# How do you factor 2x^2-3x-2?

$2 {x}^{2} - 3 x - 2 = \left(2 x + 1\right) \left(x - 2\right)$
If a polynomial with integer coefficients has a rational root of the form $\frac{p}{q}$ in lowest terms then p must be a divisor of the constant term and q a divisor of the coefficient of the highest order term.
In your case any possible rational roots of $2 {x}^{2} - 3 x - 2 = 0$ must be $\pm 2$, $\pm 1$ or $\pm \frac{1}{2}$. By trying these values you will find that $x = 2$ and $x = - \frac{1}{2}$ are zeros, so $\left(2 x + 1\right)$ and $\left(x - 2\right)$ are factors.