Let \(P(x)=x^2+\frac {x}{ 2} +b\) and \(Q(x)=x^2+cx+d\) be two polynomials with real coefficients such that \(P(x)Q(x)=Q(P(x))\) for all real \(x\). Find all real roots of \(P(Q(x))=0\)

Expanding the second relation

we get d=0 , b=-1/2 and c=1/2

therefore

now we can say

P(x) = x^2+x/2 -1/2) and Q(x)= x(x+1/2)

Now if we see Q(x) has two real roots namely x=0, and x=-1/2

and P(x) has roots  1/2 and -1 , now to check  when Q(x)=0 , but P(0) isn't 0

so we have two real roots , and also interestingly two imaginary roots since P(Q(x)) is a biquadratic equation following the fundamental theorem of algebra