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