# How do you use synthetic division to divide x^3+5x^2+x-15 by x+3?

Aug 15, 2015

#### Answer:

$\frac{{x}^{3} + 5 {x}^{2} + x - 15}{x + 3} = {x}^{2} + 2 x - 5$

#### Explanation:

Divide ${x}^{3} + 5 {x}^{2} + x - 15$ by $x + 3$.

First, you let the coefficients of each degree to be used in the division ($1 , 5 , 1 , - 15$).

Then, dividing by $x + 3 = x - \left(- 3\right)$ implies that you use $- 3$ in your upper left. So, put $- 3$, then a line (I've used a double line), and then write $\text{ 1" " 5" " 1" " -15 }$ to the right.

$\text{-3 } | |$ $\text{1 " " 5 " " 1" " -15}$
$+$
$\text{ " " }$$- - - - -$

First, bring the first $1$ down to the bottom, and multiply it by the $- 3$. Put that $- 3$ below the $5$.

$\text{-3 } | |$ $\text{1 " " 5 " " 1" " -15}$
$+$ $\text{ "" " " -3}$
$\text{ " " }$$- - - - -$
$\text{ "" " " 1}$

Then add $5 + \left(- 3\right) = 2$ Put that under the $- 3$

$\text{-3 } | |$ $\text{1 " " 5 " " 1" " -15}$
$+$ $\text{ "" " " -3}$
$\text{ " " }$$- - - - -$
$\text{ " " " " 1 " " 2}$

Multiply $- 3 \times 2 = - 6$ and put the $- 6$ under the $1$. Then add:

$\text{-3 } | |$ $\text{ 1 " " 5 " " 1 " " -15}$
$+$ $\text{ "" " " -3 " " -6}$
$\text{ " " }$$- - - - - - - -$
$\text{ " " " " 1 " " 2 " " -5}$

Now $- 3 \times - 5 = 15$, so we put $15$ under #-15 and add::

$\text{-3 } | |$ $\text{1 " " 5 " " 1 " " -15}$
$+$ $\text{ "" ""-3 " " -6 " " 15}$
$\text{ " " }$$- - - - - - - -$
$\text{ "" " " 1 " " 2 " " -5 " " 0 }$

The bottom row ignoring the last number gives us the coefficients of the quotient.
The last number on the bottom row is the remainder (and it is also $P \left(- 3\right)$).

So the division gives us:

$\frac{{x}^{3} + 5 {x}^{2} + x - 15}{x + 3} = {x}^{2} + 2 x - 5$

You can check the answer by multiplyng:

$\left(x + 3\right) \left({x}^{2} + 2 x - 5\right)$ to make sure we get ${x}^{3} + 5 {x}^{2} + x - 15$.

(I've used Synthetic Division Formatting by Truong-Son R.)