Question 27663
ok... long division:


in dividing 2345 by 18, you consider how many 18's go into "2"-000: Answer none.

Now, we extend to think of as "23"-00
--> 18's in 23? answer 1 with 5 left over.


So, now how many 18's go into 545. 
--> How many 18's into 5? Answer: zero

Now we extend again
--> how many 18's into 54? answer 3 with no remainder.


And the final 5? that is the final remainder.
--> Complete Answer is 103 and 5/18


We do the exact same process with algebraic expressions.


{{{ (8x^3 - 6x^2 + 2x)/(4x - 1)}}}


{{{ (4x - 1)|(8x^3 - 6x^2 + 2x)}}}

How many 4x's go into {{{8x^3}}}? Answer is {{{2x^2}}}.


{{{2x^2 * (4x-1)}}} is {{{(8x^3-2x^2)}}} so subtract that from the original, leaving us with {{{(-4x^2+2x)}}} --> call this the "new original"


Now repeat the process...
How many times does 4x go into {{{(-4x^2)}}}
Answer is -x.


So, {{{(-x)*(4x-1)}}} is {{{-4x^2+x}}}. Now subtract that from the "new original", leaving us with x --> call this the "new original".


Now repeat the process again...
How many times does 4x go into x
Answer is (1/4).


So {{{(1/4)*(4x-1)}}} is {{{x-(1/4)}}}. Now subtract this from the "new original", leaving us with (1/4).


This is the end.
Answer to {{{ (8x^3 - 6x^2 + 2x)/(4x - 1)}}} is {{{2x^2-x+(1/4)}}} with remainder (1/4).


The secret is to put the working out under itself, in neat columns, so you can see it all. I cannot do that here :-( However, this website 


http://www.rfbarrow.btinternet.co.uk/htmasa2/AlgDiv1.htm


shows the process perfectly.

jon.