# How do you divide (3x^3+4x^2+x+1)÷(x-1)?

Oct 11, 2015

$3 {x}^{2} + 7 x + 8 + \frac{9}{x - 1}$

#### Explanation:

You can use Remainder Theorem or Synthetic Division to do so. I will use the Synthetic Division since it is quicker and easier (but it only works for a polynomial divided by a binomial).

The first step in using theSynthetic Division is to find out what makes the binomial $0$. In this case, it is $1$. Now, you put that number in a sort of secluded area like so:

$1 \text{ } |$

Then, you determine the co-efficients for every term in your polynomial (you MUST also include co-efficients of $0$. However, there are none in this problem). Then, you make a list like this:

$1 \text{ "|color(white)(xx)3" " "4" " } 1 \textcolor{w h i t e}{\times} 1$

Then, you do this little trick where you bring down the first number ($3$), multiply by the secluded number ($1$), put the answer down under the next number ($4$), and then add. After its all done, it'll look like this:

$1 \text{ "|color(white)(xx)3" " "4" " } 1 \textcolor{w h i t e}{\times} 1$
$\textcolor{w h i t e}{\times \times \times} \downarrow \textcolor{w h i t e}{x} 3 \text{ " } 7 \textcolor{w h i t e}{\times} 8$
color(white)(xxxxx)stackrel("---------------------------------")
$\textcolor{w h i t e}{\times \times \times} 3 \text{ " "7" " } 8 | \textcolor{w h i t e}{x} 9$

Now, the remainder has the co-efficients on the third line ($3$, $7$, and $8$), making the polynomial $3 {x}^{2} + 7 x + 8$. The $9$ is the remainder (divided by the binomial).

$3 {x}^{2} + 7 x + 8 + \frac{9}{x - 1}$.