# How do you use synthetic division to divide (x^3 -8 x^2 - 25x + 203) by x-5?

May 23, 2015

Synthetic division is somewhat like long division.

Starting with $\left({x}^{3} - 8 {x}^{2} - 25 x + 203\right)$, first look for a multiplier for $\left(x - 5\right)$ that will cause give a match for the highest order term.

Choose ${x}^{2}$ as the first multiplier.

${x}^{2} \left(x - 5\right) = {x}^{3} - 5 {x}^{2}$

Subtract this from our original polynomial to get the remainder:

$\left({x}^{3} - 8 {x}^{2} - 25 x + 203\right) - \left({x}^{3} - 5 {x}^{2}\right)$

$= \left(- 3 {x}^{2} - 25 x + 203\right)$

Now choose a multiplier $\left(- 3 x\right)$ for $\left(x - 5\right)$ to match the leading term $- 3 {x}^{2}$ of the remainder...

$\left(- 3 x\right) \left(x - 5\right) = \left(- 3 {x}^{2} + 15 x\right)$

Subtract this from our remainder to get a new remainder:

$\left(- 3 {x}^{2} - 25 x + 203\right) - \left(- 3 {x}^{2} + 15 x\right) = \left(- 40 x + 203\right)$

Now choose a multiplier $\left(- 40\right)$ for $\left(x - 5\right)$ to match the leading term $- 40 x$ of our remainder...

$\left(- 40\right) \left(x - 5\right) = \left(- 40 x + 200\right)$

Subtract this from our remainder to get a new remainder:

$\left(- 40 x + 203\right) - \left(- 40 x + 200\right) = 3$

Adding our multipliers together, we find:

${x}^{3} - 8 {x}^{2} - 25 x + 203 = \left(x - 5\right) \left({x}^{2} - 3 x - 40\right) + 3$