# How do you simplify (x+3)/(x^2-2x-15)?

Sep 23, 2015

The expression simplifies into $\frac{1}{x - 5}$

#### Explanation:

You need to find the roots of the denominator: you can either use the classic $\frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$ formula, our you can use this particular one, which is good only if the coefficient of ${x}^{2}$ is $1$: in this case, the parabola is of the form ${x}^{2} + a x + b$, and you know that $a$ equals the sum of the roots, but with the opposite sign, while $b$ equals exactly the product of the roots.

So, in this case, we are looking for two roots, which sum up to $2$ and which give $- 15$ when multiplied. This numbers are obviously $5$ and $- 3$.

Now that you know the roots, you can write down your parabola in a different way: if the roots of $a {x}^{2} + b x + c$ are ${x}_{0}$ and ${x}_{1}$, then the following holds:

$a {x}^{2} + b x + c = \left(x - {x}_{0}\right) \left(x - {x}_{1}\right)$.

In your case, since ${x}_{0} = 5$ and ${x}_{1} = - 3$, you have that

${x}^{2} - 2 x - 15 = \left(x - 5\right) \left(x + 3\right)$.

You should see that you can cancel the term $\left(x + 3\right)$: your fraction can be written as

$\frac{x + 3}{{x}^{2} - 2 x - 15} = \setminus \frac{\cancel{x + 3}}{\left(x - 5\right) \setminus \cancel{\left(x + 3\right)}} = \frac{1}{\left(x - 5\right)}$