# How do you divide (x^4+x^2-3x-3)/(5x-2)?

Jan 19, 2016

Long divide the coefficients of the polynomials to find:

$\frac{{x}^{4} + {x}^{2} - 3 x - 3}{5 x - 2}$

$= \frac{1}{5} {x}^{3} + \frac{2}{25} {x}^{2} + \frac{29}{125} x - \frac{317}{625} - \frac{2509}{625 \left(5 x - 2\right)}$

#### Explanation:

I like to divide polynomials by long dividing their coefficients, including $0$'s for any missing powers of $x$ ...

This is similar to long division of numbers.

Write the dividend $1 , 0 , 1 , - 3 , - 3$ under the bar and the divisor $5 , - 2$ to the left of the bar.

Write each term of the quotient in turn, chosen to match the leading term of the running remainder when multiplied by the divisor.

So we write $\textcolor{b l u e}{\frac{1}{5}}$ which multiplied by $5 , - 2$ results in $1 , - \frac{2}{5}$, matching the leading term $1$ of the dividend.

Subtract the product from the dividend and bring down the next term of the dividend alongside the result as the running remainder.

Choose the next term $\textcolor{b l u e}{\frac{2}{25}}$ to match the leading term of this running remainder and subtract the product, etc.

Repeat until the running remainder is shorter than the divisor and there is no remaining term to bring down from the dividend. This is the final remainder.

In this particular example we find that the quotient polynomial is:

$\frac{1}{5} {x}^{3} + \frac{2}{25} {x}^{2} + \frac{29}{125} x - \frac{317}{625}$

with remainder $- \frac{2509}{625}$

That is:

$\frac{{x}^{4} + {x}^{2} - 3 x - 3}{5 x - 2}$

$= \frac{1}{5} {x}^{3} + \frac{2}{25} {x}^{2} + \frac{29}{125} x - \frac{317}{625} - \frac{2509}{625 \left(5 x - 2\right)}$

or if you prefer:

${x}^{4} + {x}^{2} - 3 x - 3$

$= \left(5 x - 2\right) \left(\frac{1}{5} {x}^{3} + \frac{2}{25} {x}^{2} + \frac{29}{125} x - \frac{317}{625}\right) - \frac{2509}{625}$