Generalization of the discriminant of a quadratic to check that a polynomial has no real roots? So I have a question that asks me what conditions must occur on the coefficients of a quadratic polynomial, so that there are no real roots. I looked around and I know if the discriminant of the quadratic formula is negative, then there are no real roots. But, I want to know how this can generalize to cases like polynomials of degree 4.
 A: A general method to find if a polynomial in one variable has real roots in an interval $(a,b] \subset \mathbb{R}$ is the Sturm's theorem.
But even for a polynomial of low degree this theorem require the construction of a chain of polynomials that is very tedious to find.
A test '' like'' the discriminant can be done for equations of degree three and four, for which we know solution formulas, but, also in this case the work to do is not trivial.
E.G.:
The $3$-degree equation:
$$
x^3+Bx^2+Cx+D=0
$$
has three real distinct solutions if, given
$$
P=\dfrac{3C-B^2}{9} \qquad Q=\dfrac{2B^3-9BC+27D}{54}
$$
the ''discriminant''
$$
\Delta=P^3+Q^2
$$
is $\Delta<0$.
A four-degree equation
$$
y^4+By^3+Cy^2+Dy+E=0
$$
can be transformend ( with $ y=x-B/4$) in:
$$
x^4+px^2+qx+r=0
$$
and has four distinct real solutions if his ''resolvent'' equation
$$
z^3+\dfrac{p}{2}z^2+\left(\dfrac{p^2}{16}-\dfrac{r}{4}\right)z-\dfrac{q^2}{64}=0
$$
has three distinct real positive solutions.
Also, we can find that, if this resolvent has one positive real root and two negative real roots, then the $4-$degree equation has two couple of complex conjugate roots, so: no real roots.
A: This is not an expression like the discriminant, but it is a test to possibly verify that a polynomial has no real roots. However, if it fails, you can't conclude anything either way.
You can show that for a polynomial $p(x) = x^n+a_{n-1}x^{n-1} + \cdots + a_0 $ with leading coefficient 1, all the roots $x_i$ satisfy $-(n+1) \max_i |a_i| \leq x_i$. You can see this question for details. Thus, $p(x-(n+1) \max |a_i|)$ must have all it's real roots be positive.  If all coefficients are positive or all coefficients are negative, by Descartes' Rule of Signs, there are no positive roots and hence no real roots.
This still leaves the possibility where there are sign changes in the coefficients of the new polynomial but there are still no positive roots. 
