How do you divide #(-x^3 - x^2-6x+5 )/((-x + 10 )#?

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Answer 1 · 2016-02-21T17:35:31

#(−x^3−x^2−6x+5)/(-x+10)# is

#x^2+11x+116-1155/(-x+10)#

To divide #−x^3−x^2−6x+5# by #-x+10#, first we observe that #-x# goes #(-x^3)/-x=x^2# times in #-x^3#.

Hence, #x^2(-x+10)-10x^2−x^2−6x+5# or
(note we have added and subtracted #10x^2) #x^2(-x+10)-11x^2−6x+5#

As #-11x^2/-x=11x#, above can be written is

#x^2(-x+10)+11x(-x+10)-110x−6x+5# or

#x^2(-x+10)+11x(-x+10)-116x+5#

and as #-116x/-x=116#, above can be written as

#x^2(-x+10)+11x(-x+10)+116x(-x+10)-1160+5# or

#x^2(-x+10)+11x(-x+10)+116x(-x+10)-1155# or

#(x^2+11x+116)(-x+10)-1155# or

Hence #(−x^3−x^2−6x+5)/(-x+10)# is

#x^2+11x+116-1155/(-x+10)#