# Can you combine \frac { 4x + 2} { 3x ^ { 2} - 6x - 24} + \frac { 9} { 3x ^ { 2} - 15x + 12} into a single term?

Aug 17, 2017

$\frac{4 {x}^{2} + 7 x + 16}{3 {x}^{3} - 9 {x}^{2} - 18 x + 24}$

#### Explanation:

First things first, we need to create a common denominator. Let's factor everything first and see if anything divides out

$\frac{2 \left(2 x - 1\right)}{3 \left({x}^{2} - 2 x - 8\right)} + \frac{3 \times 3}{3 \left({x}^{2} - 5 x + 4\right)}$

$\frac{2 \left(2 x - 1\right)}{\textcolor{g r e e n}{3} \textcolor{g r e e n}{\left(x - 4\right)} \left(x + 2\right)} + \frac{3 \times 3}{\textcolor{g r e e n}{3} \left(x - 1\right) \textcolor{g r e e n}{\left(x - 4\right)}}$

Nothing divides out, but it looks like the denominators are mostly the same, except that the right side doesn't have $\left(x + 2\right)$, and the left side is missing $\left(x - 1\right)$. Let's multiply both sides by the term they're missing (remember, multiply both the numerator and denominator)

$\frac{x - 1}{x - 1} \times \frac{4 x + 2}{3 \left(x - 4\right) \left(x + 2\right)} + \frac{9}{3 \left(x - 1\right) \left(x - 4\right) \left(x + 2\right)} \times \frac{x + 2}{x + 2}$

$\frac{4 {x}^{2} - 4 x + 2 x - 2}{3 \left(x - 4\right) \left(x + 2\right) \left(x - 1\right)} + \frac{9 x + 18}{3 \left(x - 1\right) \left(x - 4\right) \left(x + 2\right)}$

Now that the denominators match, let's combine the numerators:

$\frac{4 {x}^{2} - 2 x - 2 + 9 x + 18}{3 \left(x - 4\right) \left(x + 2\right) \left(x - 1\right)}$

That denominator could use some work...

$3 \left(x - 4\right) \left(x + 2\right) \left(x - 1\right)$

$\left(3 x - 12\right) \left(x + 2\right) \left(x - 1\right)$

$\left(3 {x}^{2} + 6 x - 12 x - 24\right) \left(x - 1\right)$

$\left(3 {x}^{2} - 6 x - 24\right) \left(x - 1\right)$

$3 {x}^{3} - 3 {x}^{2} - 6 {x}^{2} + 6 x - 24 x + 24$

$\textcolor{g r e e n}{3 {x}^{3} - 9 {x}^{2} - 18 x + 24}$

Now we have

$\frac{4 {x}^{2} + 7 x + 16}{\textcolor{g r e e n}{3 {x}^{3} - 9 {x}^{2} - 18 x + 24}}$

I hope that helps!