How to easily identify how many distinct roots a polynomial has?

Question

For example, I have $a(x) = 4 x^4 + 5 x^2 + 7 x + 2 $. $$$$ Using Descartes rule, I know $a(x)$ has maximum $2$ negative real roots. As imaginary roots come in pairs, $a(x)$ will have either $0$ or $2$ negative real roots. I also imagine the possible graph of $4 x^4 + 5 x^2 + 7 x $, it should have a single minimum. So, if the minimum value of $4 x^4 + 5 x^2 + 7 x$ is $\le -2$, then there can be a real root, otherwise both roots will be imaginary. Now I see $a(0) = 2, a(-1) = 4,$ so there can be a root in between $0$ and $1$. By trying out $0.5$, I get a root, as $a(0.5) = 0$. So I decided that $a(x)$ will have two real roots, one of which is $-1/2$. $\\$ $$$$But this interpretation proved to be wrong, when I plotted the graph. $a(x)$ has a repetitive real root at $-1/2$. In fact $ a(x) = (2 x+1)^2(x^2 -x +2)$. $$$$ I want to know if there is an easier way to identify how many distinct real roots (or if any repetitive roots) are there for a polynomial, without using any tool for graph plotting or factorizing. In the above example, if the root were $-23/245$, in place of $-1/2$, then I couldn't have been able to identify it by inspection either.

Answer 1

A complex number $x_0$ is a root of multiplicity $m>1$ of $a(x)$ if and only if $a(x)=(x-x_0)^mb(x)$ for some polynomial $b(x)$ with $b(x)\not=0$. By considering the derivative of $a(x)$ we find that $$a'(x)=m(x-x_0)^{m-1}b(x)+(x-x_0)^{m}b'(x)=(x-x_0)^{m-1}(mb(x)+(x-x_0)b'(x)).$$ Hence $x_0$ is a root of $a'(x)$ of multiplicity $m-1>0$.

So a multiple root of $a(x)$ can be found by checking whether $a(x)$ and $a'(x)$ have a common root, i.e. by finding the roots of the greatest common divisor of $a(x)$ and $a'(x)$.

In your case $a(x)=4 x^4 + 5 x^2 + 7 x + 2$, and it follows that $$\gcd(a(x),a'(x))=2x+1.$$ Therefore $x=-1/2$ is a multiple root of $a(x)$ (of multiplicity $2$).