How do you divide #(n^4+5n^3-6n+3)div(n+3)# using synthetic division?

1 Answer
May 30, 2017

Answer:

The remainder is #=-33# and the quotient is #=n^3+2n^2-6n+12#

Explanation:

Let's perform the synthetic division

#color(white)(aaaa)##-3##color(white)(aaaaa)##|##color(white)(aaaa)##1##color(white)(aaaaaa)##5##color(white)(aaaaaa)##0##color(white)(aaaaa)##-6##color(white)(aaaaaaa)##3#
#color(white)(aaaaaaaaaaaa)#_________

#color(white)(aaaa)##color(white)(aaaaaaa)##|##color(white)(aaaa)##color(white)(aaaaaa)##-3##color(white)(aaaaaa)##-6##color(white)(aaaa)##18##color(white)(aaaaa)##-36#
#color(white)(aaaaaaaaaaaa)#________

#color(white)(aaaa)##color(white)(aaaaaaa)##|##color(white)(aaaa)##1##color(white)(aaaaaaa)##2##color(white)(aaaaaa)##-6##color(white)(aaaa)##12##color(white)(aaaaa)##color(red)(-33)#

#(n^4+5n^3-6n+3)/(n+3)=(n^3+2n^2-6n+12)-33/(n+3)#

The remainder is #=-33# and the quotient is #=n^3+2n^2-6n+12#