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

1 Answer
Alan P.
Jul 9, 2015

#(2x^3-3x^2+3x-4) div (x-2) = (2x^2 - x + 1) R:(-2)#

Explanation:

enter image source here

#color(red)("[A]")##x# (the leading term of #(x-2)# goes into #2x^3# (the leading term of #(2x^3-3x^2+3x-4)#, #(2x^2)# times.

#color(red)("[B]")#Multiply #(x-2)# by #(2x^2)# and subtract from the dividend.

#color(red)("[C]")#"Bring down" the next term (#3x#); the first term of #(x-2)# goes into #(-x^2+3x)#, #(-x)# times.

#color(red)("[D]")#Multiply #(x-2)# by #(-x)# and subtract.

#color(red)("[E]")#"Bring down" the final term (#-4#); divide the first term of #(x-2)# (that is, #x#) into #(1x-4)# giving #1#; multiply #(x-2)# by #(1)#; and subtract giving the remainder.

Related questions
See all questions in Long Division of Polynomials
Impact of this question
2233 views around the world
You can reuse this answer
Creative Commons License