How do you divide #(n^3-9n-3)div(n+3)# using synthetic division?

1 Answer
Oct 26, 2016

Answer:

The division gives #(n^2-3n)-3/(n+3)#

Explanation:

Let's do the long division

#n^3##color(white)(aaaaa)##-9n-3##∣##n+3#
#n^3+3n^2##color(white)(aaaaaaa)##∣##n^2-3n#
#0-3n^2-9n#
#color(white)(aaaaa)##3n^2-9n#
#color(white)(aaaaaa)##0+0-3#

So there is a remainder of #-3#

So #(n^3-9n-3)/(n+3)=(n^2-3n)-3/(n+3)#
We can also obtain the remainder
let #f(n)=n^3-9n-3#
then #f(-3)=-27+27-3=-3#