# How do you combine 7/(x^2-x-2)+x/(x^2+4x+3)?

Jul 22, 2016

Here's what I got.

#### Explanation:

Your goal here is to find a common denominator for those two fractions. This will allow you to combine the two numerators and simplify the expression, if possible.

Take a look at the first denominator

${x}^{2} - x - 2$

If you make this equal to zero, you can use the quadratic formula

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

to find its roots. In this case, you have

${x}_{1 , 2} = \frac{- \left(- 1\right) \pm \sqrt{{\left(- 1\right)}^{2} - 4 \cdot 1 \cdot \left(- 2\right)}}{2 \cdot 1}$

${x}_{1 , 2} = \frac{1 \pm \sqrt{9}}{2} \implies \left\{\begin{matrix}{x}_{1} = \frac{1 + 3}{2} = 2 \\ {x}_{2} = \frac{1 - 3}{2} = - 1\end{matrix}\right.$

You can thus rewrite the first denominator as

${x}^{2} - x - 2 = \left(x - 2\right) \left(x + 1\right)$

Do the same for the second denominator

${x}^{2} + 4 x + 3$

to get

${x}_{1 , 2} = \frac{- 4 \pm \sqrt{{4}^{2} - 4 \cdot 1 \cdot 3}}{2 \cdot 1}$

${x}_{1 , 2} = \frac{- 4 \pm 2}{2} \implies \left\{\begin{matrix}{x}_{1} = \frac{- 4 - 2}{2} = - 3 \\ {x}_{2} = \frac{- 4 + 2}{2} = - 1\end{matrix}\right.$

The second denominator can be rewritten as

${x}^{2} + 4 x + 3 = \left(x + 3\right) \left(x + 1\right)$

The expression becomes

$\frac{7}{\left(x - 2\right) \left(x + 1\right)} + \frac{x}{\left(x + 3\right) \left(x + 1\right)}$

The common denominator will be

${\overbrace{\left(x + 1\right)}}^{\textcolor{p u r p \le}{\text{common to both denominators}}} \cdot \left(x - 2\right) \cdot \left(x + 3\right)$

Multiply the first fraction by $1 = \frac{x + 3}{x + 3}$ and the second fraction by $1 = \frac{x - 2}{x - 2}$ to get

$\frac{7}{\left(x - 2\right) \left(x + 1\right)} \cdot \frac{x + 3}{x + 3} + \frac{x}{\left(x + 3\right) \left(x + 1\right)} \cdot \frac{x - 2}{x - 2}$

$\frac{7 \cdot \left(x + 3\right)}{\left(x + 1\right) \left(x - 2\right) \left(x + 3\right)} + \frac{x \cdot \left(x - 2\right)}{\left(x + 1\right) \left(x - 2\right) \left(x + 3\right)}$

Now focus on the numerator

$7 \left(x + 3\right) + x \left(x - 2\right) = 7 x + 21 + {x}^{2} - 2 x$

$= {x}^{2} + 5 x + 21$

The expression can thus be rewritten as

$\frac{{x}^{2} + 5 x + 21}{\left(x + 1\right) \left(x - 2\right) \left(x + 3\right)}$

The numerator cannot be factored in any useful form, i.e. the quadratic

${x}^{2} + 5 x + 21 = 0$

has complex roots, so it's best to leave the answer like this.