# Question 22c29

Sep 2, 2016

${n}^{2} + 5 n + 4 \text{ rem -5}$

#### Explanation:

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

(n^3+7n^2 +14n+3)div(n+2) = ?????????
" (dividend) " div " (divisor)" = ("quotient")

$\textcolor{m a \ge n t a}{\text{step 1:}}$ The dividend must be in descending powers of r.
$\textcolor{w h i t e}{\times \times \times \times \times x} {n}^{3} + 7 {n}^{2} + 14 n + 3$
$\textcolor{w h i t e}{\times \times \times \times} \Rightarrow \textcolor{m a \ge n t a}{1 \text{ +7 +14 +3}}$

Only use the numerical coefficients in the $\textcolor{m a \ge n t a}{\text{top row}}$.

(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. $\text{ " (n+2) = 0 rArr n = color(orange)(-2) " this goes outside}$

color(white)(xxxxx) | color(brown)(1)" "+7" "+14 " "+3 color(magenta)(" step 1 top row")#
$\textcolor{w h i t e}{x . x} \textcolor{\mathmr{and} a n \ge}{- 2} \text{| darr " "color(red)(-2) " "color(blue)(-10) " } \textcolor{o l i v e}{- 8}$
$\textcolor{w h i t e}{\times \times \times} \underline{\text{ }}$
$\textcolor{w h i t e}{\times \times \times x} \textcolor{b r o w n}{1} \text{ "color(blue)(+5) " "color(olive)(+4)" "color(teal)(-5) larr " remainder!}$

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

Step 3 : Begin the division:

$\text{Bring down the " color(brown)( 1 ) " to below the line}$
$\text{multiply } \textcolor{\mathmr{and} a n \ge}{- 2} \times \textcolor{b r o w n}{1} = \textcolor{red}{- 2}$
$\text{Add } 7 + \textcolor{red}{- 2} = \textcolor{b l u e}{+ 5}$
$\text{multiply } \textcolor{\mathmr{and} a n \ge}{- 2} \times \textcolor{b l u e}{+ 5} = \textcolor{b l u e}{- 10}$
$\text{Add } 14 \textcolor{b l u e}{- 10} = \textcolor{o l i v e}{+ 4}$
$\text{multiply } \textcolor{\mathmr{and} a n \ge}{- 2} \times \textcolor{o l i v e}{4} = \textcolor{o l i v e}{- 8}$
$\text{Add } 3 + \textcolor{o l i v e}{- 8} = \textcolor{t e a l}{- 5}$

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 $\textcolor{t e a l}{- 5}$

This means that $\left(n + 2\right)$ is not a factor of ${n}^{3} + 7 {n}^{2} + 14 n + 3$

$\left({n}^{3} + 7 {n}^{2} + 14 n + 3\right) \div \left(n + 2\right) = {n}^{2} + 5 n + 4 \text{ rem -5}$