Im confuse..I read in an article that in dealing with polynomials, a quadratic equation can have either 2 real roots, 1 equal real root or 2 complex roots...but in dealing with random polynomials only two cases are possible either 2 real roots or 2 complex roots...why is that so? the article also said that using the determinants of quadratic formula b^2 - 4ac it should be equal to 0 so that there will be a multiplicity of roots but that case is rare since b^2 = 4ac is almost impossible since the coefficients is random. I run this in a mathematical software,,, and its true!! even degree polynomials has even number of real roots and odd degree polynomial has odd number of real roots,,my problem is how to prove this claim mathematically? Im not quite sure if this is always the case in all degree n...



existence of multiplicity of roots
Choose $b$ and $c$ randomly. The value $b^2/(4c)$ is now determined. Note that the probability that $b=0$ is zero, since almost all real numbers are irrational. The measure of rationals in reals is zero!


In any case, this means that the probability that a random $a$ will equal $b^2/(4c)$ is also zero.


In general, if $n=degree(f)$ even, and $f(x)$ has real random coefficients it can be factored into a product of quadratics, and the result follows from above.


If $n$ is odd, then $f(x)$ has an extra linear factor. Can this factor, say $x+b$ without loss of generality (can assume poly is monic with leading coefficient 1 by scaling without changing the nature of the roots) be complex and the overall polynomial still have real coefficients?


Let $f(x)=(x+b)(g\_0+g\_1 x+\cdots+ x^{n-1})$ Then $$f(x)=(b g\_0+ (b g\_1+g\_0)x+\cdots+ (b g\_n+g\_{n-1})x^{n-1}+x^n)$$


It seems to me if $b$ is complex, the ratios $g\_n/g\_{n-1}$ need to be complex to make the overall polynomial real, unless the coefficients are all zero and this would give a contradiction since we're not considering the zero polynomial.


Thus if we assume $g(x)$ had all its roots real, we have proved that the number of real roots are odd.


Assume some roots of $g(x)$ are complex. From the quadratic case we know then $g(x)$ splits into a product of a bunch of quadratics whose roots are complex, and a bunch of quadratics whose roots are real. Even in the case that one of these sets of quadratics is empty, the conclusion follows either by only considering the linear part (if all quadratics have complex roots) or a reduced polynomial (let $g'(x)$ be the product of the quadratics with real roots and consider $g'(x)(x+b)$ as above.


