# Polynomial Arithmetic

Let P(x) be a real polynomial of degree less than or equal to 4, such that there are at least 5 distinct solutions to P(x) = 5. Find P(5).

I've been trying to wrap my head around this supposedly basic problem, but I keep running into a blank wall. I can get at most 4 distinct solutions to P(x) = 5 for a given degree 4 polynomial, but I keep coming to the conclusion that 5 distinct solutions is impossible. For example, I used P(x) = x^4 - 2x^2 + 5.5. Given this P(x) it is indeed possible to find 4 distinct x's for which P(x) = 5. What am I missing here? Is there perhaps a complex solution, and if so, how would I find it?

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## 1 Answer

Hint: What if there were much more than five distinct solutions to $P(x)=5$? Can you think of how this could be possible? Try to think of lower degree polynomials, rather than degree four.

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Wow, Ok, I think I might have been the world's biggest idiot. Correct me if I am wrong, but a polynomial of degree zero is still a polynomial. If that statement is true, then all that remains to be done is allow P(x) = 5, and thus all the requirements on P are met, and I can confidently say that P(5) = 5. –  user79790 Aug 12 '13 at 23:57
@user79790 That's pretty much it. The key point is a non-zero $n$-degree polynomial can have at most $n$ roots, but $P(x)-5$ has $5$ roots. Since $P(x)-5$ is of degree less than or equal to $4$ it must be zero. –  JSchlather Aug 13 '13 at 0:16