How do you use the remainder theorem to find the remainder for the division #(n^4-n^3-10n^2+4n+24)div(n+2)#?

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Answer 1 · 2016-10-10T05:22:53

#f(-2) = 0""#
There is no remainder and #(n+2)# is a factor of #n^4 -n^3-10n^2 +4n+24#

Let #f(n) =n^4 -n^3-10n^2 +4n+24#

Setting # n+2 = 0" "# gives #n=-2#

Substitute #n=-2" into " f (n)""# to find the remainder.

#f(-2) = (-2)^4 -(-2)^3-10(-2)^2 +4(-2)+24#

#f(-2) = 16+8-40-8+24 =0#

As this is 0, it means there is no remainder.

This tells us that #n+2# is a factor of #n^4 -n^3-10n^2 +4n+24# and therefore divides into it exactly without leaving a remainder.