# How do you divide (x^4-81)/(x-3) using synthetic division?

Sep 28, 2016

$\left({x}^{4} - 81\right) \div \left(x - 3\right) = {x}^{3} + 3 {x}^{2} + 9 x + 27$

#### Explanation:

In synthetic division we only use the numerical coefficients.

Make sure that they are in descending order of the powers of x.
In this case we have $\text{ "x^4" "x^3" "x^2" "x^1" } {x}^{0}$

Write a $0$ for any missing term.

If $x - 3 = 0 \rightarrow x = 3$

$\textcolor{red}{3}$ is used at the side.
color(white)(xxxxxxxxxxxxx)" "(x^4" "x^3" "x^2" "x^1" " x^0)

Coefficients :$\rightarrow \textcolor{w h i t e}{\times} \text{ "1" "0" "0" "0" } - 81$

color(white)(xxxxxxxxxxxx)color(red)(3)color(white)(xx)ul(darr " "color(blue)(3)" "color(purple)(9)" "color(mediumpurple)(27) " "color(darkcyan)(81)
$\textcolor{w h i t e}{\times \times \times \times \times \times x} \textcolor{w h i t e}{\times x} 1 \text{ "color(lime)(3)" "color(orange)(9)" "color(magenta)(27)" } 0 \leftarrow$ remainder

$\textcolor{w h i t e}{\times \times \times \times \times \times x} \text{ "x^3" "x^2" "x^1" } {x}^{0}$

Steps: Follow the colours...
bring down the 1.
color(red)(3)xx1 = color(blue)(3)" add " 0+color(blue)(3) = color(lime) (3)
$\textcolor{red}{3} \times \textcolor{\lim e}{3} = \textcolor{p u r p \le}{9} \text{ add } 0 + \textcolor{p u r p \le}{9} = \textcolor{\mathmr{and} a n \ge}{9}$
$\textcolor{red}{3} \times \textcolor{\mathmr{and} a n \ge}{9} = \textcolor{m e \mathrm{di} u m p u r p \le}{27} \text{ add } 0 + \textcolor{m e \mathrm{di} u m p u r p \le}{27} = \textcolor{m a \ge n t a}{27}$
$\textcolor{red}{3} \times \textcolor{m a \ge n t a}{27} = \textcolor{\mathrm{da} r k c y a n}{81} \text{ add } - 81 + \textcolor{\mathrm{da} r k c y a n}{81} = 0 \leftarrow$ the remainder

The numbers in the bottom row are now the numerical coefficients of the quotient.

${x}^{4} \div x = {x}^{3}$ So the first term will be ${x}^{3}$

The quotient is:

$1 {x}^{3} + \textcolor{\lim e}{3} {x}^{2} + \textcolor{\mathmr{and} a n \ge}{9} x + \textcolor{m a \ge n t a}{27} \text{ rem } 0$