Polynomial with $k$ positive and $k+1$ negative zeroes

Question

A polynomial of degree $n$ is written in standard form. All numerical coefficients are positive. It has $k$ positive zeroes and $k+1$ negative zeroes. $0$ is not a zero of the polynomial. What can we deduce about $n$?

The answer in the book said

$n=1$

I, however, got the answer

$n=1,5,9,13,17,21,\dots$

How do you eliminate the leftover cases ($5,9,13,\dots)$?

EDIT: Note, as per the comments, that $n$ is meant to be $2k+1$.

Presumably you've considered the last coefficient to eliminate the cases $3,7,11...$. Perhaps consider another coefficient? — preferred_anon, Nov 03 '15 at 10:06
Let the polynomial be $P(x)$.
If all numerical coefficients are positive, what can we say about $P(x)$ when $x>0$? — Element118, Nov 03 '15 at 10:17
@GerryMyerson Yes, $n=2k+1$ as we have $n$ roots, of which $k$ are positive and $k+1$ negative. — ghosts_in_the_code, Nov 03 '15 at 15:43
Then please edit that information into the question. The way the question is currently written, the polynomial $(x+1)(x^2+1)$ satisfies the conditions with $n=3$ and $k=0$. — Gerry Myerson, Nov 03 '15 at 22:20

score 2 · Accepted Answer · answered Nov 03 '15 at 11:16

2

If a polynomial $f(X)$ has positive coefficients, and $0$ is not a root, it has only negative roots. Hence $k=0$. Thus is decomposition into a product of irreducible factors over $\mathbf R$ has the form: $$f(X)=c(X+a)p_1(X)\dotsm p_r(X),$$ where the $p_i(X)$ are quadratic polynomials with a negative discriminant. Thus the degree $n$ is odd.

answered Nov 03 '15 at 11:16

Bernard

175,478

What about $(x-100)^2(x+1)^3$? It has 3 negative and 2 positive roots. $k=2$ and $n=5$ – ghosts_in_the_code Nov 03 '15 at 14:07
I supposed the O.P. meant distinct roots. And this one has a negative coefficient. – Bernard Nov 03 '15 at 14:27
What about $(x+100)^3(x-1)^2$ – ghosts_in_the_code Nov 03 '15 at 15:38
As I said, I supposed the hypotheses were about distinct roots.. – Bernard Nov 03 '15 at 15:40
It could also be written as $(x+99.9)(x+100)(x+100.1)(x-0.99)(x-1.01)$. This now has $5$ distinct roots. – ghosts_in_the_code Nov 03 '15 at 15:42
It also has a negative coefficient. If a polynomial has positive coefficients, it can't have a positive root. – Bernard Nov 03 '15 at 15:51
Ok, I got it. If all coefficients are positive, and you multiply these with powers of positive numbers (roots), you still get a positive answer (and not zero). Thanks! – ghosts_in_the_code Nov 03 '15 at 16:05

Polynomial with $k$ positive and $k+1$ negative zeroes

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