# How do you divide ( 6x^3 + 10x^2 + x + 8)/(2x^2 + 1)?

Jan 2, 2016

Long divide the coefficients to find quotient $3 x + 5$ with remainder $- 2 x + 3$.

#### Explanation:

I like to long divide the coefficients like this:

...not forgetting to include a $0$ in the divisor to stand for the missing $x$ term.

In this example we find the quotient is $3 x + 5$ with remainder $- 2 x + 3$

The process is similar to long division of numbers:

Write the dividend ($6 , 10 , 1 , 8$) under the bar.

Write the divisor ($2 , 0 , 1$) to the left of the bar.

Start writing the quotient, term by term, choosing each successive term to match the leading term of the running remainder:

The first term of the quotient is $\textcolor{b l u e}{3}$, so that when multiplied by $2 , 0 , 1$, the resulting first term matches the leading $6$ of the dividend.

We then write out the product $6 , 0 , 3$ of $3$ and $2 , 0 , 1$ under the dividend and subtract it to give $10 , - 2$. We bring down the next term from the dividend alongside it to give our running remainder.

We choose the next term of the quotient $\textcolor{b l u e}{5}$, so that when multiplied by $2 , 0 , 1$, the resulting first term matches the leading term $10$ of our running remainder.

We then write out the produce $10 , 0 , 5$ of $5$ and $2 , 0 , 1$ under the running remainder and subtract it to give $- 2 , 3$.

There are no more terms to bring down from the dividend, so this is our final remainder.

Then interpret the coefficient sequences by applying them to the appropriate powers of $x$ to find:

$\frac{6 {x}^{3} + 10 {x}^{2} + x + 8}{2 {x}^{2} + 1} = 3 x + 5 + \frac{- 2 x + 3}{2 {x}^{2} + 1}$

Or if you prefer:

$6 {x}^{3} + 10 {x}^{2} + x + 8 = \left(3 x + 5\right) \left(2 {x}^{2} + 1\right) + \left(- 2 x + 3\right)$