# How do you use polynomial division to determine whether x=3 is a root of x^3+x^2+15-x?

Nov 24, 2015

Divide $\left({x}^{3} + {x}^{2} - x + 15\right) \div \left(x - r\right)$ with long polynomial division.

If you can divide without a remainder, $x = r$ is a root.
If you have a remainder, it's not a root.

#### Explanation:

Well, just to check if $x = 3$ is a root of your term, it would be easier to plug $x = 3$ and see if the result is $0$. :-)

But of course, it is also possible to determine the outcome with polynomial division.

To do so, you must divide $\left({x}^{3} + {x}^{2} - x + 15\right)$ by $\left(x - 3\right)$.

I know that in some countries, a different notation for long division is being used but I will use the one that I'm familiar with and I hope that it will be no problem for you to rewrite it in your notation if necessary.

$\textcolor{w h i t e}{\times} \left({x}^{3} + {x}^{2} \textcolor{w h i t e}{\times} - x \textcolor{w h i t e}{\times} + 15\right) : \left(x - 3\right) = {x}^{2} + 4 x + 11$
$\textcolor{w h i t e}{i} - \left({x}^{3} - 3 {x}^{2}\right)$
$\textcolor{w h i t e}{\times x} \frac{\textcolor{w h i t e}{\times \times \times}}{\textcolor{w h i t e}{x}}$
$\textcolor{w h i t e}{\times \times \times x} 4 {x}^{2} \textcolor{w h i t e}{x} - x$
$\textcolor{w h i t e}{\times \times x} - \left(4 {x}^{2} - 12 x\right)$
$\textcolor{w h i t e}{\times \times \times x} \frac{\textcolor{w h i t e}{\times \times \times \times}}{\textcolor{w h i t e}{x}}$
$\textcolor{w h i t e}{\times \times \times \times \times \times} 11 x \textcolor{w h i t e}{x} + 15$
$\textcolor{w h i t e}{\times \times \times \times \times} - \left(11 x \textcolor{w h i t e}{x} - 33\right)$
$\textcolor{w h i t e}{\times \times \times \times \times \times} \frac{\textcolor{w h i t e}{\times \times \times \times}}{\textcolor{w h i t e}{x}}$
$\textcolor{w h i t e}{\times \times \times \times \times \times \times \times \times} 48$

At the end of the computation, there is the remainder $48$.

As you have a remainder, it means that $x = 3$ is not a root of ${x}^{3} + {x}^{2} - x + 15$.