# How do you divide ( -3x^3+12x^2-7x-6 )/((x + 1)(4x+12) )?

Dec 28, 2017

$- 3 \frac{x}{4} + 6 - \frac{47 x + 39}{2 \left({x}^{2} + 4 x + 3\right)}$

#### Explanation:

Expand $\left(x + 1\right) \left(4 x + 12\right)$ into $4 {x}^{2} + 16 x + 12$

Now you can long divide $\frac{- 3 {x}^{3} + 12 {x}^{2} - 7 x - 6}{4 {x}^{2} + 16 x + 12}$.

First, divide the leading coefficients, that is $- 3 {x}^{3}$ and $4 {x}^{2}$ into $- 3 \frac{x}{4}$. Now multiply $\left(4 {x}^{2} + 16 x + 12\right)$ by $- 3 \frac{x}{4}$ to get $- 3 {x}^{3} - 12 {x}^{2} - 9 x$. Subtract $- 3 {x}^{3} - 12 {x}^{2} - 9 x$ from $- 3 {x}^{3} + 12 {x}^{2} - 7 x - 6$ to get the remainder, that is $24 {x}^{2} + 2 x - 6$.

Therefore,
$\frac{- 3 {x}^{3} + 12 {x}^{2} - 7 x - 6}{4 {x}^{2} + 16 x + 12} = - 3 \frac{x}{4} + \frac{24 {x}^{2} + 2 x - 6}{4 {x}^{2} + 16 x + 12}$.

Now, divide $\frac{24 {x}^{2} + 2 x - 6}{4 {x}^{2} + 16 x + 12}$ with the same steps above. First, divide the leading coefficients to get $6$. Multiply $4 {x}^{2} + 16 x + 12$ by $6$ to get $24 {x}^{2} + 96 x + 72$. Subtract $24 {x}^{2} + 96 x + 72$ from $24 {x}^{2} + 2 x - 6$ to get the second remainder, $- 94 x - 78$.

Therefore,

$\frac{24 {x}^{2} + 2 x - 6}{4 {x}^{2} + 16 x + 12} = 6 + \frac{- 94 x - 78}{4 {x}^{2} + 16 x + 12}$.

$- 3 \frac{x}{4} + 6 + \frac{- 94 x - 78}{4 {x}^{2} + 16 x + 12}$
$- 3 \frac{x}{4} + 6 - \frac{47 x + 39}{2 \left({x}^{2} + 4 x + 3\right)}$