# How do you use synthetic division to divide (2x^3 + x^2 - 2x + 2)  by x+2?

Jun 28, 2015

(Formatting copied from another answer by Truong-Son R.)

How do you use synthetic division to divide $\left(2 {x}^{3} + {x}^{2} - 2 x + 2\right)$ by $x + 2$?

First, you let the coefficients of each degree be used in the division ($2 , 1 , - 2 , 2$).

Then, dividing by $x + 2$ implies that you use $- 2$ in your upper left (if it was $x - 2$, put $2$). So, draw the bottom and right sides of a square, put $- 2$ inside it, and then write $\text{2" " 1" " -2" " 2}$ to the right.

$\text{-2} | |$ $\text{2 " "1 " "-2 " "2 }$
$+$
$\text{ " }$$- - - - -$

First, bring the $2$ down to the bottom, and multiply it by the $- 2$. Put that $- 4$ below $1$.

$\text{-2} | |$ $\text{2 " "1 " "-2 " "2 }$
$+$ $\text{ " " "" "-4}$
$\text{ " }$$- - - - -$
$\text{ " " "2}$

$\text{-2} | |$ $\text{2 " "1 " "-2 " "2 }$
$+$ $\text{ " " "" "-4}$
$\text{ " }$$- - - - -$
$\text{ " " "2" "" " -3}$

Repeat a few times once you've figured out the simple pattern ($\text{divisor"*"new sum}$, put result under next-lowest degree term, add to get another $\text{new sum}$, repeat).

$\text{-2} | |$ $\text{2 " "1 " "-2 " "2 }$
$+$ $\text{ " " "" "-4" "" " 6}$
$\text{ " }$$- - - - -$
$\text{ " " "2" "" " -3" "" " 4}$

$\text{-2} | |$ $\text{2 " "1 " "-2 " "2 }$
$+$ $\text{ " " "" "-4" "" " 6" " -8}$
$\text{ " }$$- - - - -$
$\text{ " " "2" "" " -3" "" " 4"" -6}$

You know you can stop when you reach the far right and you have no spot left below the original dividend ($2 , 1 , - 2 , 2$). to insert a product.

Since you started with a cubic, the answer is a quadratic. Thus, you have:

$2 {x}^{2} - 3 x + 4$ $\left(r = - 6\right)$

Indeed, if you multiply them together and add the remainder, you get the original back:

$\left(2 {x}^{2} - 3 x + 4\right) \left(x + 2\right) = 2 {x}^{3} - 3 {x}^{2} + 4 {x}^{2} + 4 x - 6 x + 8 + \left(- 6\right) = 2 {x}^{3} + {x}^{2} - 2 x + 2$

So you would have:

$\frac{2 {x}^{3} + {x}^{2} - 2 x + 2}{x + 2} = 2 {x}^{2} - 3 x + 4 + \frac{- 6}{x + 2}$

or

$\frac{2 {x}^{3} + {x}^{2} - 2 x + 2}{x + 2} = 2 {x}^{2} - 3 x + 4 - \frac{6}{x + 2}$