(2) P(x)=a₀+a₁x+a₂x²............a_nx ⁿ, a_n # 0 be such that P(x²)=(P(x))^2. than value of P(x) =

P(1²) = [ P(1) ]²

Let n = 1

a0 = a0²

a0 = 1

LEt n = 2

a0  + a1 = a0² + a1² + 2.a0.a1

1 + a1 = 1 + a1² + 2a1

-a1 = a1²

a1 = 0, -1

as an # 0 ... so only solution is a1 = -1

so P(1) for n = 2 is 1 - 1 = 0

Similarly by solving ahead, we observe that:

P(x) = 0