1. Let $P(x)=x^{4}+a x^{3}+b x^{2}+c x+d, a, b, c, d$ be constants, and $P(1)=2000, P(2)=4000$, $P(3)=6000$, try to calculate $P(9)+P(-5)$.
1. Let $Q(x)=P(x)-2000 x$, from the problem we know $Q(1)=0, Q(2)=0, Q(3)=0$. Therefore, we can let $Q(x)=(x-1)(x-2)(x-3)(x-r)$ (by the Factor Theorem). Thus, $P(9)+P(-5)=Q(9)+Q(-5)+2000 \times 4=8 \cdot 7 \cdot 6 \cdot(9-r)+6 \cdot$ $7 \cdot 8 \cdot(5+r)+8000=6 \cdot 7 \cdot 8 \cdot 14+8000=12704$