f(x)=(k²-3k+2)x²+2x+m+1
f(-x)=(k²-3k+2)x²-2x+m+1
-f(x)=-(k²-3k+2)x²-2x-m-1
(k²-3k+2)x²-2x+m+1=-(k²-3k+2)x²-2x-m-1
(k²-3k+2)x²+m+1=-(k²-3k+2)x²-m-1
m+1=-m-1
m=-1
k²-3k+2=-k²+3k-2
k²-3k+2=0
(k-1)(k-2)=0
k=1 k=2