# How do you simplify [(x^3 - 1)/(x + 4)][(2x - 7)/(x^2 + 3x + 1)]?

Nov 20, 2017

$\frac{2 {x}^{4} - 7 {x}^{3} - 2 x + 7}{{x}^{3} + 7 {x}^{2} + 13 x + 4}$

#### Explanation:

We multiply the numerators with numerators and denominators with denominators.

First, let's look at the numerators.

We multiply the numerators like this:
$\left({x}^{3} - 1\right) \left(2 x - 7\right)$ and now we need to simplify using the rainbow, FOIL, box method, or whatever way you want to do it.

${x}^{3} \cdot 2 x = 2 {x}^{4}$

${x}^{3} \cdot - 7 = - 7 {x}^{3}$

$- 1 \cdot 2 x = - 2 x$

$- 1 \cdot - 7 = 7$

So if we put them all together, we will get $2 {x}^{4} - 7 {x}^{3} - 2 x + 7$.

Now multiply the denominators:
$\left(x + 4\right) \left({x}^{2} + 3 x + 1\right)$

$x \cdot {x}^{2} = {x}^{3}$

$x \cdot 3 x = 3 {x}^{2}$

$x \cdot 1 = x$

$4 \cdot {x}^{2} = 4 {x}^{2}$

$4 \cdot 3 x = 12 x$

$4 \cdot 1 = 4$

Again, let's put them all together, and we get ${x}^{3} + 3 {x}^{2} + x + 4 {x}^{2} + 12 x + 4$

We still have to combine the "like terms":
${x}^{3} + 7 {x}^{2} + 13 x + 4$

So our final answer is: $\frac{2 {x}^{4} - 7 {x}^{3} - 2 x + 7}{{x}^{3} + 7 {x}^{2} + 13 x + 4}$