# polynomial factorisation in $\mathbb R[X]$

I read that every polynomial in $\mathbb R[X]$ factorizes in a product of linear and quadratic polynomials. Do we have a result stating that every polyonomial of degree $\geq 3$ has at least a real root?

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Every polynomial of odd degree has at least one real root. – Isaac Solomon Mar 27 '12 at 8:33

The factorization follows from the following two facts:

a) Every complex polynomial in $\mathbb{C}[X]$ can be factored into linear factors by the fundamental theorem of algebra; b) If a polynomial with real coefficients has a complex root $z$ then also the complex conjugate $\bar{z}$ is a root.

So a real polynomial with a complex root $z$ has the quadratic factor $(X-z)(X-\bar{z})$ with real coefficients.

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Here by complex root you mean a complex with non zero imaginary part; ${\mathbb C} -{\mathbb R}$. Because for example the polynomial $p(x)=(x-1)(x-2)(x-3)$ does not have conjugate roots – palio Mar 27 '12 at 9:30
It is very frequent that "complex" stands for "complex and not real". This always perplexes me : should this usage be suppressed in the name of precision or would suppression be nitpicking? I really don't know. – Georges Elencwajg Mar 27 '12 at 11:28
I prefer "non-real" when the distinction is necessary. – Matthew Pressland Mar 27 '12 at 14:31

No - the polynomial $x^4+1$ has no real roots.

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Then why every polynomial in $\mathbb R[X]$ factorizes in a product of linear and quadratic polynomials, so the partial decomposition has always a denominator of the form $(ax+b)^n$ or $(ax^2+bx+c)^n$, so we have only powers of linear or quadratic polynomials – palio Mar 27 '12 at 8:41
The point is that some factor only as products of quadratics, and these quadratics may have no real roots. As pointed out in other answers, if the degree is even then at least one of the factors must be linear, and then you have a real root. In this particular case, $x^4+1=(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)$, so it factors into quadratics, each of which has no real roots. It is a common mistake to assume that a polynomial with no roots in a field $K$ is irreducible over $K$. – Matthew Pressland Mar 27 '12 at 10:35
Oops, I meant "if the degree is odd" above, not even. – Matthew Pressland Mar 27 '12 at 10:44

No. For every positive integer $n$, the polynomial $x^{2n}+1$ has no real root: for every real number $x$, $x^{2n}=(x^n)^2\ge 0$, so $x^{2n}+1>0$ for all $x\in\Bbb R$.

On the other hand, every real polynomial of odd degree has at least one real root. To see this, suppose that $p(x)$ is a polynomial of odd degree. If the leading coefficient of $p(x)$ is positive, then $$\lim_{x\to\infty}p(x)=\infty\quad\text{ and }\quad\lim_{x\to -\infty}p(x)=-\infty\;,$$ while if the leading coefficient is negative, then $$\lim_{x\to\infty}p(x)=-\infty\quad\text{ and }\quad\lim_{x\to -\infty}p(x)=\infty\;.$$ In either case we can find real numbers $a$ and $b$ such that $p(a)<0$ and $p(b)>0$, and the intermediate value theorem then ensures that $p(x)$ has a zero somewhere between $a$ and $b$.

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