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.
