How do you divide #-x ^ { 3} + 3x + 6# by # x + 3#?

1 Answer
Tony B
May 16, 2017

Due formatting issues on this site I am showing division by using its reverse (multiplication)

#-x^2+3x-3+15/(x+3)#

Explanation:

Really this is the same as long division but you write parts of the process in different places.

NOTE: I use place holder to make lining up easier. Example: #0x^2#

#" "-x^3+0x^2+3x+6#
#color(magenta)(-x^2)(x+3)->color(white)(.)ul( -x^3-3x^2) larr" Subtract"#
#" "0+3x^2+3x+6#
#color(white)()color(magenta)(+3x)(x+3) ->" "ul(3x^2+9x) larr" Subtract"#
#" "0-3x+6#
#color(magenta)(-3)(x+3)->" "ul( -3x-9)larr" Subtract"#
#" "0color(magenta)(+15larr" Remainder")#

#color(magenta)(-x^2+3x-3+15/(x+3))#

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