To prove a polynomial $P(x)$ has a real zero.

Question

Q. Suppose that $P(x)$ is a polynomial with real coefficients such that for some positive real number $c,d$ and for all natural numbers $n$, we have $$c|n|^3 \le |P(n)|\le d|n|^3$$

Prove that $P(x)$ has a real zero.

Here I am not really sure how I shall prove this, shall I take any random polynomial as for example a quadratic equation/cubic equation to prove this? And if I do so, it won't be "general" you see? How will it prove that other polynomials satisfy this or not

Note: This is a question from subjective test and here they didn't gave any hint, so I would need to prove it by writing it down step wise, any help would be greatly appreciate.

gt6989b · Accepted Answer · 2017-05-08T15:49:51.110

7

HINT

You can prove that $P$ is a cubic. Then, the fundamental theorem of Algebra implies it has at least 1 real root.

UPDATE

To prove $P$ is cubic, assume that it is not. Then it should have a term of higher degree (which leads to a contradiction with the upper bound) or the max degree term is strictly smaller than 3 (which leads to a contradiction with the lower bound).

edited May 08 '17 at 15:49

answered May 08 '17 at 15:41

gt6989b

54,422

How am I supposed to prove the function is cubic? – Iti Shree May 08 '17 at 15:45
@ItiShree see the update, you need to fill in how to get the contradictions, but i think they are pretty evident. – gt6989b May 08 '17 at 15:50

To prove a polynomial $P(x)$ has a real zero.

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