Question

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

Answer 1

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.

`color(green)("==============================")`
Step 1: `4x^3 divide x = 4x^2`

Used the `4x^3` from `4x^3+2x-6` and the `x` from`x-1`

So the `color(red)("first")` part of your answer is `color(blue)(4x^2)`

`color(green)("============================")`
Step 2: Find the remainder

`4x^2 times (x-1) = 4x^3 - 4x`
This is then subtracted so we have:

`4x^3 + 2x -6` .......Original equation
`(4x^3 -4x) -` ...... Subtract
~~~~~~~~~~~~~
`color(white)("xxxxxx") 6x - 6`. This is the first remainder

`color(green)("=================================")`

Step 3.

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

So the `color(red)("second")" "` part of the answer is `color(blue)(+6)`

`6(x-1) =6x-6` which is subtracted from the most recent remainder giving:

`4x^3 + 2x -6` .......Original equation
`(4x^3 -4x) -` ...... Subtract
~~~~~~~~~~~~~
`color(white)("xxxxxx") 6x - 6`. This is the remainder
`color(white)("xxxxxx") ( 6x - 6) -`. Subtract
~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(white)("xxxxxxx") 0 + 0" "` which is the second remainder.

The zeros mean that we have an exact division

In this case
`(4x^3 +2x-6)/(x-1) =4x^2+6`

Suppose we had ended up with a remainder that the `x " in " (x-1)` could not be divided into. In that case we would express the whole of that remainder as a fraction with `(x-1)` as the denominator.

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

`4x^2 + 2/(x-1) + 6`

Hope this helps. It takes a lot of practice.