Pol degree 3, at most 3 real roots?

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 : $$f(x)=ax^3+bx^2+cx+d$$. 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.

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

Now, let's furhter suppose that $$x_1 \ not \ equal \ to \ x_2$$

So this way we have two distinct critical points.

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

Now if a>0 $$x_1$$ will be a local max, and $$x_2$$ a local min, wheras if a<0, it will be vice-versa.
Now let's further suppose, for our purporse, that

$$f(x_1)*f(x_2)<0$$. So from Intermediate Value Theorem we know that $$\exists k \in(x_1,x_2)$$ 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 $$n_1,n_2 \in [x_1,x_2]$$ such that n1 and n2 are different.

$$f(n_1)=f(n_2)=0$$. 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

$$\exists r \in(n_1,n_2) \subset [x_1,x_2]$$ 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.

First of all let me say that is an impressive argument!

But you made alot of suppositions in the beginning that need to either be addressed by either arguing that they are WLOG (without loss of generality) or considering the other cases. I'm talking about when you said $$f(x_1)f(x_2) < 0$$. Maybe I just missed the reason why you can assume that, I don't know.

Also the argument is long enough, that it would have been faster to just find the roots explicitly, it's a classic result just not standard topic in math classes.

Well, i know i have made a lot of suppositions, but since the question was asking to show that a polynomial of degree 3 has at most 3 roots, all my suppositions lead to this "perfect" case. In other words i made all the supositions in order to get me to the most rare case when a polynomial would have 3 real roots, i don't know if i am making myself clear at this point. SO that's why i supposed that $$f(x_1)f(x_2)<0$$ because if this wouldn't hold, then defenitely there would not be possible for that polynomial to have 3 real roots, since the graph would not cross the x axis betwheen x1 and x2. So, i made all those suppositions, in order to lead me to the most perfect case where $$f(x)=ax^3+bx^2+cx+d$$ has 3 real roots.
I don't know if i am making myself clear enough?
Moreover i assumed that $$f(x_1)(f(x_2)<0$$ in order to use the Intermediate Value Theorem that would guarantee me that there actually exists a point c in the interval (x1,x2) such that f(c)=0. Otherwise i would not be guaranteed that such a point exists there....and as a result i would fail to show that a pol. of degree 3 has at most 3 real roots. Because at that case i would end up just with two roots.

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Also the argument is long enough, that it would have been faster to just find the roots explicitly, it's a classic result just not standard topic in math classes.
Yeah i could just find it's roots, but then part b) of the question was to generalize this method to a polynomial of degree n, so the method of finding it's roots explicitly would defenitely not work.

And i did generalize this method to a polynomial of degree n, but as you may predict it took me about three full pages A4 format. Because i had to use induction there twice, that's why i did not post the whole thing here. Because if what i did for case of degree 3 holds, then i think even for case n should hold.

Edit: I actually lied about the first part, since i dont actually know how to solve a 3rd degree polynomial in the general case, with the exception when i can figure out some tricks in there.

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Okay I didn't know about part (b).

Exploring the case where there are three real roots doesn't actually prove what you want to prove.

Just because you can find a special case where there are three real roots, doesn't mean that there is no case where there are more than three real roots!

I suggest you say suppose that there are more than three real roots, then show that the polynomial is either constant or higher degree than you are considering. That sounds like proof by contradiction, but it's actually proof by contrapositive.

And then the generalization would be done by induction.

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Vid
The generalization would be a fairly simple induction.

A derivative of a (k+1)-polynomial is a k-polynomial. By induction the derivative has at most k real roots.
The real roots of k can be ordered s.t. c1<=c2<=...<=ck
The k+1 polynomial has a root whenever p(ci) has a different sign then p(c(i+1)) because of IVP. k+1 also has a root if lim x->-inf has a different sign from p(c1) and similarly for positive infinity Thus, the optimal case is when the sequence;
p(-inf), p(c1), p(c2), ... , p(ck), p(inf) alternates signs.
Since the IVP guarantees a root between each of these, we have k+1 roots.

HallsofIvy
Homework Helper
Suppose your third order equation had more than 3 roots. say 4. Write out its linear factors and multiply. What do you get?

Suppose your third order equation had more than 3 roots. say 4. Write out its linear factors and multiply. What do you get?
We get a fourth order equation, right? Which would be a contradiction, since we supposed that our equation is of degree 3, right?

I'll try to prove this using DavidWhitbeck's later as well. HOwever, i still don't see why my proof, although a lill bit tedious and long, doesn't hold? Moreover, i think that we are expecte to somewhere use Rolles theorem to show this too.

DavidWhitbeck argued that what i did there doesn't guarantee us that there is no other root, in other words that that eq of degree 3 doesn't have more than 3 real roots. In contrary i argue that i proved by contradiction why that eq cannot have another root, that is cannot have more than 3 roots.
In order for my proof to be complete, i know that i should consider other cases as well, but not going into details, every other case, besides the one that i elaborated and typed here, leads us to less than 3 roots of that equation. Like if we supposed that f(x1)f(x2)>0, then we defenitely would have only 2 roots. Or if i supposed that D=0, or D<0, then defenitely the eq would have only one real root or no real roots at all, respectively.

So, my question now is, although my proof is long and boring, is it mathematically correct??
I know that math looks for the most nice and short proofs, in case this is possible, and i will follow your advice in doing that, because it looks like i get faster to the result.

Thnx a lot again!

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The generalization would be a fairly simple induction.

A derivative of a (k+1)-polynomial is a k-polynomial. By induction the derivative has at most k real roots.
The real roots of k can be ordered s.t. c1<=c2<=...<=ck
The k+1 polynomial has a root whenever p(ci) has a different sign then p(c(i+1)) because of IVP. k+1 also has a root if lim x->-inf has a different sign from p(c1) and similarly for positive infinity Thus, the optimal case is when the sequence;
p(-inf), p(c1), p(c2), ... , p(ck), p(inf) alternates signs.
Since the IVP guarantees a root between each of these, we have k+1 roots.
Yeah, what i did to generalize this, is simmilar to what you did here, besides that in my case these c1,c2,...ck-1, are critical points, and i argued that whenever we have alternating signs like you said we are gonna have k roots.
This is what i did for the case of degree 3 though.

Whenever f(-infty),f(x1),f(x2),f(infty) , alternate signs we are gonna have at most 3 real roots. Here we cannot plug infty inside the function, but just to illustrate the point.

What HallsofIvy posted was what I had in mind.

But if you add in those other cases that you just mentioned it should fill in the remaining holes in your proof.

What HallsofIvy posted was what I had in mind.

But if you add in those other cases that you just mentioned it should fill in the remaining holes in your proof.
Well, i actually did all those cases also when i did it on my paper, but just didn't post all of them here, i thought they were trivial.
So, it means that with these other cases, one could take my proof as a valid one, right?

There should be a much simpler proof.

Given any polynomial P(x) of degree n>= 1 with root r (i.e. P(r) = 0), we can write it as (x-r)Q(x) where Q(x) has degree exactly n-1. (This needs to be proved obviously. I don't know how to prove it, but I know that it is true, and it seems like it should be provable without using the result. The degree n-1 part follows immediately, but it is not clear how to prove that you can factor P)

The fact that it can have at most n roots falls out of the fact that if Q is of degree 0, it has no roots (the 0 polynomial is not defined as having any degree, and if Q(x) = c != 0, then there is no number x such that Q(x) = 0)

Anyway, this is entirely true in a much greater context than the real numbers, so it's necessarily the case that no calculus is required to prove any of it. For instance, it is even true if your coefficients are matrices.

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mathwonk
Homework Helper
this result follows over any field immediately from two algebraic facts:

1) if a polynomial f(X) has a root c in a field, then f(X) = (X-c)g(X) where g hAS DEGREE one LESS THAN F.

2) if the product of two field elements is zero, one of them is zero.

this result follows over any field immediately from two algebraic facts:

1) if a polynomial f(X) has a root c in a field, then f(X) = (X-c)g(X) where g hAS DEGREE one LESS THAN F.

2) if the product of two field elements is zero, one of them is zero.

I forgot to mention at the beggining, that this is a question on a calculus I topic, so i guess we are supposed to prove this only with calculus tools.

P.S. I need to wait untill next semester,when i will be taking Abstract Algebra to prove it that way...lol !!!

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Suppose your third order equation had more than 3 roots. say 4. Write out its linear factors and multiply. What do you get?

However here, it is true that this will tell us that a polynomial of degree 3 cannot have 4 real roots, but it defenitely does not guarantee us that it will have 3 roots either.

However here, it is true that this will tell us that a polynomial of degree 3 cannot have 4 real roots, but it defenitely does not guarantee us that it will have 3 roots either.
However, how you have posed the problem, this is exactly what you want. Your problem is to show that a polynomial of degree 3 has at most 3 roots i.e. that it has <= 3 roots; that it cannot have 4 or more roots

Any proof that tries to show that it has exactly 3 roots will somewhere be flawed because there are many polynomials of degree 3 that have only 1 root (though every polynomial of degree 3 has at least 1. Why?)

For instance x^3 + 1 only has -1 as a root (as it factors into (x-1)(x^2+x+1). The other roots are complex)

You also do not need to show that any roots at all exist because even though it is not possible for a polynomial of degree 3 to have no roots, 0 <= 3, so if it had 0 roots, this would fall under the scope of the problem.

But don't we need first to show that a 3rd degree polynomial can sometimes have 3 roots, and after that claim that it cannot have more than that, like i did? Because i did not claim that it has exactly 3 roots, i just was trying to prove that at some case it can have 3 real roots, but there is no case where it can have more than that.

Or...... yeah, i think i got your point!!!

MOreover,

How would you rate this problem if it was in a Calculus I exam?

1.Very Hard
2.Hard
3.Not that hard(Intermediate)
3. Easy
4.Very Easy

???

HallsofIvy
Homework Helper
Depends on what you are allowed to use. If you can use the theorem that "If a, b, c are zeroes of polynomial P(x), then P(x)= (x-a)(x-b)(x-c)Q(x) where Q(x) is a polynomial" then it is very easy.

HallsofIvy
Homework Helper
However here, it is true that this will tell us that a polynomial of degree 3 cannot have 4 real roots, but it defenitely does not guarantee us that it will have 3 roots either.
Yes, it won't because that is not true. The original problem was to show that a polynomial of degree 3 can have at most 3 real roots.

Depends on what you are allowed to use. If you can use the theorem that "If a, b, c are zeroes of polynomial P(x), then P(x)= (x-a)(x-b)(x-c)Q(x) where Q(x) is a polynomial" then it is very easy.

Well, if you are not allowed to use that theorem, but in contrary are required at some point to use Rolle's theorem in your proof.

HallsofIvy
Homework Helper
Suppose there exist 4 distinct roots to a cubic equation. Those 4 values have three intervals between them. Apply Rolles theorem on each interval to show there exist a point in each interval where f '(x)= 0. In other words, there exist at least three roots to the equation f '(x)= 0. But f '(x) is quadratic. By the quadratic formula, that equation cannot have more than 2 roots. (Or, just do the same thing again to show that a quadratic equation cannot have more than 2 roots.)

Suppose there exist 4 distinct roots to a cubic equation. Those 4 values have three intervals between them. Apply Rolles theorem on each interval to show there exist a point in each interval where f '(x)= 0. In other words, there exist at least three roots to the equation f '(x)= 0. But f '(x) is quadratic. By the quadratic formula, that equation cannot have more than 2 roots. (Or, just do the same thing again to show that a quadratic equation cannot have more than 2 roots.)
Oh and that naturally suggests how to do the inductive argument for the general case! Cool beans.

lurflurf
Homework Helper
I am fond of this proof.
Suppose we have 4 polynomials and 4 points
polynomials p,q,r,s
points w,x,y,z
p(w)=q(x)=r(y)=s(z)=1
each polynomial being 0 on the other points
clearly
ap+bq+cr+ds
a,b,c,d numbers
is the general cubic
supose f is cubic with
f(w)=f(x)=f(y)=f(z)=0
f can be written
f=ap+bq+cr+ds
where
f(w)=a
f(x)=b
f(y)=c
f(z)=d
thus a=b=c=d=0
but then f is the zero polynomial and not cubic
-><-
QED