f(n)=n4+9n3+14n2+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).
Bonus
By the rational roots theorem, any rational zeros of f(n) are expressible in the form pq 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 92. 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 92 and therefore non-zero.
Hence f(n) has no rational zeros, so no rational linear factors.
