# How do you use synthetic division to divide (x^2 + 13x + 40) ÷ (x + 8)?

May 16, 2015

Doing synthetic division is rather like doing long division.

First look for a multiplier for $\left(x + 8\right)$ that will match the highest term ${x}^{2}$. That multiplier must be $x$:

$x \left(x + 8\right) = {x}^{2} + 8 x$

Subtract the right hand side from the original ${x}^{2} + 13 x + 40$ to find the remainder:

$\left({x}^{2} + 13 x + 40\right) - \left({x}^{2} + 8 x\right) = 5 x + 40$

Now look for a multiplier for $\left(x + 8\right)$ that will match the highest remaining term $5 x$. That multiplier must be $5$:

$5 \left(x + 8\right) = 5 x + 40$

Subtract the right hand side from our last remainder $5 x + 40$ to find the remainder:

$\left(5 x + 40\right) - \left(5 x + 40\right) = 0$

Bingo! It divides perfectly.

Adding together the multipliers $x$ and $5$ that we found we get

$\left({x}^{2} + 13 x + 40\right) \div \left(x + 8\right) = x + 5$