Is there a theorem like this in number theory or analysis? Given a cubic polynomial $$ax^3 + bx^2 + cx + d = 0$$
What conditions on $a,b,c,d$ ensure that the polynomial has only one real root?
 A: You can calculate the discriminant, but this is a purely algebraic approach: 
$$\Delta=b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$$


*

*$\Delta>0$ : The equation has three real roots.

*$\Delta<0$ : The equation has one real root and  two cojugate complex roots

*$\Delta=0$ : All three roots of the equation are real but at least one root has a multiplicity higher than one.
You can also show that there is only one root by using analysis. First you show that every cubic equation has a real root by using Intermediate Value Theorem and saying that $f(x)=ax^3+bx+cx+d$ has the same sign as $a$ for sufficiently large $x$ and the opposite sign of $a$ for a sufficiently large negative number so there must be at least one root between these two values because $f(x)$ changes sign between these numbers and $f(x)$ is continuous. Then you can apply Rolle's theorem and say if the equation has more than one roots, then at some point its derivative must become zero by Rolle's theorem, then you apply Rolle's theorem and get a second degree equation that has no real roots, so that contradicts the hypothesis that there are more than one root.
So, let's apply what I said to $p(x)=ax^3+bx^2+cx+d$:
$p'(x) = 3ax^2+2bx+c$
$$\displaystyle \Delta = 4b^2-4(3a)(c)=4b^2-12ac<0$$
SECOND EDIT:
You can count the multiplicity of the roots of a polynomial equation by taking derivatives of it. Once you find a root of the equation, you take first, second, third, ..., nth derivative and see if they become zero when you plug in the root. Then the multiplicity of the root is $n+1$ if $f^{(n)}(x)=0$ and $f^{(n+1)} \neq 0$. So, in your case, if $\alpha$ is the root of your equation, you should add the condition that $p'(\alpha) \neq 0$.
A: Similarly to the quadratic discriminant, there is a cubic discriminant, and a discriminant for higher order polynomials that can tell you information about the number of real roots. In the cubic case, the discriminant is $18abcd -4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2$, and the cubic has one real root when this is negative.
A: Hint: Use Rolle's theorem to consider what you can say about the roots of the derivative if the cubic polynomial has more than one distinct real root.
A: Perhaps a more analytic approach is as follows. WLOG by scaling it is enough to consider the case $a=1$. To compute points where the derivative vanishes we need to solve
$3x^2+2bx+c=0$
i.e. $x=\frac{-b\pm\sqrt{b^2-3c}}{3}$. If $b^2<3c$ then there are no real roots to this equation and hence the derivative is always greater than zero (on the real line) so we must have one real root. If $b^2=3c$ there is exactly one $x$ where the derivative vanishes and hence again only one real root. If $b^2>3c$ the there are two distinct $x$ where the derivative vanishes and we have three roots iff one of these $x$ is greater than $0$ and one is less than $0$. This corresponds to three roots in this case iff $c<0$.
In summary, we have three (distinct) real roots iff $b^2>3c$ and $c<0$. Or after scaling back $b^2/a^2>3c/a$ and $c/a<0$.
