How do you divide #(5x^3-7x^2-4x+1)/(3x-1) #?

#(3x-1)# will go into #(5x^3-7x^2-4x+1)# a total of #5/3x^2# times.

#5/3x^2 (3x - 1) = (5x^3 - 5/3x^2)#

This will leave a remainder of:

#" "" "" "" ""5/3x^2#
#" "" "" "" ""----------------------------"#
#3x-1" "| 5x^3-7x^2-4x+1#
#" "" "" "-(5x^3 - 5/3x^2)#
#" "" "" "" ""--------------"#
#" "" "" "" "" "" "-16/3x^2#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

So now we're left with #-16/3x^2-4x+1#.

#(3x-1)# will go into this #-16/9x# times.

#-16/9x(3x - 1) = (-16/3x^2 + 16/9x)#

This will leave a remainder of:

#" "" "" "" ""5/3x^2 - 16/9x#
#" "" "" "" ""----------------------------"#
#3x-1" "| 5x^3-7x^2-4x+1#
#" "" "" "-(5x^3 - 5/3x^2)#
#" "" "" "" ""-----------------"#
#" "" "" "" "" "" "-16/3x^2 -4x+1#
#" "" "" "" "-(-16/3x^2+16/9x)#
#" "" "" "" "" ""-------------------------"#
#" "" "" "" "" "" "" "" "" "-52/9x#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

So now we're left with #-52/9x + 1#.

#(3x-1)# will go into this #-52/27# times.

#-52/27(3x-1) = (-52/9x + 52/27)#

This will leave a remainder of:

#" "" "" "" ""5/3x^2 - 16/9x - 52/27#
#" "" "" "" ""----------------------------"#
#3x-1" "| 5x^3-7x^2-4x+1#
#" "" "" "-(5x^3 - 5/3x^2)#
#" "" "" "" ""-----------------"#
#" "" "" "" "" "" "-16/3x^2 -4x+1#
#" "" "" "" "-(-16/3x^2+16/9x)#
#" "" "" "" "" ""-------------------------"#
#" "" "" "" "" "" "" "" "" "-52/9x + 1#
#" "" "" "" "" "" "" "-(-52/9x + 52/27)#
#" "" "" "" "" "" "" ""-------------------------"#
#color(white)"MMMMMMMMMMMMMM-"-25/27#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Since the degree of this remainder is smaller than the degree of our divisor, we will just divide it by the divisor as the last term in our answer:

#5/3x^2 - 16/9x - 52/27 -(25/27)/(3x-1)#

#5/3x^2 - 16/9x - 52/27 -25/(27(3x-1))#

Final Answer