Extra Polynomial

We can use Bezout's polynomial remainder theorem by defining an extra polynomial Q(x) = P(x) - 10x (1).

We know 1;2;3 are solutions for Q(x). Hence Q(x) = (x-1)(x-2)(x-3)(x-m) (m is another solution for Q(x), since Q(x) is a fourth degree polynomial).

Substituting Q(x) in (1) gives us P(x)= (x-1)(x-2)(x-3)(x-m) + 10x.

Now we can easily calculate $\frac{P(12)+P(-8)}{10}$ (While calculating, you may notice m is eliminated). The answer is $\boxed{1984}$

Note that the $x$ -values we're interested in $(-8,1,2,3,12)$ are symmetric about $2$ . So define an extra polynomial $Q(x)=P(x+2)$ . The information we have is: $Q(-1)=10$ ; $Q(0)=20$ ; $Q(1)=30$ .

If we let $Q(x)=x^4+Ax^3+Bx^2+Cx+D$ , these become $1-A+B-C+D=10$ , $D=20$ , $1+A+B+C+D=30$

Substituting for $D$ and summing the other two equations gives $B=-1$ .

Now, $P(-8)=Q(-10)$ and $P(12)=Q(10)$ . In general, $Q(-x)+Q(x)=2x^4+2Bx^2+2D$ ; substituting in for $x=10$ , $B=-1$ , $D=20$ gives $Q(-10)+Q(10)=20000-200+40=19840$ , so the required answer is $\boxed{1984}$ .