<s>[INST] What is the sum of the coefficients? If $h(x) = x^4+ax^3+bx^2+cx+d$ then what is $a+b+c+d$?

I try:
\begin{align}
x=2: 2^4+2^3a+2^2b+2c+d = 3 &\implies 8a+4b+2c+d = -13 \label{I} \tag{I}\\
x=-2:-2^4-2^3a-2^2b-2c+d=3 &\implies  -8a+4b-2c+d = -13 \label{II} \tag{II}\\
\eqref{I} + \eqref{II} \colon 8b +2d = -26 &\implies \boxed{4b+d = -13} \ em \tag{I}\\
8a-13+2c=-13 &\implies 8a+2c = 0 \implies\boxed{4a+c = 0}
\end{align}
I stop here..don't find another equation
 [/INST] Hint
I believe there is a better approach for this problem.
Considering the graph, the polinomial $h(x)-3$ has a double root at $-2$ and $2$. So, it can be writen as
$$h(x)-3=k(x-2)^2(x+2)^2=k(x^2-4)^2.$$
In addition, $h(1)$ give you the sum of the coefficients of $h(x)$.
Can you finish?
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