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