# How do you factor 2n^2+5n-3?

Jun 15, 2015

$2 {n}^{2} + 5 n - 3 = \left(n + 3\right) \left(2 n - 1\right)$

#### Explanation:

Let $f \left(n\right) = 2 {n}^{2} + 5 n - 3$

By the rational roots theorem, if $f \left(n\right) = 0$ has rational roots then they are all of the form $\frac{p}{q}$ in lowest terms, where $p$ is a divisor of $3$ and $q$ is a divisor of $2$.

Moreover, since $2$ only factors as $1 \times 2$ (or $- 1 \times - 2$), one of the two corresponding linear factors must have $q = \pm 1$, so $\frac{p}{q}$ is an integer.

As a result, one of $\pm 1$ or $\pm 3$ must be a root of $f \left(n\right) = 0$...

$f \left(1\right) = 2 + 5 - 3 = 4$
$f \left(- 1\right) = 2 - 5 - 3 = - 6$
$f \left(3\right) = 18 + 15 - 3 = 30$
$f \left(- 3\right) = 18 - 15 - 3 = 0$

So $n = - 3$ is a root and $\left(n + 3\right)$ is a factor.
The other factor must be $\left(2 n - 1\right)$ in order that the coefficient of ${n}^{2}$ is $2$ and the constant term is $- 3$ when these two factors are multiplied.

It's actually quicker to do than to write these words, but we find:

$2 {n}^{2} + 5 n - 3 = \left(n + 3\right) \left(2 n - 1\right)$