Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$2x^{2}-3x-2$$
Factor the expression by grouping. First, the expression needs to be rewritten as $2x^{2}+ax+bx-2$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-3$$ $$ab=2\left(-2\right)=-4$$
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 $-4$.
$$1,-4$$ $$2,-2$$
Calculate the sum for each pair.
$$1-4=-3$$ $$2-2=0$$
The solution is the pair that gives sum $-3$.
$$a=-4$$ $$b=1$$
Rewrite $2x^{2}-3x-2$ as $\left(2x^{2}-4x\right)+\left(x-2\right)$.
$$\left(2x^{2}-4x\right)+\left(x-2\right)$$
Factor out $2x$ in $2x^{2}-4x$.
$$2x\left(x-2\right)+x-2$$
Factor out common term $x-2$ by using distributive property.
$$\left(x-2\right)\left(2x+1\right)$$
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 $ax^{2}+bx+c=0$.
$$2x^{2}-3x-2=0$$
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 $x=\frac{3±5}{4}$ when $±$ is plus. Add $3$ to $5$.
$$x=\frac{8}{4}$$
Divide $8$ by $4$.
$$x=2$$
Now solve the equation $x=\frac{3±5}{4}$ when $±$ is minus. Subtract $5$ from $3$.
$$x=-\frac{2}{4}$$
Reduce the fraction $\frac{-2}{4}$ to lowest terms by extracting and canceling out $2$.
$$x=-\frac{1}{2}$$
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 $-\frac{1}{2}$ for $x_{2}$.
