I know that in some countries, the long division of polynoms is being written in a different way. I will use the one that is familiar for me though and hope that you can convert it into your notation easily. :-)
Let me explain to you how to do the long division - and if you already know, you can skip to the end of the answer to see the whole division there.
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1) First, check that the the terms in your numerator and denominator are ordered by the power of #x# - this is already the case for you.
2) Now, take the first term - the one with the biggest power - from the numerator and divide it by the first term - the one with the biggest power - from the denominator.
In your case, it's #2x^3 -: x = 2x^2#, so you get:
#color(white)(xii)(color(red)(2x^3) - 3x^2 color(white)(x)+ 3x - 4) -: (color(red)(x)-2) = color(red)(2x^2) #
3) As next, you need to backwards multiply your new result (#2x^2#) with the denominator, so compute #2x^2 * (x-2) = 2x^3 - 4x^2# and subtract it from your numerator:
#color(white)(xii)(2x^3 - 3x^2 color(white)(x) + color(grey)(3x) - 4) -: (color(blue)(x-2)) = color(blue)(2x^2) #
#-(color(blue)(2x^3 - 4x^2))#
#color(white)(x)(color(white)(xxxxxxxx))/()#
#color(white)(xxxxxxx) x^2 + color(grey)(3x)#
4) As expected, you don't have a #x^3# term anymore, and your new term with the highest power of #x# in the numerator is #x^2#, so you need to divide #x^2# by #x#:
#color(white)(xii)(2x^3 - 3x^2 color(white)(x) + 3x - 4) -: (color(red)(x)-2) = 2x^2 + color(red)(x)#
#-(2x^3 - 4x^2)#
#color(white)(x)(color(white)(xxxxxxxx))/()#
#color(white)(xxxxxxx) color(red)(x^2) + 3x#
5) Again, do the backward multiplication and subtract the result:
#color(white)(xii)(2x^3 - 3x^2 color(white)(x) + 3x color(white)(x) color(grey)(- 4)) -: (color(blue)(x-2)) = 2x^2 + color(blue)(x)#
#-(2x^3 - 4x^2)#
#color(white)(x)(color(white)(xxxxxxxx))/()#
#color(white)(xxxxxxx) x^2 + 3x#
#color(white)(xxxiix) -(color(blue)(x^2 - 2x))#
#color(white)(xxxxxx)(color(white)(xxxxxx))/()#
#color(white)(xxxxxxxxxxx) 5x color(white)(x)color(grey)(- 4)#
6) As next, divide #5x# by #x#...
#color(white)(xii)(2x^3 - 3x^2 color(white)(x) + 3x color(white)(x) - 4) -: (color(red)(x)-2) = 2x^2 + x + color(red)(5)#
#-(2x^3 - 4x^2)#
#color(white)(x)(color(white)(xxxxxxxx))/()#
#color(white)(xxxxxxx) x^2 + 3x#
#color(white)(xxxiix) -(x^2 - 2x)#
#color(white)(xxxxxx)(color(white)(xxxxxx))/()#
#color(white)(xxxxxxxxxxx) color(red)(5x) color(white)(x)- 4#
7)... and perform the backward multiplication and subtraction...
#color(white)(xii)(2x^3 - 3x^2 color(white)(x) + 3x color(white)(x) - 4) -: (color(blue)(x-2)) = 2x^2 + x + color(blue)(5)#
#-(2x^3 - 4x^2)#
#color(white)(x)(color(white)(xxxxxxxx))/()#
#color(white)(xxxxxxx) x^2 + 3x#
#color(white)(xxxiix) -(x^2 - 2x)#
#color(white)(xxxxxx)(color(white)(xxxxxx))/()#
#color(white)(xxxxxxxxxxx) 5x color(white)(x)- 4#
#color(white)(xxxxxxxxi) -(color(blue)(5x color(white)(x)- 10))#
#color(white)(xxxxxxxxxxx)(color(white)(xxxxxxx))/()#
#color(white)(xxxxxxxxxxxxxxxxi) 6#
8) At this point, as you can't divide #6 -: x# anymore, you have finished your division and have the remainder #6# at the end:
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Solution:
#(2x^3 - 3x^2+3x-4)/(x-2) = 2x^2 + x + 5 + (6)/(x-2)#
