# How do you divide (x^4 - 7x^3 + 2x^2 + 9x)/(x^3-x^2+2x+1)?

Jan 13, 2016

Long divide the coefficients to find:

$\frac{{x}^{4} - 7 {x}^{3} + 2 {x}^{2} + 9 x}{{x}^{3} - {x}^{2} + 2 x + 1} = x - 6 + \frac{- 6 {x}^{2} + 4 x + 6}{{x}^{3} - {x}^{2} + 2 x + 1}$

That is, the quotient is $x - 6$ with remainder $- 6 {x}^{2} + 4 x + 6$

#### Explanation:

I like to long divide the coefficients, not forgetting to include $0$'s for any missing powers of $x$. In this example, that means the constant term of the dividend...

The process is similar to long division of numbers.

Write the dividend ($1 , - 7 , 2 , 9 , 0$) under the bar and the divisor ($1 , - 1 , 2 , 1$) to the left.

Choose the first term $\textcolor{b l u e}{1}$ of the quotient so that when multiplied by the divisor it matches the leading term of the dividend.

Write out the product of this first term of the quotient and the divisor below the dividend and subtract it.

Bring down the next term ($0$) of the dividend alongside it.

Choose the next term $\textcolor{b l u e}{- 6}$ of the quotient so that when multiplied by the divisor it matches the leading term of the remainder.

Write out the product of this second term of the quotient and the divisor below the remainder and subtract it.

The resulting remainder $- 6 , 4 , 6$ is shorter than the divisor and there are no more terms to bring down from the dividend. So this is our final remainder.

In this example, we find the quotient is $x - 6$ and remainder $- 6 {x}^{2} + 4 x + 6$