Pre-cal: polynomial..a value of k such that the function has three zeroes?

Question

f(x)= 1/5 (x^5 - 20x^3 + 64x) + k

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is there a value of k such that the function has three zeroes?

Answer 1

Yes.

Graph it.  Clearly for:

| x | > 10 there are only three zeroes.

For | x | < 9 there are five zeroes.

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Since the question only asked if there was such a value k, I simply gave an approximate range.  I didn't try to solve for the value of k where it switches from five to three zeroes.