F(p(x))=p(x+1)-p(x)+x^2p(0)
P3の基底1,x,x^2,x^3 とP2の基底1,x,x^2

F(1)=x^2
F(x)=1
F(x^2)=x^2+2x+1-x^2=2x+1 F(x^3)=x^3+3x^2+3x+1-x^3=3x^2+3x+1

(1 x x^2 x^3)->(1 x x^2)

0 1 1 1
0 0 2 3
1 0 0 3

要転置?