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

1 Answer
Dec 15, 2015

#x^3 - 3/2 x^2 + 9/4 x - 11/8# with remainder #41/8#

Explanation:

I know that there are in some countries, a different notation of long polynomial division is being used. Let me use the notation that I'm most familiar with, I hope that it will be no problem for you to convert it into your prefered notation.

#color(white)(xii)(2x^4 color(white)(xxxxxxxxxxx)+ 4x + 1) -: (2x + 3) = x^3 - 3/2 x^2 + 9/4 x - 11/8#
# - (2x^4+ 3x^3)#
# color(white)(xx)(color(white)(xxxxxxx))/()#
# color(white)(xxxxx)- 3x^3 #
# color(white)(xx) - (-3x^3 - 9/2 x^2) #
# color(white)(xxxx)(color(white)(xxxxxxxxxxx))/()#
# color(white)(xxxxxxxxxxxx)9/2 x^2 + 4x#
# color(white)(xxxxxxxxx)-(9/2 x^2 + 27/4 x)#
# color(white)(xxxxxxxxxxx)(color(white)(xxxxxxxxxx))/()#
# color(white)(xxxxxxxxxxxxxxx) -11/4 x + color(white)(i) 1#
# color(white)(xxxxxxxxxxxx) -(-11/4 x -33/8)#
# color(white)(xxxxxxxxxxxxxx)(color(white)(xxxxxxxxxxx))/()#
# color(white)(xxxxxxxxxxxxxxxxxxxxxxx)41/8#

Thus, your quotient is

#x^3 - 3/2 x^2 + 9/4 x - 11/8#

and your remainder is #41/8#.

In total,

#(x^4 + 4x + 1)/(2x + 3) = x^3 - 3/2 x^2 + 9/4 x - 11/8 + 41/(8(2x+3))#