# How do you divide (5x^3-2x^2+5x-16)div(x-1) using synthetic division?

##### 1 Answer
Aug 25, 2016

$\left(5 {x}^{3} - 2 {x}^{2} + 5 x - 16\right) \div \left(x - 1\right)$
$= 5 {x}^{2} + 3 x + 8 \text{ rem -8}$

#### Explanation:

The method is very easy, but the process is a bit difficult to explain.
Follow the colours.

(5x^3-2x^2+5x-16)div(x-1) = ?????????
" (dividend) " div " (divisor)" = ("quotient")

$\textcolor{m a \ge n t a}{\text{step 1:}}$ The dividend must be in descending powers of x.
$\textcolor{w h i t e}{\times \times \times \times \times} 5 {x}^{3} - 2 {x}^{2} + 5 x - 16$
$\textcolor{w h i t e}{\times \times \times \times} \Rightarrow \textcolor{m a \ge n t a}{5 \text{ -2 +5 -16}}$

Write the numerical coefficients in the $\textcolor{m a \ge n t a}{\text{top row}}$.

(If there are any powers missing, leave a space or fill in a zero).

$\textcolor{\mathmr{and} a n \ge}{\text{Step 2}}$: Make the divisor = 0. $\text{ " (x-1) = 0 rArr x = color(orange)(1) " this goes outside}$

Step 3 : Begin the division:

$\text{Bring down the " color(brown)( 5 ) " to below the line}$
$\text{multiply } \textcolor{\mathmr{and} a n \ge}{1} \times \textcolor{b r o w n}{5} = \textcolor{red}{5}$
$\text{Add } - 2 + \textcolor{red}{5} = \textcolor{b l u e}{+ 3}$
$\text{multiply } \textcolor{\mathmr{and} a n \ge}{1} \times \textcolor{b l u e}{+ 3} = \textcolor{b l u e}{3}$
$\text{Add } 5 \textcolor{b l u e}{+ 3} = \textcolor{o l i v e}{+ 8}$
$\text{multiply } \textcolor{\mathmr{and} a n \ge}{1} \times \textcolor{o l i v e}{8} = \textcolor{o l i v e}{8}$
$\text{Add } - 16 + \textcolor{o l i v e}{8} = \textcolor{t e a l}{- 8}$

That's it Folks!

color(white)(xxxxx) | color(brown)(5)" "-2" " 5 " "-16 color(magenta)(" step 1 top row")
$\textcolor{w h i t e}{x \ldots . x} \textcolor{\mathmr{and} a n \ge}{1} \text{| darr " "color(red)(5) " "color(blue)(3) " } \textcolor{o l i v e}{8}$
$\textcolor{w h i t e}{\times \times \times} \underline{\text{ }}$
$\textcolor{w h i t e}{\times \times \times x} \textcolor{b r o w n}{5} \text{ "color(blue)(+3) " "color(olive)(+8)" "color(teal)(-8) larr " the remainder!}$

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

We have now found the numerical coefficients of the terms in the quotient (answer)

We divided an expression with ${x}^{3}$ by an expression with $x$,
so the first term will be ${x}^{3} / x = {x}^{2}$

The quotient is $5 {x}^{2} + 3 x + 8 - 8$

The last value is the remainder. In this case it is $\textcolor{t e a l}{- 8}$

This means that $\left(x - 1\right)$ is not a factor of $5 {x}^{3} - 2 {x}^{2} + 5 x - 16$

$\left(5 {x}^{3} - 2 {x}^{2} + 5 x - 16\right) \div \left(x - 1\right) = 5 {x}^{2} + 3 x + 8 \text{ rem -8}$