How do you divide #(x^3-7x-6) / (x+1)#?

2 Answers
Jul 1, 2018

The quotient is #=x^2-x-6# and the remainder is #=0#

Explanation:

Perform a long division

#color(white)(aaaa)##x^3+0x^2-7x-6##color(white)(aaaa)##|##x+1#

#color(white)(aaaa)##x^3+x^2##color(white)(aaaaaaaaaaaaa)##|##x^2-x-6#

#color(white)(aaaaa)##0-x^2-7x#

#color(white)(aaaaaaa)##-x^2-x#

#color(white)(aaaaaaa)##-0-6x-6#

#color(white)(aaaaaaaaaaa)##-6x-6#

#color(white)(aaaaaaaaaaaa)##-0-0#

The quotient is #=x^2-x-6# and the remainder is #=0#

#(x^3+0x^2-7x-6)/(x+1)=x^2-x-6#

Jul 1, 2018

#x^2-x-6#

Explanation:

Given: #(x^3-7x-6)/(x+1)#

Using place keepers of no value. Example: #0x^2#

#color(white)("ddddd.ddddd.d")x^3+0x^2-7x-6#
#color(red)(+x^2)(x+1) ->ul(x^3+x^2 larr" Subtract") #
#color(white)("dddddddddddd.")0 -x^2-7x-6#
#color(red)(-x)(x+1)->color(white)("dd.")ul(-x^2-x larr" Subtract")#
#color(white)("dddddddddddddddd")0-6x-6#
#color(red)(-6)(x+1)->color(white)("dddddd.")ul(-6x-6larr" Subtract") #
#color(white)("dddddddddddddddddddd")0+0 larr" Remainder"#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#(x^3-7x-6)/(x+1)=color(red)(x^2-x-6)#