# How do you divide (x^3+3x^2-2x+5) div(x+1) and identify any restrictions on the variable?

##### 2 Answers
Oct 25, 2017

Long division

${x}^{2} + 2 x - 4 + \frac{9}{x + 1} , x \ne - 1$

#### Explanation:

(I am using the square root function to format it as close as possible to a long division sign, but it is not a square root.)

We cannot divide by $0$, so the divisor must not equal $0$. Thus, our variable is restricted to:

$x + 1 \ne 0$
$x \ne - 1$

The first step to long division is to fill in any terms of $x$ missing in the dividend, but in this case ${x}^{3} + 3 {x}^{2} - 2 x + 5$ is not missing any terms of $x$.

Next, you divide the first term in the dividend by the first term in the divisor, in this case $x + 1$.

${x}^{3} / x = {x}^{2}$

$\textcolor{w h i t e}{x \otimes \otimes o} \textcolor{red}{{x}^{2}}$
$x + 1 \sqrt{{x}^{3} + 3 {x}^{2} - 2 x + 5}$

The quotient of this division is multiplied by all terms in divisor, and this product is then subtracted from original dividend. Then drop down the rest of the original function.

$\textcolor{w h i t e}{x \otimes \otimes o} {x}^{2}$
$x + 1 \sqrt{{x}^{3} + 3 {x}^{2} - 2 x + 5}$
$\textcolor{w h i t e}{x o i} \textcolor{red}{- \underline{\left({x}^{3} + {x}^{2}\right)}} \textcolor{w h i t e}{x o} \textcolor{red}{\downarrow} \textcolor{w h i t e}{x o} \textcolor{red}{\downarrow}$
$\textcolor{w h i t e}{x \otimes \otimes \otimes i} \textcolor{red}{2 {x}^{2} - 2 x + 5}$

The first term of the new function $\left(2 {x}^{2}\right)$ is divided by first term of dividend, similar to the first division. The quotient is again multiplied by all terms of dividend and subtracted from the new function. The remaining terms are again dropped down.

$\textcolor{w h i t e}{x \otimes \otimes o} {x}^{2} \textcolor{red}{+ 2 x}$
$x + 1 \sqrt{{x}^{3} + 3 {x}^{2} - 2 x + 5}$
$\textcolor{w h i t e}{x o i} - \underline{\left({x}^{3} + {x}^{2}\right)} \textcolor{w h i t e}{x o} \downarrow \textcolor{w h i t e}{x o} \downarrow$
$\textcolor{w h i t e}{x \otimes \otimes \otimes i} 2 {x}^{2} - 2 x + 5$
$\textcolor{w h i t e}{x \otimes o i i} \textcolor{red}{- \underline{\left(2 {x}^{2} + 2 x\right)}} \textcolor{w h i t e}{i} \textcolor{red}{\downarrow}$
$\textcolor{w h i t e}{x \otimes \otimes \otimes \otimes o} \textcolor{red}{- 4 x + 5}$

One more division, in the same fashion as the 2 previous ones.

$\textcolor{w h i t e}{x \otimes \otimes o} {x}^{2} + 2 x \textcolor{red}{- 4}$
$x + 1 \sqrt{{x}^{3} + 3 {x}^{2} - 2 x + 5}$
$\textcolor{w h i t e}{x o i} - \underline{\left({x}^{3} + {x}^{2}\right)} \textcolor{w h i t e}{x o} \downarrow \textcolor{w h i t e}{x o} \downarrow$
$\textcolor{w h i t e}{x \otimes \otimes \otimes i} 2 {x}^{2} - 2 x + 5$
$\textcolor{w h i t e}{x \otimes o i i} - \underline{\left(2 {x}^{2} + 2 x\right)} \textcolor{w h i t e}{i} \downarrow$
$\textcolor{w h i t e}{x \otimes \otimes \otimes \otimes o} - 4 x + 5$
$\textcolor{w h i t e}{x \otimes \otimes \otimes} \textcolor{red}{- \underline{\left(- 4 x - 4\right)}}$
$\textcolor{w h i t e}{x \otimes \otimes \otimes \otimes \otimes \otimes \otimes} \textcolor{red}{9}$

When the remaining difference is a term with no $x$ term, you perform the final step, which is to write this number divided by the divisor, and to append it to the existing quotient. This is the remainder.

${x}^{2} + 2 x - 4 + \textcolor{red}{\frac{9}{x + 1}} , x \ne - 1$

This is the final answer.

Oct 25, 2017

${x}^{2} + 2 x - 4 + \frac{9}{x + 1}$

#### Explanation:

$\text{one way is to use the divisor as a factor in the numerator}$

$\text{consider the numerator}$

$\textcolor{red}{{x}^{2}} \left(x + 1\right) \textcolor{m a \ge n t a}{- {x}^{2}} + 3 {x}^{2} - 2 x + 5$

$= \textcolor{red}{{x}^{2}} \left(x + 1\right) \textcolor{red}{+ 2 x} \left(x + 1\right) \textcolor{m a \ge n t a}{- 2 x} - 2 x + 5$

$= \textcolor{red}{{x}^{2}} \left(x + 1\right) \textcolor{red}{+ 2 x} \left(x + 1\right) \textcolor{red}{- 4} \left(x + 1\right) \textcolor{m a \ge n t a}{+ 4} + 5$

$= \textcolor{red}{{x}^{2}} \left(x + 1\right) \textcolor{red}{+ 2 x} \left(x + 1\right) \textcolor{red}{- 4} \left(x + 1\right) + 9$

$\text{quotient "=color(red)(x^2+2x-4)," remainder } = + 9$

$\Rightarrow \frac{{x}^{3} + 3 {x}^{2} - 2 x + 5}{x + 1}$

$= {x}^{2} + 2 x - 4 + \frac{9}{x + 1} \to \left(x \ne - 1\right)$