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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).
color(white)()
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.
