Zeroes of derivative of cubic polynomial If $p$ is a polynomial of degree 3, then between any two consecutive zeroes of p, can there be two zeroes of its derivative ( $p'$) ?
Suppose if there are two zeroes of $p'$ between two consecutive zeroes of $p$, then the graph needs to have structure which looks like the pic below

From the graph, it is clear that $p'$ has $3$ zeroes, which is not possible for cubic polynomial.
I tried to prove this by using Rolle's theorem. I was only successful in showing that cubic polynomial has no constant term. I would like to know if there is any other way to prove this apart from graphical solution.
 A: If by "two roots between" you mean two distinct roots strictly between two roots of $p(x)$, you are correct: $p'(x)$ cannot have two distinct roots between consecutive roots of $p(x)$, if $p(x)$ is a cubic polynomial with real coefficients.
Say two  roots of $p(x)$ are $x=a$ and $x=b$, $a\lt b$, and the two roots of $p'(x)$ are $r_1$ and $r_2$, with $a\lt r_1\lt r_2\lt b$. We will show that $p(x)$ has a zero strictly between $a$ and $b$, so that they are not "consecutive roots".
Since $p'(x)$ is a quadratic, it follows that $p'(x)$ changes signs at $x=r_1$ and at $x=r_2$. Replacing $p(x)$ with $-p(x)$ if necessary, we may assume that $p'(x)$ is positive to the left of $r_1$, negative between $r_1$ and $r_2$, and positive to the right of $r_2$.
That means that $p(x)$ is increasing on $(-\infty,r_1)$, decreasing on $(r_1,r_2)$, and increasing on $(r_2,\infty)$. Thus, $p(r_1)\gt 0$ (since $p(a)=0$ and $a\lt r_1$); and $p(r_2)\lt  0$ (because $p(b)=0$ and $r_2\lt b$). By the intermediate value theorem, it follows that $p(x)$ has a zero between $r_1$ and $r_2$, hence $p(x)$ has a zero between $a$ and $b$.

On the other hand, if you allow one of the roots of $p'(x)$ to coincide with either $a$ or $b$, then you can take $a=0$, $b=1$, and $p(x) = x(x-1)^2$ as a counterexample; here you get $p'(x) = (x-1)^2 + x(x-1) = (x-1)(x-1+x) = (x-1)(2x-1)$, which has roots at $x=\frac{1}{2}$ and $x=1$, both "between" $a$ and $b$ (since we are allowing them to equal $b$ in this scenario).
If you allow the two roots of $p'(x)$ to coincide, that is, to be a double root, then you are fine again unless all the roots for both $p(x)$ and $p'(x)$ coincide. By doing a horizontal shift we may assume that the double root of $p'(x)$ is at $0$, so that $p'(x) = Ax^2$ for some nonzero $A$; then $p(x)=ax^3+c$ for some $c$, and this polynomial has a single real root at $x=\sqrt[3]{-\frac{c}{a}}$; but if $c\neq 0$ then the roots of $p'(x)$ are not "between" the roots of $p(x)$. And if $c=0$, then all roots are the same, which of course is possible.
