# How do you use synthetic division to divide ( x^4 - 2x^3 + x - 8 ) / ( x - 1 )?

Oct 21, 2015

Long divide the coefficients to find:

${x}^{4} - 2 {x}^{3} + x - 8 = \left(x - 1\right) \left({x}^{3} - {x}^{2} - x\right) - 8$

$\frac{{x}^{4} - 2 {x}^{3} + x - 8}{x - 1} = {x}^{3} - {x}^{2} - x - \frac{8}{x - 1}$

#### Explanation:

I like to lay it out similar to long division, but just writing the coefficients...

There are other possible layouts, but I find this easiest to work with (especially when the divisor is more complex than linear).

Notice the $0$ coefficient for ${x}^{2}$ in the dividend.

So write the dividend $1 , - 2 , 0 , 1 , - 8$ under the bar and the divisor $1 , - 1$ to the left.

Choose the first term $1$ for the quotient so that when multiplied by the divisor it results in a tuple $1 , - 1$ whose first term matches the first term of the divisor.

Write down the product of the first term of the quotient and the divisor under the dividend and subtract it, resulting in a remainder $- 1$.

Bring down the next term $0$ from the dividend alongside this remainder.

Choose the next term $- 1$ for the quotient so that when multiplied by the divisor results in a tuple $- 1 , 1$ whose first term matches the first term of the remainder.

Continue in similar fashion until we run out of terms to bring down from the dividend.

At this point we arrive at a final quotient $1 , - 1 , - 1 , 0$ meaning ${x}^{3} - {x}^{2} - x$ and a remainder $- 8$.

So:

${x}^{4} - 2 {x}^{3} + x - 8 = \left(x - 1\right) \left({x}^{3} - {x}^{2} - x\right) - 8$

or if you prefer:

$\frac{{x}^{4} - 2 {x}^{3} + x - 8}{x - 1} = {x}^{3} - {x}^{2} - x - \frac{8}{x - 1}$