# Thread: polynomial having real zero...

1. ## polynomial having real zero...

Suppose that $\displaystyle p(x)$ is a polynomial with real coefficients such that for some positive real numbers $\displaystyle c,d$and for all natural numbers $\displaystyle n$ ,we have $\displaystyle c|n|^{3}\leq |p(n)|\leq d|n|^{3}$.
Prove that $\displaystyle p(x)$ has a real zero.

HELP!

2. i think we have to somehow use the concepts of limits
or something like this.....divide the whole inequality by 4,and take the limits tending to infinity

3. Do natural numbers include zero? (There are different conventions.) If yes, then the claim follows immediately from the inequality.

Edit: Regardless of the question above, p must be a cubic polynomial. This is because the limit of the ratio of two polynomials when x tends to infinity is determined by the degrees of the polynomials and their leading coefficients. Namely, if the degree of the nominator is less than that of the denominator, the limit is 0. If it is greater, the limit is plus/minus infinity. If the degrees are the same, the limit is the ratio of the leading coefficients. Let us know if you don't know why this is so.

So, if the degree of p were anything other than 3, the inequalities could not hold for all n. And cubic polynomials always have a real root.