# How do you divide ( x^3+4x^2-7x-6 )/((x + 1)(x + 10) )?

$\left(x - 7\right) + \frac{60 x + 64}{{x}^{2} + 11 x + 10} = \left(x - 7\right) + \frac{4 \left(15 x + 16\right)}{\left(x + 1\right) \left(x + 10\right)}$

#### Explanation:

There are no factors of the numerator that can help us, and so we're left to do this via long division. Let's first expand the denominator:

$\left(x + 1\right) \left(x + 10\right) = {x}^{2} + 11 x + 10$

And now for the long division:

$\textcolor{w h i t e}{\frac{{x}^{2} + 11 x + 10}{\textcolor{b l a c k}{{x}^{2} + 11 x + 10 \text{)}}} \frac{{x}^{3} + 4 {x}^{2} - 7 x - 6}{\textcolor{b l a c k}{\overline{{x}^{3} + 4 {x}^{2} - 7 x - 6}}}}$

${x}^{2}$ goes into ${x}^{3}$ $x$ times:

$\textcolor{w h i t e}{\frac{{x}^{2} + 11 x + 10}{\textcolor{b l a c k}{{x}^{2} + 11 x + 10 \text{)}}} \frac{\textcolor{b l a c k}{x +} 4 {x}^{2} - 7 x - 6}{\textcolor{b l a c k}{\overline{{x}^{3} + 4 {x}^{2} - 7 x - 6}}}}$
$\textcolor{w h i t e}{\frac{{x}^{2} + 11 x + 10}{{x}^{2} + 11 x + 10 \text{)}} \frac{\textcolor{b l a c k}{{x}^{3} + 11 {x}^{2} + 10 x \textcolor{w h i t e}{- 6}}}{\textcolor{b l a c k}{\overline{0 {x}^{3} - 7 {x}^{2} - 17 x - 6}}}}$

${x}^{2}$ goes into $- 7 {x}^{2}$ $- 7$ times:

$\textcolor{w h i t e}{\frac{{x}^{2} + 11 x + 10}{\textcolor{b l a c k}{{x}^{2} + 11 x + 10 \text{)}}} \frac{\textcolor{b l a c k}{x - 7} {x}^{2} - 7 x - 6}{\textcolor{b l a c k}{\overline{{x}^{3} + 4 {x}^{2} - 7 x - 6}}}}$
$\textcolor{w h i t e}{\frac{{x}^{2} + 11 x + 10}{{x}^{2} + 11 x + 10 \text{)}} \frac{\textcolor{b l a c k}{{x}^{3} + 11 {x}^{2} + 10 x \textcolor{w h i t e}{- 6}}}{\textcolor{b l a c k}{\overline{\textcolor{w h i t e}{0 {x}^{3}} - 7 {x}^{2} - 17 x - 6}}}}$
$\textcolor{w h i t e}{\frac{{x}^{2} + 11 x + 10}{{x}^{2} + 11 x + 10 \text{)}} \frac{\textcolor{b l a c k}{\textcolor{w h i t e}{{x}^{3}} - 7 {x}^{2} - 77 x - 70}}{\textcolor{b l a c k}{\overline{\textcolor{w h i t e}{0 {x}^{3}} + 0 {x}^{2} + 60 x + 64}}}}$

This gives us:

$\frac{{x}^{3} + 4 {x}^{2} - 7 x - 6}{{x}^{2} + 11 x + 10} = \left(x - 7\right) + \frac{60 x + 64}{{x}^{2} + 11 x + 10} = \left(x - 7\right) + \frac{4 \left(15 x + 16\right)}{\left(x + 1\right) \left(x + 10\right)}$