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#f(n) = n^4+9n^3+14n^2+50n+9#

By the remainder theorem, #(n-a)# is a factor of #f(n)# if and only if #f(a) = 0#.

So in our case #(n+8)# is a factor if #f(-8) = 0#.

Note that if #n# is even, then all of the first four terms are even, but the constant term is odd. So #f(n)# is odd and thus non-zero.

So #(n+8)# is not a factor of #f(n)#.

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**Bonus**

By the rational roots theorem, any *rational* zeros of #f(n)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #9# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible *rational* zeros are:

#+-1, +-3, +-9#

When #n=+1# we have #3# odd terms and two even ones, so the sum is odd and therefore non-zero.

When #n=+-3# then first three terms and the last (constant) term are all divisible by #9#, but the fourth term is not. So #f(n)# is not divisible by #9# and therefore non-zero.

When #n=+-9# then the first three terms are all divisible by #9^2#. The sum of the last two terms is #9(1+-50)#, which is a multiple of #9# and possibly #27#, but not #81#. So #f(n)# is not divisible by #9^2# and therefore non-zero.

Hence #f(n)# has no *rational* zeros, so no rational linear factors.