# polynomial question

• Oct 30th 2009, 04:23 PM
dannyboycurtis
polynomial question
Hi So I have to prove the following:
If $p(x)=a_{n}x^n+...+a_{1}x+a_{0}$ with $a_{0}<0$ and $a_{n}>0$ and n is an even positive integer, then prove that p has at least two distinct roots.

So my proof looks a little like this so far:
p is continuous on $\mathbb{R}$.
$\lim_{x\to +\infty}p(x)= +\infty$ and $\lim_{x\to -\infty}p(x)= -\infty$.
Hence $\exists \alpha, \beta \in \mathbb{R}$ such that:
$p(\alpha)<0.
...
At this point Im stuck. I found, via graphing example polynomials, that the negative $a_{0}$ values cause the polynomial to have two distinct roots, but how to state this mathematically to finish the proof?
Any suggestions would be appreciated thanks.
• Oct 30th 2009, 04:38 PM
Enrique2
In your conditions the result is not true, just consider $p(x)=x^3-1$

If n is odd or n even and $a_n<0$, $p(0)<0$, $\lim_{x\to-\infty}p(x)=\lim_{x\to\infty}p(x)=\infty$.Thus there exists $\alpha<0,\beta>0$ such that
$p(0)<0, and $p(0)<0.

Now you only need Bolzano.
• Oct 30th 2009, 04:55 PM
dannyboycurtis
the hypothesis states that n is even,
p(x) = x^3 - 1 does not satisfy the hypothesis
• Oct 30th 2009, 05:03 PM
Enrique2
You are right!!!

I was translating odd an even to spanish strangely!

OK, if n is even, then with your hypothesis $\lim_{x\to-\infty}p(x)=\lim_{x\to-\infty}p(x)=+\infty$ thus the rest of the argument given in my post works!
• Oct 31st 2009, 01:42 AM
HallsofIvy
Quote:

Originally Posted by dannyboycurtis
Hi So I have to prove the following:
If $p(x)=a_{n}x^n+...+a_{1}x+a_{0}$ with $a_{0}<0$ and $a_{n}>0$ and n is an even positive integer, then prove that p has at least two distinct roots.

So my proof looks a little like this so far:
p is continuous on $\mathbb{R}$.
$\lim_{x\to +\infty}p(x)= +\infty$ and $\lim_{x\to -\infty}p(x)= -\infty$.
Hence $\exists \alpha, \beta \in \mathbb{R}$ such that:
$p(\alpha)<0.
...
At this point Im stuck. I found, via graphing example polynomials, that the negative $a_{0}$ values cause the polynomial to have two distinct roots, but how to state this mathematically to finish the proof?
Any suggestions would be appreciated thanks.

Use the intermediate value property: If f(x) is a continous function on [a,b], then it takes on all values between f(a) and f(b).

f(0)< 0 while, for some large x, f(x)> 0.

f(0)< 0 while for some larger negative x, f(x)> 0.
• Nov 1st 2009, 06:23 PM
dannyboycurtis
how could I show that if n was odd, p would NOT have an odd number of distinct roots?
• Nov 2nd 2009, 03:17 AM
HallsofIvy
You wouldn't. It is not true. For example, $f(x)= x^3- x$ has odd degree and an odd number of distinct roots.

If you mean "if n is odd, p may not have an odd number of distinct roots", use the counter example $p(x)= x^3$.
• Nov 2nd 2009, 04:56 PM
dannyboycurtis
sorry I miswrote that question,
how would I prove that if n was odd, that p would not have 2 distinct roots?
• Nov 3rd 2009, 03:18 AM
HallsofIvy
Again, you don't prove it. It's not true. For example, $x^3- x= 0$ has roots x= 0, x= 1, x= -1, three distinct roots, so it would certainly also be correct to say it has 2 distinct roots.

If you mean "does not have exactly two distinct roots" it is still not true. $x^3- 2x^2+ x= 0$ has exactly x= 0 and x= 1 as distinct roots.

What is true is that a polynomial, of odd degree, with real coefficients cannot have exactly two (or any even number) real roots (distinct or not). But that is not anything like what you have been saying.