How do you divide #4x^3 + 3x + 5# by #2x-3#?

1 Answer

#(4x^3+3x+5)/(2x-3)=2x^2+3x+6+23/(2x-3)#

Explanation:

First rearrange from the highest degree term to the lowest

both dividend #(4x^3+3x+5)=(4x^3+0*x^2+3x+5)# and divisor #(2x-3)#

#" " " " " " " "underline(2x^2+3x+6" " " " " " " " ")#
#2x-3" "|~4x^3+0*x^2+3x+5#
#" " " " " " " "underline(4x^3-6x^2" " " " " " " " " ")#
#" " " " " " " " " " " " " "6x^2+3x+5#
#" " " " " " " " " " " " " "underline(6x^2-9x" " " " " )#
#" " " " " " " " " " " " " " " " " " "12x+5#
#" " " " " " " " " " " " " " " " " " "underline(12x-18)#
#" " " " " " " " " " " " " " " " " " " " " " " 23" " "#the remainder

so that the final answer

#"dividend"/"divisor"="quotient"+"remainder"/"divisor"#

#(4x^3+3x+5)/(2x-3)=2x^2+3x+6+23/(2x-3)#