Not so easy

It is given that the quadratic polynomial $p(x) \ge 0$ for all real $x$ and that $p(1) = 0$ . This means that the polynomial is of the form: $\space p(x) = a(x-1)^2$ .

Since $\space p(2) = a(2-1)^2 = 2 \quad \Rightarrow a = 2$ .

Therefore, $\space p(0) = 2(0-1)^2 = 2\space$ and $\space p(3) = 2(3-1)^2 = 8$ ,

$\quad \Rightarrow p(0) + p(3) = 2 + 8 = \boxed{10}$ .

let f ( x ) = a x ^(2)+b x +c then

f (1)=a+b+c=0 -------- (1)

f (2)=4a+2b+c=2 ------- (2)

Since p (x)>=0 , hence D >=0

=> b^(2) - 4ac>=0

=> (a+c)^(2) - 4ac >=0 ( Using eq 1 )

=> (a-c)^(2) >=0

which implies either a=c or a>c

taking a=c and solving equation 1 and 2

we get,

a=c=2 ; and b= -4

substitute the values in p (x) we get desired ans which is 10 :)