let P(x) is a polynomial function Given $P(x^2)= x^2(x^2+1) P(x)$ and $P(2)=3$, Find $P(3)$ Question: let P(x) is a polynomial function
Given $P(x^2)= x^2(x^2+1) P(x)$ and 
$P(2)=3$, Find $P(3)$
If I put $x=2$, we get $P(4)$ which is not required.
I tried to find $P(x)$ to let it general polynomial to find its roots but unable to find
Please help!
 A: We're looking for a polynomial that satisfies:
$$
\phantom{(\ast)} \qquad P(x^2) = x^2 (x^2 + 1) P(x) . \qquad (\ast)
$$
Hint An outline of one possible solution:


*

*As a few people have suggested in the comments, use $(\ast)$ to compute that $\deg P = 4$.

*Use $(\ast)$ to deduce that $P$ is even.

*Use $(\ast)$, again, to show that $0$ is a root of $P$; since $P$ is even, $0$ is at least a double root of $P$. If $r$ is one of the remaining $2$ roots, by evenness so is $-r$, so any polynomial $P$ satisfying $(\ast)$ has the form $$P(x) = a x^2 (x + r) (x - r)$$
for some $a, r$.

*Use $(\ast)$ yet again, to show that $P(1) = 0$, so we may as well take $r = 1$.

*Substitute our expression for $P$ into the condition $P(2) = 3$ to determine $a$.

*Compute $P(3)$.

A: Elaborating on some details:
$P(0)=P(0^2)=0^2(0^2+1)P(0)=0$
$P(1^1)=P(1)=1^2(1^2+1)P(1)$ so $P(1)=2P(1)$ implying $P(1)=0$
$0=P(1)=P((-1)^2)=(-1)^2((-1)^2+1)P(-1)=2P(-1)$ implying $P(-1)=0$
So, we have found three roots so far.

As for the degree of $P$, if we were to suppose $P(x)$ were a degree $n$ polynomial, it follows that $P(x^2)$ should be a degree $2n$ polynomial.
As $\underbrace{P(x^2)}_{\text{degree }2n}=\underbrace{x^2(x^2+1)}_{\text{degree }4}\times \underbrace{P(x)}_{\text{degree }n}$ it follows that $2n=4+ n$ and so $n=4$

By the fundamental theorem of algebra, it follows that $P(x)$ can be written in the form:
$$P(x)=x(x-1)(x+1)(ax+b)$$
for some values $a,b$
Let us try to find out those values of $a,b$ using $P(2)$ and $P(4)$.
From $P(2)$ we get that $3=P(2)=2(2-1)(2+1)(2a+b)$ and so $2a+b=\frac{1}{2}$.  That is a good start but not quite enough to finish, so we use one more.
From $P(4)$ we get $P(4)=P(2^2)=2^2(2^2+1)P(2)=4\cdot 5\cdot 3 = 60$ and that $P(4)=4(4-1)(4+1)(4a+b)$ and so $4a+b=1$
Subtracting the first result away from the second, this gives $2a=\frac{1}{2}$ and $a=\frac{1}{4}$, thus $b=0$
As a result, we have that $$P(x)=x(x-1)(x+1)(\frac{1}{4}x+0)$$ or rewritten as you had arrived at in the comments $$P(x)=\frac{x^2(x^2-1)}{4}$$
From here, finding $P(3)$ is a matter of just plugging in the value and we get $P(3)=\frac{9\cdot 8}{4}=18$
A: Here's a solution which is not so efficient as some other methods in this case, but which illustrates


*

*a more general concept, and

*that you need not actually compute $P$ in order to determine $P(3)$.


Recall the (forward) finite difference operator $$\Delta[f](x) := \Delta_1 [f](x) = f(x + 1) - f(x) ,$$ which formally behaves like a derivative in some ways. The corresponding Taylor-like series is
$$f(x) \sim \sum_{k = 0}^\infty \frac{1}{k!} \Delta^k[f](a) (x - a)_k ,$$
where $y_k$ denotes the falling factorial $y_k := y (y - 1) \cdots (y - k + 1)$.
If $f$ is a polynomial $\Delta f$ is a polynomial of lower degree (or the same degree, $-\infty$, if $f = 0$), so $\Delta^k[f] = 0$ for all $k > \deg f$, and in particular the above series terminates at order $k = \deg f$.
The iterated difference operators are given by
$$\Delta_1^k [f](a) = \sum_{i = 0}^k {k \choose i} (-1)^{k - i} f(a + i) .$$
Now, since $P(x^2)$ and $x^2 (x^2 + 1)$ are both even in $x$, so is $P$. So we can compute, e.g., $P(3) = P(-3)$ by evaluating the series at $a = -2, x = -3$, and to do so we only need to know $P(-2), P(-1), \ldots, P(-2 + \deg P)$. Since $x - a = -1$, $(x - a)_k = (-1)_k = (-1)^k k!$, and so
$$\phantom{(\ast)} \qquad P(3) = P(-3) = \sum_{k = 0}^{\deg P} \sum_{i = 0}^k {k \choose i} (-1)^i f(-2 + i) \qquad (\ast) .$$

 We can do this efficiently for $P$: Comparing the degrees of both sides of $$\phantom{(\ast\ast)} \qquad P(x^2) = x^2 (x^2 + 1) P(x) \qquad (\ast)$$ gives that $\deg P = 4$, so to compute $P(3)$ we only need to know $P(0), P(-1) = P(1), P(-2) = P(2) = 3$. The constraint $(\ast)$ implies that $P(-1) = P(0) = P(1) = 0$, so all that remains to do is to substitute in $(\ast\ast)$. Now, reindexing the double summation gives $$P(3) = P(-3) = \sum_{i = 0}^k (-1)^i f(-2 + i) \sum_{k = 0}^4 {k \choose i} .$$ But the only nonzero values among $f(-2), \ldots, f(2)$ are $f(\pm 2) = 3$, so only the $i = 0, 4$ terms contribute, leaving $$\boxed{P(3) = f(-2) \sum_{k = 0}^4 {k \choose 0} + f(2) \sum_{k = 0}^4 {k \choose 4} = 5 \cdot f(2) + 1 \cdot f(2) = 6 f(2) = 6 \cdot 3 = 18} .$$

