# Thread: Polynomial roots

1. ## Polynomial roots

I've just noticed that every polynomial of degree $\displaystyle 2k+1$, with $\displaystyle k \ne 0$, has to at least have one real root (because complex numbers happen in pairs).
This is true, right? It's shown in my A-level textbook that if $\displaystyle z = x+iy$ is a root of a polynomial equation with real coefficients, then $\displaystyle z = x-iy$ is
also a root of the polynomial equation. But I doubt showing just that would be sufficient. How can it be shown?

2. Originally Posted by TheCoffeeMachine
I've just noticed that every polynomial of degree $\displaystyle 2k+1$, with $\displaystyle k \ne 0$, has to at least have one real root (because complex numbers happen in pairs).
This is true, right? It's shown in my A-level textbook that if $\displaystyle z = x+iy$ is a root of a polynomial equation with real coefficients, then $\displaystyle z = x-iy$ is
also a root of the polynomial equation. But I doubt showing just that would be sufficient. How can it be shown?
2k+1 is odd; therefore, [iMATH]x^{2k+1}[/iMATH] must have at least one real solution. Actually induction may work here.

For roots of the polynomial, we only have 2 options either $\displaystyle x\in\mathbb{R}$ or $\displaystyle x\in\mathbb{C}$. If $\displaystyle x\in\mathbb{C}$, then $\displaystyle x=a\pm b\mathbf{i}$.

3. Originally Posted by dwsmith
2k+1 is odd; therefore, $\displaystyle x^{2k+1}$ must have at least one real solution.
But your conclusion is exactly what I'm trying to prove. I've noticed it while sketching odd
degree polynomials, but I have written it in that way (because I thought it might be better).
And thanks, I don't know I was thinking when I put the restriction that $\displaystyle k\ne 0$.

4. Since complex answers come in pairs, they are of the form 2p which is even.

$\displaystyle x^{2p+1}=x^{2p}x$

The roots of $\displaystyle x^{2p}$ are complex.

Since $\displaystyle x$ is of degree 1, x must be a real solution.

5. Originally Posted by TheCoffeeMachine
I've just noticed that every polynomial of degree $\displaystyle 2k+1$, with $\displaystyle k \ne 0$, has to at least have one real root (because complex numbers happen in pairs).
This is true, right? It's shown in my A-level textbook that if $\displaystyle z = x+iy$ is a root of a polynomial equation with real coefficients, then $\displaystyle z = x-iy$ is
also a root of the polynomial equation. But I doubt showing just that would be sufficient. How can it be shown?
Hello friend,

Your hypothesis is true. Namely that if $\displaystyle p(x)=\alpha_{2n+1} x^{2n+1}+\cdots+\alpha_0,\text{ }n\geqslant 1$ then $\displaystyle p(x)=0$ for some $\displaystyle x\in\mathbb{R}$

There are many proofs of this. One, as you mentioned is that by the fundamental theorem of algebra $\displaystyle p:\mathbb{C}\to\mathbb{C}$ has exactly $\displaystyle 2n+1$ zeros (counting multiplicity), $\displaystyle \alpha_0,\cdots,\alpha_{2n+1}\in\mathbb{R}$ then we know that $\displaystyle p(x)=(x-\beta_1)\cdots(x-\beta_{2n+1}),\text{ }\beta_{1}\cdots,\beta_{2n+1}\in\mathbb{C}$. Now, it is fairly easy to show that if all of the betas are non-real that $\displaystyle p$ must have an imaginary coefficient contradictory to our hypothesis. Thus, $\displaystyle p$ must have a real zero.

Thinking more along the lines of analysis $\displaystyle p:\mathbb{R}\to\mathbb{R}$ is continuous. But, clearly (why?) $\displaystyle \lim_{x\to\infty}p(x)=-\lim_{x\to-\infty}p(x)$ and thus $\displaystyle p(x)<0$ at some point and $\displaystyle p(x)>0$ at some point and the intermediate value theorem implies the existence of a zero.

6. Originally Posted by TheCoffeeMachine
I've just noticed that every polynomial of degree $\displaystyle 2k+1$, with $\displaystyle k \ne 0$, has to at least have one real root (because complex numbers happen in pairs).
This is true, right? It's shown in my A-level textbook that if $\displaystyle z = x+iy$ is a root of a polynomial equation with real coefficients, then $\displaystyle z = x-iy$ is
also a root of the polynomial equation. But I doubt showing just that would be sufficient. How can it be shown?
It is easily proved using the Intermediate Value Theorem.