# Thread: polynomial question

1. ## polynomial question

Hi So I have to prove the following:
If $\displaystyle p(x)=a_{n}x^n+...+a_{1}x+a_{0}$ with $\displaystyle a_{0}<0$ and $\displaystyle a_{n}>0$ 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 $\displaystyle \mathbb{R}$.
$\displaystyle \lim_{x\to +\infty}p(x)= +\infty$ and $\displaystyle \lim_{x\to -\infty}p(x)= -\infty$.
Hence $\displaystyle \exists \alpha, \beta \in \mathbb{R}$ such that:
$\displaystyle p(\alpha)<0<p(\beta)$.
...
At this point Im stuck. I found, via graphing example polynomials, that the negative $\displaystyle a_{0}$ 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.

2. In your conditions the result is not true, just consider $\displaystyle p(x)=x^3-1$

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

Now you only need Bolzano.

3. the hypothesis states that n is even,
p(x) = x^3 - 1 does not satisfy the hypothesis

4. You are right!!!

I was translating odd an even to spanish strangely!

OK, if n is even, then with your hypothesis $\displaystyle \lim_{x\to-\infty}p(x)=\lim_{x\to-\infty}p(x)=+\infty$ thus the rest of the argument given in my post works!

5. Originally Posted by dannyboycurtis
Hi So I have to prove the following:
If $\displaystyle p(x)=a_{n}x^n+...+a_{1}x+a_{0}$ with $\displaystyle a_{0}<0$ and $\displaystyle a_{n}>0$ 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 $\displaystyle \mathbb{R}$.
$\displaystyle \lim_{x\to +\infty}p(x)= +\infty$ and $\displaystyle \lim_{x\to -\infty}p(x)= -\infty$.
Hence $\displaystyle \exists \alpha, \beta \in \mathbb{R}$ such that:
$\displaystyle p(\alpha)<0<p(\beta)$.
...
At this point Im stuck. I found, via graphing example polynomials, that the negative $\displaystyle a_{0}$ 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.

6. how could I show that if n was odd, p would NOT have an odd number of distinct roots?

7. You wouldn't. It is not true. For example, $\displaystyle f(x)= x^3- x$ 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 $\displaystyle p(x)= x^3$.

8. sorry I miswrote that question,
how would I prove that if n was odd, that p would not have 2 distinct roots?

9. Again, you don't prove it. It's not true. For example, $\displaystyle x^3- x= 0$ 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. $\displaystyle x^3- 2x^2+ x= 0$ 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.