Rolle's Theorem - Polynomial Question

Question

By using Rolle’s theorem, prove that if $p(x)$ is a polynomial with real coefficients then the equation

$0 = (x^2−x)^2p'''(x)+6x(2x^2−3x+1)p''(x)+6(6x^2−6x+1)p'(x)+12x(2x−1)p(x)$

has a solution in $(0,1)$.

So I tried to find a function which differentiates to the expression above and satisfies the conditions of Rolle's theorem in the interval $[0,1]$, but I got to

$(x^2−x)^2p''(x)+4x(2x^2−3x+1)p'(x)+2(6x^2−6x+1)p(x)$

which doesn't quite differentiate to the correct function. Even if the coefficient of $p(x)$ should have been $12(2x-1)$, the above doesn't have the same value at $0$ and $1$, so I don't think you could apply Rolle's.

Any suggestions, am I making a mistake, or is the mistake in the question?

Answer 1

It looks like there's a mistake in the question and the coefficient of $p(x)$ ought to be $12(2x-1)$. With that coefficient:

Let $h(x) = (x^2-x)^2p(x) = q(x)p(x)$. We see that $h(0) = h(1) = 0$, so by Rolle's theorem there is a $c\in (0,1)$ with $h'(c) = 0$. Now

$$h'(x) = q(x)p'(x) + q'(x)p(x),$$

from which we can read off $h'(0) = h'(1) = 0$. Then Rolle's theorem says that there are $d\in (0,c)$ and $e \in (c,1)$ with $h''(d) = h''(e) = 0$. And another application of Rolle's theorem asserts the existence of $f\in (d,e)\subset (0,1)$ with $h'''(f) = 0$.

Then we compute

\begin{align} h''(x) &= q(x)p''(x) + 2q'(x)p'(x) + q''(x)p(x), \\ h'''(x) &= q(x)p'''(x) + 3q'(x)p''(x) + 3q''(x)p'(x) + q'''(x)p(x) \\ &= (x^2-x)^2p'''(x) + 6x(2x^2-3x+1)p''(x) + 6(6x^2-6x+1)p'(x) + 12(2x-1)p(x). \end{align}