# How do you divide  s+s^2 -8 by x+4?

Oct 17, 2015

$- 2 + \frac{{s}^{2} + s + 2 x}{x + 4}$

Another way of writing it: $- 2 + \frac{{s}^{2} + s}{x + 4} + \frac{2 x}{x + 4}$

#### Explanation:

I do not know how to set up long division on this site so I will do my best without it:

Based on the assumption you mean: $\left({s}^{2} + s - 8\right) \div \left(x + 4\right)$

There is nothing to associate $x$ with $s$ so the only thing we can reasonably divide are the constants of (-8) and 4. We would then be left over with some form of unsolvable fraction which would have to be written 'as is'.

Write $\left({s}^{2} + s - 8\right) \div \left(x + 4\right) \text{ as } \frac{{s}^{2} + s - 8}{x + 4}$

considering just the constants: $\left(- 8\right) \div 4 = - 2$

So the first part of our solution is -2

If the division gives -2 then we have to subtract
$- 2 \times \left(x + 4\right) = - 2 x - 8$ ................ ( 1 )
from the original expression to determine what is left. This in turn will also have to be divided by $\left(x + 4\right)$. This process is like the old remainder system you would have been taught some years back.

so we have in effect a remainder of:

$\frac{\left({s}^{2} + s - 8\right) - \left(- 2\right) \left(x + 4\right)}{x + 4}$

$\frac{\left({s}^{2} + s - 8\right) + 2 \left(x + 4\right)}{x + 4}$ ....... ( 2 )

Adding the remainder to the first part we have:

$\text{part (1) + part(2)} \to - 2 + \frac{\left({s}^{2} + s - 8\right) + 2 \left(x + 4\right)}{x + 4}$

This would give:

$- 2 + \frac{{s}^{2} + s + 2 x}{x + 4}$

or $- 2 + \frac{{s}^{2} + s}{x + 4} + \frac{2 x}{x + 4}$

If my solution is correct then multiplying it by $\left(x + 4\right)$ should give us the original expression. This would take the form:
$\left(x + 4\right) \left[- 2 + \frac{{s}^{2} + s + 2 x}{x + 4}\right]$
$\implies - 2 x - 8 + {s}^{2} + s + 2 x$
which gives us: ${s}^{2} + s - 8$