设f(x)=ax²+bx+c
则f(x-1)+f(2x+1)
=a(x-1)²+b(x-1)+c+a(2x+1)²+b(2x+1)+c
=5ax²+x(2a+3b)+2a+2c
=5x²+2x
对比系数得:
5a=5
2a+3b=2
2a+2c=0
解得：a=1,b=0,c=-1
故f(x)=x²-1