Btukfyl

I can't seem to solve this problem, can anyone help me please? The problem is:

Let real numbers $a$,$b$ and $c$, with $a ≤ b ≤ c$ be the 3 roots of the polynomial $p(x)=x^3 + qx^2 + rx + s$. Show that if we divide the interval $[b, c]$ into six equal parts, then one of the root of $p'(x)$ (the derivative of the polynomial $p(x)$) will be in the 4th part.

What I did was:

Because we know the roots of $p(x)$ are $a$,$b$ and $c$, we can write the polynomial $p(x)$ like this: $p(x) = (x-a)(x-b)(x-c)$

So we have $p(x) = x^3 + qx^2 + rx + s = (x-a)(x-b)(x-c)$

We find the value of $q$,$r$ and $s$:
$q = -(a+b+c)$

$r = (ab + ac + bc)$

$s = -abc$

We have that $p'(x) = 3x^2 + 2qx + r$
We want to find the roots of $p'(x)$, so if we apply the quadratic formula, we get:
$(-2q ± 2*sqrt{q^2 - 3r})/6$

Because we know the value of q,r and s, we can rewrite this expression like this:
$(2(a+b+c) ± 2sqrt{a^2 + b^2 + c^2 - ab - bc -ca})/6$

But I am stuck here, any help would be great. Thank you in advance.

Since $p(x) = (x-a)(x-b)(x-c)$,

$$begin{align}p'(x) &= (x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)\
&= (x-b)(x-c) + (x-a)(2x - (b+c))end{align}$$
At $x = frac{b+c}{2}$, we have

$$p'(x) = frac{c-b}{2}cdotfrac{b-c}{2} + (x-a)cdot 0 = -frac{(c-b)^2}{4} le 0$$

At $x = frac{b+2c}{3}$, we have

$$begin{align}p'(x) &= frac{2(c-b)}{3}cdotfrac{b-c}{3}
+ left(frac{b+2c}{3} - aright)cdotfrac{c-b}{3}\
&= -frac{2(c-b)^2}{9} + left(frac{2(c-b)}{3} + (b-a)right)cdotfrac{c-b}{3}\
&= frac{(b-a)(c-b)}{3}\&ge 0end{align}$$
This implies $p'(x)$ has a root in $left[frac{b+c}{2},frac{b+2c}{3}right]$.

I have an inelegant brute force solution. Since the problem is translation invariant, we may assume $a=0$ to simplify calculations so that $0leq bleq c$. It should be clear that the root of $p'(x)$ we are interested in is the largest of the two i.e. the one with the plus sign. We need to prove that
$$dfrac{3b+3c}{6}leqdfrac{2(b+c)+2sqrt{b^2+c^2-bc}}{6}leqdfrac{2b+4c}{6}$$
On the left we obtain
$$b+cleq 2sqrt{b^2+c^2-bc}$$
where both sides are non-negative, squaring gives
$$b^2+c^2+2bcleq 4(b^2+c^2-bc)Rightarrow 3b^2+3c^2-6bcgeq 0$$
which holds. On the right we have
$$2sqrt{b^2+c^2-bc}leq 2c$$
so that
$$b^2+c^2-bcleq c^2Rightarrow b(b-c)leq 0$$
which holds since $0leq bleq c$.