# Question #e7b5e

Nov 9, 2015

${x}^{2} + 4 x + 3$

#### Explanation:

Here's what the set up looks like...
$1 | 1 + 3 - 1 - 3$

The $1$ on the left is the "root" of the bottom, or the number that x would be to make it equal $0$. Or you can just think of it is the opposite of the number after $x$.

All of the numbers after are the coefficients in front of the powers of $x$ in decreasing order.

Now the process. Draw leave some space and draw a line underneath like so...

$1 | 1 + 3 - 1 - 3$

$- - - - - - -$

Add down and multiply up diagonally by the number in the top left (I hope this makes sense)

$1 | 1 + 3 - 1 - 3$
$\textcolor{w h i t e}{X l l} 0 \textcolor{w h i t e}{X X X X X X X X X}$$< - - -$start with a zero here
$- - - - - - -$
$\textcolor{w h i t e}{X l l} 1 \textcolor{w h i t e}{X X X X X X X X X}$ $< - - -$ add down

$1 | 1 + 3 - 1 - 3$
$\textcolor{w h i t e}{X l l} 0 \textcolor{w h i t e}{l l l l} 1 \textcolor{w h i t e}{X X X X X X X}$$< - - -$multiply up by 1
$- - - - - - -$
$\textcolor{w h i t e}{X l l} 1 \textcolor{w h i t e}{X l} 4 \textcolor{w h i t e}{X X X X X X}$ $< - - -$ add down

Here's the rest:
$1 | 1 + 3 - 1 - 3$
$\textcolor{w h i t e}{X l l} 0 \textcolor{w h i t e}{l l l l} 1 \textcolor{w h i t e}{l l l a} 4 \textcolor{w h i t e}{l l l a} 3$
$- - - - - - -$
$\textcolor{w h i t e}{X l l} 1 \textcolor{w h i t e}{X l} 4 \textcolor{w h i t e}{i i i a} 3 \textcolor{w h i t e}{i i i a} 0$

Now use those numbers on the bottom as coefficients for a polynomial with degree one lower than the original numerator:

$1 {x}^{2} + 4 x + 3 + \frac{0}{x - 1}$ (this last term is the remainder, and this is how you write it)

Clean it up...
${x}^{2} + 4 x + 3$