Information about the roots of a polynomial without their calculation

Question

Suppose I have a polynomial (of any order) and I'm not able to calculate the roots. Is there a way to get at least some information about the roots such as how many of them are complex, negative or positive? For example, I can safely identify the behavior (and therefore roots' character) of $f(x)=ax+b$ or even a quadratic expression just by inspection.

I'm aware of Descartes' sign rule http://en.wikipedia.org/wiki/Descartes%27_rule_of_signs but it apparently provides only an upper bound on the number of positive/negative roots. Is there something more general giving an exact number (of positive roots) preferably without using the methods of calculus?

Answer 1

If $P=\sum_{k=0}^n a_kX^k$, then the roots of $P$ are bounded by $\displaystyle \max\{1,\sum_{k=0}^{n-1}\left|\frac{a_k}{a_n}\right|\}$

Note that the proof adapts to complex polynomials as well.

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Suppose $P=\sum_{k=0}^n a_kX^k$ has complex roots $\alpha_1,\ldots,\alpha_n$.

Let $i\in \{1,\ldots,n\}$

$\displaystyle P(\alpha_i)=0=\sum_{k=0}^n a_k\alpha_i^k$

Two cases come up:

• $|\alpha_i|\leq 1$

• $|\alpha_i|> 1$

In this last case, write $\displaystyle |\alpha_i|=|\sum_{k=0}^{n-1}\frac{a_k}{a_n}\alpha_i^{k-n+1}|\leq \sum_{k=0}^{n-1}\left|\frac{a_k}{a_n}\alpha_i^{k-n+1}\right|$

By assumption, $\forall k\in\{0,\ldots,n-1\},|\alpha_i^{k-n+1}|\leq 1$

Hence $\displaystyle |\alpha_i|\leq \sum_{k=0}^{n-1}\left|\frac{a_k}{a_n}\right|$

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Therefore, if $P=\sum_{k=0}^n a_kX^k$, then the roots of $P$ are bounded by $\displaystyle \max\{1,\sum_{k=0}^{n-1}\left|\frac{a_k}{a_n}\right|\}$

Answer 2

A couple of ideas:

1)Rational roots theorem

Given $p(x):=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ with all $a_i$ integer and $a_0, a_n \neq 0$ and $r=c/d$ a root, then :

i)$c|a_0$ and

ii) $d|a_n$,

where $x|y$ means " x divides y" , i.e., $y=cx$ for some integer $c$.

http://en.wikipedia.org/wiki/Rational_root_theorem,

2) An odd-degree polynomial has at least one Real root. When your polynomial has integer coefficients:

Because if $z=a+ib$ is a root of $p(x)$, so is its conjugate $w=a-ib$, so Complex roots come in pairs, so an odd-degree polynomial must have at least one Real root.