I like to just long divide the coefficients, not forgetting to include #0#'s for missing powers of #x#. In addition, I'm not keen on fractions, so I will choose to multiply the numerator by #2# first, then divide by #2# at the end.
#2 xx (4x^4+4x^3-8x+2) = 8x^4+8x^3+0x^2-16x+4#
This process is similar to long division of decimal numbers.
Write the dividend #(8, 8, 0, -16, 4)# under the bar and the divisor #(2, -3, 1)# to the left of the bar.
Choose the first term #color(blue)(4)# of the quotient so that when multiplied by the divisor it matches the leading term of the dividend.
Write the product #(8, -12, 4)# under the dividend and subtract it to get a remainder. Bring down the next term #-16# of the dividend alongside it to give the running remainder.
Choose the next term #color(blue)(10)# of the quotient so that when multiplied by the divisor it matches the first term of the running remainder.
Write the product #(20, -30, 10)# under the running remainder and subtract it to get the new running remainder. Bring down the next term #4# of the dividend alongside it.
Choose the final term #color(blue)(13)# of the quotient as before, write down the product #(26, -39, 13)# and subtract it to give the final remainder #color(red)("("13, -9")")#.
So:
#8x^4+8x^3-16x+4 = (2x^2-3x+1)(4x^2+10x+13)+(13x-9)#
Dividing by #2#:
#4x^4+4x^3-8x+2#
#= (2x^2-3x+1)(2x^2+5x+13/2) + (13/2x-9/2)#
In other words:
#(4x^4+4x^3-8x+2)/(2x^2-3x+1)#
#=(2x^2+5x+13/2) + (13/2x-9/2)/(2x^2-3x+1)#
