# How do you divide (n^3+6n^2+4n-2)div(n+1) using synthetic division?

Nov 3, 2016

$\left({n}^{3} + 6 {n}^{2} + 4 n - 2\right) \div \left(n + 1\right)$

$= {n}^{2} + 5 n - 1 \text{ } r e m - 1$

#### Explanation:

The method is very easy, but the process is a bit difficult to explain.

(n^3+6n^2+4n-2)div(n+1) = ?????????
" (dividend) " div " (divisor)" = ("quotient")

$\textcolor{m a \ge n t a}{\text{step 1:}}$ The dividend must be in descending powers of n.
$\textcolor{w h i t e}{\times \times \times \times \times x} {n}^{3} + 6 {n}^{2} + 4 n - 2$
$\textcolor{w h i t e}{\times \times \times \times x} \Rightarrow \textcolor{m a \ge n t a}{1 \text{ +6 +4 -2}}$

In the dividing use only the numerical coefficients $\textcolor{m a \ge n t a}{\text{(top row)}} \downarrow$.

(If there are any missing, leave a space or fill in a zero).

$\textcolor{\mathmr{and} a n \ge}{\text{Step 2}}$: Make the divisor = 0. $\text{ " (n+1) = 0 rArr n = color(orange)(-1) " this goes outside the box}$

Step 3 : Begin the division - see details below....

color(white)(xxxxx) | color(brown)(1)" "+6" "+4 " "-2 color(magenta)(" step 1 top row")
$\textcolor{w h i t e}{x . x} \textcolor{\mathmr{and} a n \ge}{- 1} \text{| darr " "color(red)(-1) " "color(blue)(-5) " } \textcolor{o l i v e}{+ 1}$
$\textcolor{w h i t e}{\times \times \times} \underline{\text{ }}$
$\textcolor{w h i t e}{\times \times \times x} \textcolor{b r o w n}{1} \text{ "color(blue)(+5) " "color(olive)(-1)" "color(teal)(-1) larr " the remainder!}$

$\textcolor{w h i t e}{\times \times . \times} \uparrow \text{ "uarr " } \uparrow$
$\textcolor{w h i t e}{\times \times \times x} {n}^{2} \text{ "n^1 " } {n}^{0}$

Dividing details

$\text{Bring down the " color(brown)( 1 ) " to below the line}$
$\text{multiply } \textcolor{\mathmr{and} a n \ge}{- 1} \times \textcolor{b r o w n}{1} = \textcolor{red}{- 1}$
$\text{Add } + 6 \textcolor{red}{- 1} = \textcolor{b l u e}{+ 5}$
$\text{multiply } \textcolor{\mathmr{and} a n \ge}{- 1} \times \textcolor{b l u e}{+ 5} = \textcolor{b l u e}{- 5}$
$\text{Add } 4 \textcolor{b l u e}{- 5} = \textcolor{o l i v e}{- 1}$
$\text{multiply } \textcolor{\mathmr{and} a n \ge}{- 1} \times \textcolor{o l i v e}{- 1} = \textcolor{o l i v e}{+ 1}$
$\text{Add } - 2 + \textcolor{o l i v e}{1} = \textcolor{t e a l}{- 1}$

That's it Folks!

We have now found the numerical coefficients of the terms in the quotient (answer)

We divided an expression with ${n}^{3}$ by an expression with $n$,
so the first term will be ${n}^{3} / n = {n}^{2}$

The last value is the remainder. In this case it is $\textcolor{t e a l}{- 1}$

This means that $\left(n + 1\right)$ is not a factor of ${n}^{3} + 6 {n}^{2} + 4 n - 2$

$\left({n}^{3} + 6 {n}^{2} + 4 n - 2\right) \div \left(n + 1\right) = {n}^{2} + 5 n - 1 \text{ rem -1}$