Factor theorem: if x-k is a factor of p(x), then p(k) = 0. The same can be said in reverse: if p(k) = 0, then x-k is a factor of p(x).
In this case, x-1 = x-k so k = 1. We'll plug this into f(x) below
f(x) = 14x^99 - 65x^56 + 51
f(1) = 14(1)^99 - 65(1)^56 + 51 ... replace every x with 1
f(1) = 14(1) - 65(1) + 51
f(1) = 14 - 65 + 51
f(1) = -51 + 51
f(1) = 0
The result is 0, so h(x) = (x-1) is a factor of f(x) = 14x^99-65x^56+51
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Answer: Choice A) yes
