Polynomial roots

**TheCoffeeMachine** · May 13th 2010, 02:36 PM

I've just noticed that every polynomial of degree $2k+1$ , with $k \ne 0$ , 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 $z = x+iy$ is a root of a polynomial equation with real coefficients, then $z = x-iy$ is
also a root of the polynomial equation. But I doubt showing just that would be sufficient. How can it be shown?

**dwsmith** · May 13th 2010, 02:49 PM

Originally Posted by TheCoffeeMachine

I've just noticed that every polynomial of degree $2k+1$ , with $k \ne 0$ , 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 $z = x+iy$ is a root of a polynomial equation with real coefficients, then $z = x-iy$ 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 $x\in\mathbb{R}$ or $x\in\mathbb{C}$ . If $x\in\mathbb{C}$ , then $x=a\pm b\mathbf{i}$ .

**TheCoffeeMachine** · May 13th 2010, 03:00 PM

Originally Posted by dwsmith

2k+1 is odd; therefore, $x^{2k+1}$ 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 $k\ne 0$ .

**dwsmith** · May 13th 2010, 03:11 PM

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

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

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

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

**Drexel28** · May 17th 2010, 12:18 AM

Originally Posted by TheCoffeeMachine

I've just noticed that every polynomial of degree $2k+1$ , with $k \ne 0$ , 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 $z = x+iy$ is a root of a polynomial equation with real coefficients, then $z = x-iy$ 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 $p(x)=\alpha_{2n+1} x^{2n+1}+\cdots+\alpha_0,\text{ }n\geqslant 1$ then $p(x)=0$ for some $x\in\mathbb{R}$

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

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

**mr fantastic** · May 17th 2010, 04:28 AM

Originally Posted by TheCoffeeMachine

I've just noticed that every polynomial of degree $2k+1$ , with $k \ne 0$ , 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 $z = x+iy$ is a root of a polynomial equation with real coefficients, then $z = x-iy$ 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.

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