How do you factor 2x^2-x-15 ?

Aug 8, 2015

$\left(2 x + 5\right) \left(x - 3\right)$

Lol I hope you aren't trying to find your homework answers online to copy!

Explanation:

Firstly, consider the first term. Since it's $2 {x}^{2}$, there are only two linear factors that it can be 'broken down' into: $x$ and $2 x$. Thus we can re-write the expression $2 {x}^{2} - x - 15$ as $\left(2 x + a\right) \left(x + b\right)$ where $a$ and $b$ are real numbers to be determined

Thus
$2 {x}^{2} - x - 15 = \left(2 x + a\right) \left(x + b\right)$
$2 {x}^{2} - x - 15 = 2 {x}^{2} + \left(a + 2 b\right) x + a b$

Consider the last term. In order to get a negative number, we need one positive and one negative number. Thus, if $a$ is positive, $b$ is negative and vice versa.

Now lets think about the factors that make up $15$.
$15 = 1 \cdot 15 = 3 \cdot 5$

Now guess-and-check (not much to do) which combination ($1$ and $15$ or $3$ and $5$) will give $- 1$ (coefficient of center term), that is, which $a$ and $b$ will fit the expression $a + 2 b = - 1$. Remember that if $a$ is positive, $b$ is negative and so on. With some luck you should get:

$5 + 2 \left(- 3\right) = - 1$, $a = 5$, $b = - 3$

Thus,
$2 {x}^{2} - x - 15 = \left(2 x + a\right) \left(x + b\right) = \left(2 x + 5\right) \left(x - 3\right)$