1.

Find p(0)for polynomial :p(y)=ysquare-y+1

Answer» 1 ONE
-1
P(y)=y2-y+1Put y=0P(0)=(0)2-0+1P(0)= 1
p(y) = y2 - y + 1p(0) = (0)2 - 0 + 1p(0) = 1


Discussion

No Comment Found