Perhaps this is similar to the Lagrange polynomial case.
The Chebyshev polynomials, $T_0, \cdots T_n$ are mutually orthogonal with inner product taken as a sum over the zeros of $T_{n+1}$. The zeros are easily calculated because $T_k(\cos \theta) = \cos k\theta $ (here $x = \cos \theta$). The polynomials are also easily obtained:
$$ T_0 = 1, T_1(x) = x, T_{n+1}(x) = 2x T_{n}(x) - T_{n-1}(x) $$
If interested in differentiation, the derivative of each $T_k$ is simply $k$ time the corresponding Chebyshev polynomial of the second kind (see Wikipedia).
For your problem if you precompute the first $n$ Chebyshev polynomials and save their coefficients, evaluate you $p(x)$ at the $n+1$ zeros and take inner product against the $n+1$ polynomials $ T_0, \cdots, T_{n} $ you quickly obtain coefficients for $p$ in the basis $T_0, \cdots, T_n$. Then use you pre-computed table to give you the coefficients for $p(x)$ in terms of $1, x, \cdots, x^n$. You will need to calculate the norms of each $T_0, \cdots, T_n $ under the inner product - I can't remember if there is a simple form for it or not.
Postscript: I checked. The sum inner product over the $n+1$ zeros of $T_{n+1}$, $\omega_0, \cdots \omega_n$ satisfies $$\langle T_i, T_j \rangle = \sum_{r=0}^n T_i(\omega_r)T_j(\omega_r) = \left\{ \array{0 & i\neq j \\ n+1 & i=j=0 \\ (n+1)/2 & i=j > 0} \right. $$
Best Answer
[ EDIT ] The original answer is in the P.S. at the bottom of this post. Immediately following is a different and complete answer.
For $P(x) = 2x^4 - ax^3 + 19x^2 - 20x + 12$ to have a double root, it must have that root in common with its derivative $P'(x) = 8 x^3 -3 a x^2 +38 x - 20$. Eliminating $a$ between the two equations $P(x)=P'(x)=0$ gives:
$$ Q(x) = 2 x^4 - 19 x^2 + 40 x - 36 = 0 $$
Therefore any double root of $P$ must be a root of $Q$. It is easily verified that any integer root of $Q$:
must be a divisor of $36$;
must be even since the polynomial has exactly one odd coefficient;
if a multiple of $3$ it must be a multiple of $9$ otherwise the equality fails $\bmod 9$.
Given that the double root is assumed to be a positive integer, the above leaves $x=2,4,18,36 \;$ $(\dagger)\;$ as the only eligible roots. By direct verification, $2$ is the only one which is in fact a root of $Q(x)$.
Substituting $x=2$ back in $P(x)=0$ gives $a=10$. Given how the derivation worked, $x=2$ is known at this point to be a double root of $P$ for $a=10$ so no further verification is necessary.
$\dagger\;$ Note: if $a$ is assumed to be an integer, then the common root would have to be a divisor of $\gcd(12, 20, 36) = 4$ which would limit the eligible values to just $2,4$.
[ EDIT ] P.S. Below is the original answer.
Partial but quick solution: let $2x^4 - ax^3 + 19x^2 - 20x + 12 = (2 x^2 + b x + c)(x^2 - 2k x + k^2)$.
Assume that $c \in \mathbb{Z}\;$ ("assume" because $c$ is not otherwise required to be an integer), then it follows that $k^2 \mid 12$ with integer solutions $k = \pm 1, \pm 2$. The polynomial has no negative roots by the rule of signs, which leaves $k=1,2$ as candidates.
$k=1$ implies $2 - a + 19 - 20 + 12 = 0$ so $a=13$. But the derivative $8 x^3 - 39 x^2 + 38 x - 20$ does not vanish at $x=1$ so $1$ is not a double root.
following the same steps, $k=2$ gives $a=10$ which can be verified to satisfy all conditions.
Note that the above finds one solution quickly, but does not prove that it's the only solution.