<s>[INST] Function related polynomial problem 
My approach is as follow $f\left( x \right) = 2{x^4} + 3{x^3} - 3{x^2} - 6x + a$
$f'\left( x \right) = 8{x^3} + 9{x^2} - 6x - 6$
$f''\left( x \right) = 24{x^2} + 18x - 6 = 6\left( {4{x^2} + 3x - 1} \right)$
The roots of $f''(x)$ are $-1$ and $\frac{1}{4}$, how do I arrive at the proper conclusion.
$f\left( 1 \right) = a - 4$
$f\left( 2 \right) = 32 + 24 - 12 - 12 + a = a + 32$
$f\left( 1 \right).f\left( 2 \right) < 0$
$\left( {a - 4} \right)\left( {a + 32} \right) < 0;a \in \left( { - 32,4} \right)$
I found this question on the  ineternet.
 [/INST] It's easy to show that there are no other roots on the $(1,2)$ besides $y$. Indeed, from the positive sign of the second derivative, we see that $f'$ is increasing on the interval $(1,2)$. Since $f'(1) > 0$, we obtain that $f$ is an increasing function that's why it can have only one root on the $(1,2)$, namely $y$.
From the mean value theorem we have
$$
\left|f(y) - f\left( \frac{p}{q} \right) \right| = |f'(\xi)| \cdot \left| y - \frac{p}{q} \right|.
$$
$\xi$ is a point on the interval between $y$ and $p/q$. This gives
$$
\left| y - \frac{p}{q} \right| = \frac{\left| f\left( \frac{p}{q} \right) \right|}{|f'(\xi)|}.
$$
Now we need to estimate the right-hand side.
Denote
$$
M = \max_{x \in [1,2]} |f'(x)|
$$
and let's see what happens when we plug $\frac{p}{q}$ in our polynomial
$$
f\left( \frac{p}{q} \right) = \frac{2p^4}{q^4} + \frac{3p^3}{q^3} - \frac{3p^2}{q^2} - \frac{6p}{q} + a = \frac{2p^4 + 3p^3q - 3p^2q^2 - 6pq^3 + aq^4}{q^4}.
$$
The numerator is a non-zero integer (it's not equal to $0$ since $p/q$ is not a root). That's why its absolute value is at least one and
$$
\left| f\left( \frac{p}{q} \right) \right| \ge \frac{1}{q^4}.
$$
This gives us the needed estimate
$$
\left| y - \frac{p}{q} \right| \ge \frac{1}{Mq^4}.
$$
It should be noted that this result can be easily generalised to the polynomial of any degree $n$. It's actually a key argument in the Liouville theorem. A precise statement of this theorem can be found for example here.
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