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:

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#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.

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