Consider the function is, f(x)=ax2+bx+c From the given condition, f(0)+f(1)=0 c+a+b+c=0 a+b+2c=0.......(1) Again, f(−2)=0 a(−2)2+b(−2)+c=0 4a−2b+c=0 4a−2b+c=0.......(2) Solving equation (1) and (2), we get
a
5
=
b
7
=
c
−6
=k a=5k,b=7k,c=−6k The function becomes f(x)=k(5x2+7x−6) 5x2+7x−6=0 (x−2)(5x−3)=0 x=−2,