression by grouping. First, the expression needs to be rewritten as $2x^{2}+ax+bx-6$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-1$$ $$ab=2\left(-6\right)=-12$$
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 $-12$.
$$1,-12$$ $$2,-6$$ $$3,-4$$
Calculate the sum for each pair.
$$1-12=-11$$ $$2-6=-4$$ $$3-4=-1$$
The solution is the pair that gives sum $-1$.
$$a=-4$$ $$b=3$$
Rewrite $2x^{2}-x-6$ as $\left(2x^{2}-4x\right)+\left(3x-6\right)$.
$$\left(2x^{2}-4x\right)+\left(3x-6\right)$$
Factor out $2x$ in the first and $3$ in the second group.
$$2x\left(x-2\right)+3\left(x-2\right)$$
Factor out common term $x-2$ by using distributive property.
$$\left(x-2\right)\left(2x+3\right)$$
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 $ax^{2}+bx+c=0$.
$$2x^{2}-x-6=0$$
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 $x=\frac{1±7}{4}$ when $±$ is plus. Add $1$ to $7$.
$$x=\frac{8}{4}$$
Divide $8$ by $4$.
$$x=2$$
Now solve the equation $x=\frac{1±7}{4}$ when $±$ is minus. Subtract $7$ from $1$.
$$x=-\frac{6}{4}$$
Reduce the fraction $\frac{-6}{4}$ to lowest terms by extracting and canceling out $2$.
$$x=-\frac{3}{2}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $2$ for $x_{1}$ and $-\frac{3}{2}$ for $x_{2}$.
