# How do you divide (4x^3+2x-6) /(x-1)?

Oct 30, 2015

See explanation
Bit long, but the process takes quite a bit of getting used to!

#### Explanation:

You asked how to do it so I am explaining the process:

You divide the sequential $x$ parts in the numerator into sequential $x$ parts of the denominator. Each stage leaves a remainder for which the process is repeated.

Demonstration within the context of this question.

$\textcolor{g r e e n}{\text{==============================}}$
Step 1: $4 {x}^{3} \div i \mathrm{de} x = 4 {x}^{2}$

Used the $4 {x}^{3}$ from $4 {x}^{3} + 2 x - 6$ and the $x$ from$x - 1$

So the $\textcolor{red}{\text{first}}$ part of your answer is $\textcolor{b l u e}{4 {x}^{2}}$

$\textcolor{g r e e n}{\text{============================}}$
Step 2: Find the remainder

$4 {x}^{2} \times \left(x - 1\right) = 4 {x}^{3} - 4 x$
This is then subtracted so we have:

$4 {x}^{3} + 2 x - 6$ .......Original equation
$\left(4 {x}^{3} - 4 x\right) -$ ...... Subtract
~~~~~~~~~~~~~
$\textcolor{w h i t e}{\text{xxxxxx}} 6 x - 6$. This is the first remainder

$\textcolor{g r e e n}{\text{=================================}}$

Step 3.

Again divide the $6 x$ in the previous remainder by the $x$ in $x - 1$ giving $6$.

So the color(red)("second")" " part of the answer is $\textcolor{b l u e}{+ 6}$

$6 \left(x - 1\right) = 6 x - 6$ which is subtracted from the most recent remainder giving:

$4 {x}^{3} + 2 x - 6$ .......Original equation
$\left(4 {x}^{3} - 4 x\right) -$ ...... Subtract
~~~~~~~~~~~~~
$\textcolor{w h i t e}{\text{xxxxxx}} 6 x - 6$. This is the remainder
$\textcolor{w h i t e}{\text{xxxxxx}} \left(6 x - 6\right) -$. Subtract
~~~~~~~~~~~~~~~~~~~~~~~~~~
color(white)("xxxxxxx") 0 + 0" " which is the second remainder.

The zeros mean that we have an exact division

In this case
$\frac{4 {x}^{3} + 2 x - 6}{x - 1} = 4 {x}^{2} + 6$

Suppose we had ended up with a remainder that the $x \text{ in } \left(x - 1\right)$ could not be divided into. In that case we would express the whole of that remainder as a fraction with $\left(x - 1\right)$ as the denominator.

Suppose that had ended up with a remainder of just 2. Then in that case the answer would be:

$4 {x}^{2} + \frac{2}{x - 1} + 6$

Hope this helps. It takes a lot of practice.