Let $f(x)$ be a 3rd degree polynomial such that $f(x^2)=0$ has exactly $4$ distinct real roots Problem : 
Let $f(x)$ be a 3rd degree polynomial such that $f(x^2)=0$ has exactly four distinct real roots, then which of the following options are correct : 
(a) $f(x) =0$ has all three real roots 
(b) $f(x) =0$  has exactly two real roots 
(c) $f(x) =0$ has only one real root 
(d) none of these 
My approach : 
Since $f(x^2)=0 $ has exactly four distinct real roots. Therefore the remaining two roots left [ as $ f(x^2)=0$ is a degree six polynomial].
How can we say that the remaining two roots will be real . This will not have one real root ( as non real roots comes in conjugate pairs). 
So, option (c) is incorrect. I think the answer lies in option (a) or (b) . But I am not confirm which one is correct. Please suggest.. thanks..

Both $f(x)=(x-1)(x-4)^2$ (two real roots including one double root), and $f(x)=(x+1)(x-1)(x-4)$ (three real roots) fulfit the conditions of the problem, with $f(x^2)=0$ having roots in $x=\pm1$ and $x=\pm2$.
