Proving a polynomial has 3 roots, what does it means?

Hi,

I am asked to prove a 4th degree polynomial:

p(x) = x^4+x^3-3

Has at least 2 roots...

And that some other equation doesn't roots in interval (-oo,-2)

and that it has 3 roots in (-oo,+oo)

what does it means that an equation has roots? or to find it?

like with the quadratic formula that you find the +- x's ? (is has either 1 or 2 right?)

Is there a simple way to find roots of such equations (with x^3 or higher degree)?

Thanks (again).

Re: Proving a polynomial has 3 roots, what does it means?

Hey ryu1.

Hint: Consider the second derivative being 0 and how that correlates with the information about the roots.

Also do the roots have to be real or can they be complex as well?

Re: Proving a polynomial has 3 roots, what does it means?

Quote:

what does it means that an equation has roots? or to find it?

A root of a polynomial $\displaystyle f(x)$ is a number $\displaystyle a$ such that $\displaystyle f(a)=0$. For example the roots of a quadratic equation is given by the quadratic formula. For a more concrete example, the number $\displaystyle a=1$ is a root for the polynomial $\displaystyle f(x)=x^3-1$ since $\displaystyle f(a)=a^3-1=1-1=0$.

Quote:

Is there a simple way to find roots of such equations (with x^3 or higher degree)?

Not in general. There are some truly monstrous equations for general polynomials of degree 3 and 4. However, it has been proven that such an formula is impossible for degrees 5 and higher (not just unknown, but cannot possibly exist). So your best bet in these questions is to look for properties for your specific polynomial, not a general one. I cannot guarantee I can give you an answer which you can understand without knowing how much you know, but here is an attempt.

Quote:

I am asked to prove a 4th degree polynomial:

p(x) = x^4+x^3-3

Has at least 2 roots...

This approach is different from chiro's, so you can choose what you like. First, we note that $\displaystyle p(x)\to \infty $ as $\displaystyle x\to \pm \infty$, so there is some number $\displaystyle N>0$ such that $\displaystyle f(-N)>0$ and $\displaystyle f(N)>0$. Next, we see that $\displaystyle p(x)$ is negative at some point, for example $\displaystyle p(0)=-3$. Applying the intermediate value theorem on $\displaystyle [-N, 0]$ and $\displaystyle [0,N]$ gives the result.