theory of equations using transformations

Question

i think this case is not possible. If you take $x$ as $5$, and -$5$, $P(27)$ will have two values. so i think the ans is D.

Answer 1

No such polynomial exists because the degree of $P(x^2+2)$ is even but the degree of $x^{17}+\cdots$ is odd.

Answer 2

Indeed, there will be no such polynomial $P$. As you pointed out correctly, plugging in $x=-5$ and $x=5$ will give you different answers, but $P(27)$ will only be one number.

Another way to see this is if $P(y) = \sum_{k=0}^n a_k y^k$ is any polynomial and you consider the new polynomial $Q(x)=P(x^2+2)=\sum_{k=0}^n a_k (x^2+2)^k$, it will only contain even powers of $x$, but yours is supposed to have odd ones.