$Determine all Polynomial $p(x)$ with real coefficients such that $p(2)=12$\\\\ and $p(x^2)=x^2(x^2+1)p(x),\forall x\in R.$
62first
If p(x) is of degree n
then p(x^2) is of degree 2n
so 2n=4+n
so n=4
so p(x) is a degree 4 polynomial
LHS is an even function.. so RHS has to be even.. infact should have no odd powers of x
hence p(x) is of the form ax^4+bx^2+c
also p(0)=0
so we can say that c=0
p(1)=2p(1)
so p(1)=0
so a+b+c=1
but c=0 so a+b=1
also,
p(2)=12
so 16a+4b=12
4a+b=3
a+b=1
solving we get a=2/3 and b=1/3
Try to check the result for any calculation mistakes :)
341As Nishant sir has noted, we must have a polynomial of fourth degree. Let us see if we can extract some information regarding its roots.
Setting x=0, we see that P(0) = 0
Setting x=i, we see that P(-1) =0
Again setting x=-1, we see that P(1) =0
Now, we cannot have another root as if \alpha is a root then so is \alpha^2. So the 4th root is either 0 or 1.
Let P(x) = ax(x^2-1)(x-\alpha)
Then substituting, we see that \alpha=0
From P(2)=12, we have a=1. Hence P(x)=x^2(x^2-1)
