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I was trying to prove the statement with FTA as in the question.

The idea of some proofs I found [1] was by stating that the root of the polynomial P(x) must go in conjucate pairs. Since P(x) has only non real roots this will be contradictory by FTA as P(x) has precisely as many roots (counted with multiplicity) as its degree while the number of roots P(x) has is even.

But how can we be sure about the multiplicity of the complex roots? If the parity of the multiplicity of a root is different to its conjucate, the sum of the multiplicity count of this pair is not even and the proof won't hold..

How can we fix this?

Reference:[1] Proof that every polynomial of odd degree has one real root

Angae MT
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fesOsorPro
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2 Answers2

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If the parity of the multiplicity of a root is different to its conjucate,

Well that's simply a false premise. Did you try to find even one example like that?

What you should do is show when $(x-r)^k$ is a factor of a polynomial with real coefficients, $(x-\overline{r})^k$ is also a factor of the polynomial and vice versa. That implies $r$ and $\overline{r}$ have the same multiplicity as roots of a polynomial with real coefficients.

KCd
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I assume you know the complex conjugate root theorem of polynomials. That is, if $z$ is a zero to $f(x)$, then $\bar{z}$ is also a zero. This theorem only tells the existence, but none are related to its multiplicities. Therefore you need some efforts to show that the multiplicity of $(x-z)$ and $(x-\bar{z})$ are same in $f(x)$.

Let $\alpha$ be the multiplicity of $(x-z)$ in $f(x)$, $\alpha\in\mathbb{N}$. Then by factor theorem, $f(x)=g(x)(x-z)^\alpha$ with $g(z)\ne0$. By complex-conjugate root theorem, we know $g(\bar{z})=0$, so we can write $$f(x)=h(x)(x-z)^{\alpha-1}(x-z)(x-\bar{z})\quad\text{with $h(z)\ne0$}$$ Now we can inductively apply the complex conjugation root theorem on $h(x)(x-z)^{\alpha-1}$, to show that it will have a factor of $(x-\bar{z})$, for $\alpha$ times in total. Note that we cannot go for more than $\alpha$ times because those $h(x)$ polynomials does not have $(x-z)$ as factor. This shows the multiplicities of $(x-z)$ and $(x-\bar{z})$ are the same.

Angae MT
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