From the given #(-3x^3+12x^2-14x-6)/(x+1)#

We have our

dividend #-3x^3+12x^2-14x-6#

divisor #x+1#

Let us begin the

#color(red)("Long Division Method")#

#" " " " " "underline(-3x^2+15x-29)larr#the quotient

#x+1|~-3x^3+12x^2-14x-6#

#" " " " " "underline(-3x^3-3x^2" " " " " " " " " " ")#

#" " " " " " " " " 0"" +15x^2-14x-6#

#" " " " " " " " " " " "underline(15x^2+15x" " " " ")#

#" " " " " " " " " " " " " "0"" -29x-6#

#" " " " " " " " " " " " " " " " underline(-29x-29)#

#" " " " " " " " " " " " " " " " " " " 0""+23 larr#the remainder

We write our final answer

#(-3x^3+12x^2-14x-6)/(x+1)=-3x^2+15x-29+23/(x+1)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(red)("Another method is by Synthetic Division")#

Our trial divisor comes from the divisor #x+1#. Simply equate this to zero

#x+1=0# and #x=-1#

Use the numerical coefficients of the dividend

#" "x^3" " " "" "x^2" " " " " "x^1" " " " " "x^0#

#-3" " " "+12" " " "-14" " " "-6" " "|~-1larr#trial divisor

#underline(" " " " " " " "+3" " " "-15" " " "+29" " )#

#-3" " " "+15" " " "-29" " " "+23larr #the remainder

The last set of numbers #-3, +15, -29# are the numerical coefficients of #x^2, x^1, x^0# repectively so that our quotient is

#-3x^2+15x-29#

since the remainder is #+23#, we write the last term #(+23)/(x+1)#

Our final answer is

#(-3x^3+12x^2-14x-6)/(x+1)=-3x^2+15x-29+23/(x+1)#

God bless....I hope the explanation is useful.