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## Main Question or Discussion Point

**Pol degree 3, at most 3 real roots??**

Well, i am trying to prove the following, indeed i think i have proved it, but it is just that it looks a little bit long, and i was wondering whether if first of all i have done it right, and second if there is any other method proving this using calculus tools.

**Problem:**Show that a polynomial of degree 3 has at most three real roots.

__Proof:__Let : [tex] f(x)=ax^3+bx^2+cx+d[/tex]. Hence, f is continuous and differentiable on the whole real line. Also its derivative, since it is also a polynomial function of a smaller degree, it is also diff. and cont. on the whole real line.

First let's find all extremums of this function.

[tex]f'(x)=3ax^2+2bx+c=0=>x_1=\frac{-b+\sqrt{b^2-3ac}}{3a},x_2=\frac{-b-\sqrt{b^2-3ac}}{3a}[/tex]

Now, let's furhter suppose that [tex]x_1 \ not \ equal \ to \ x_2[/tex]

So this way we have two distinct critical points.

this basically is possible when [tex] b^2-3ac>0[/tex] so for our purporse lets suppose that this is the case.

Now if a>0 [tex]x_1[/tex] will be a local max, and [tex]x_2[/tex] a local min, wheras if a<0, it will be vice-versa.

Now let's further suppose, for our purporse, that

[tex]f(x_1)*f(x_2)<0[/tex]. So from Intermediate Value Theorem we know that [tex]\exists k \in(x_1,x_2)[/tex] such that f(k)=0. So k is a real root so far.

Also since we showed that the function f, has max two and only two extremums, in our case, it means that the graph of f(x) must cross x-axis on the left of x1 and on the right of x2 (where x1<x2) once and only once. That is once on the left and once on the right. Because if it crossed the x-axis more than once than we would have another cr.point, and also another extremum, which is not possible.

Now, i'll use Rolle's Theorem, and proof by contradiction to show that there also are no more roots between x1 and x2.

Let [tex] n_1,n_2 \in [x_1,x_2][/tex] such that n1 and n2 are different.

[tex]f(n_1)=f(n_2)=0[/tex]. In other words, let n1 and n2 be two distinct roots between x1 and x2.

Now, since f suffices all rolles' theorems conditions we have

[tex]\exists r \in(n_1,n_2) \subset [x_1,x_2][/tex] such that

f'(r)=0.

But this is not possible, since this implies that r is a cr. point that lies between x1 and x2, and we showed above that there cannot be more than two critical points. So the contradiction we arrived at, implies that our assumption that there are two distinct roots between x1 and x2 is wrong. So n1=n2.

This is what we needted to prove.

I don't know whether everything that i have done there follows correctly.

I also tried to generalize this one, and prove that a polynomial of degree n has at most n real roots. I pretty much followed the same pattern as here, but furthermore i used induction first to show that every polynomial of degree n first has n-1 extremumes, then i reasoned as in case of degree 3. I know that i could have proved this one using a different way, also by induction but by supposing that first a polynomial of degree n-1 has at most n-1 roots, and then proving for the case of degree n, but i wanted to use the same methodology as i applied for degree 3.

SO, is what i have done correct?

Thnx in return.