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