# Find the polynomial P(x) with real coefficients such that P(2)=12 and P(x^2)=x^2(x^2+1)P(x) for each x in RR?

Oct 28, 2016

$P \left(x\right) = {x}^{4} - {x}^{2}$

#### Explanation:

First, observe that the degree of $P \left({x}^{2}\right)$ will be twice the degree of $P \left(x\right)$, and that the degree of ${x}^{2} \left({x}^{2} + 1\right) P \left(x\right)$ will be $4$ greater than the degree of $P \left(x\right)$. Taken together, we know that $P \left(x\right)$ is a polynomial of degree $4$.

Further observations:

$P \left({0}^{2}\right) = 0 \left(1\right) P \left(0\right) = 0$

Thus $x$ is a factor of $P \left(x\right)$.

$P \left(1\right) = P \left({1}^{2}\right) = 1 \left(2\right) \left(P \left(1\right)\right)$

$\implies P \left(1\right) = 2 P \left(1\right)$

$\implies P \left(1\right) = 0$

Thus $\left(x - 1\right)$ is a factor of $P \left(x\right)$

$0 = P \left(1\right) = P \left({\left(- 1\right)}^{2}\right) = 1 \left(2\right) P \left(- 1\right)$

$\implies P \left(- 1\right) = 0$

Thus $\left(x + 1\right)$ is a factor of $P \left(x\right)$

With those, we can write $P \left(x\right)$ as, for some $a , c \in \mathbb{R}$,

$P \left(x\right) = c x \left(x - 1\right) \left(x + 1\right) \left(x - a\right) = c x \left({x}^{2} - 1\right) \left(x - a\right)$

Plugging in $P \left(2\right) = 12$, we get

$c \left(2\right) \left(3\right) \left(2 - a\right) = 12$
$\implies 12 c - 6 a c = 12$
$\implies 2 c - a c = 2$

Plugging in $P \left(4\right) = P \left({2}^{2}\right)$, we get

$P \left(4\right) = 4 \left(5\right) \left(12\right) = 240$
$\implies c \left(4\right) \left(15\right) \left(4 - a\right) = 240$
$\implies 240 c - 60 a c = 240$
$\implies 4 c - a c = 4$

Solving the system $\left\{\begin{matrix}2 c - a c = 2 \\ 4 c - a c = 4\end{matrix}\right.$, we arrive at

$\left\{\begin{matrix}a = 0 \\ c = 1\end{matrix}\right.$

$\therefore P \left(x\right) = {x}^{2} \left({x}^{2} - 1\right) = {x}^{4} - {x}^{2}$

Oct 28, 2016

$P \left(x\right) = {x}^{4} - {x}^{2}$

#### Explanation:

For $x \ne 0$ we have $P \left(x\right) = \frac{P \left({x}^{2}\right)}{{x}^{2} \left({x}^{2} + 1\right)}$ so

$P \left(x\right) = P \left(- x\right)$ an even function. Also

$\mathrm{de} g \left(P \left({x}^{2}\right)\right) = 2 \mathrm{de} g \left(P \left(x\right)\right) = 4 + \mathrm{de} g \left(P \left(x\right)\right)$ so

$\mathrm{de} g \left(P \left(x\right)\right) = 4$

If $\mathrm{de} g \left(P \left(x\right)\right) = 4$ and $P \left(x\right)$ is even then

$P \left(x\right) = a {x}^{4} + b {x}^{2} + c$ but

$P \left(0\right) = 0$ so $c = 0$. Now

$P \left(2\right) = {2}^{4} a + {2}^{2} b = 12$ and
$P \left({\left(\sqrt{2}\right)}^{2}\right) = 2 \left(2 + 1\right) \left({2}^{2} a + 2 b\right) = 12$

solving for $a , b$ we get $a = 1$ and $b = - 1$ so

$P \left(x\right) = {x}^{4} - {x}^{2}$

Oct 28, 2016

${x}^{2} \left({x}^{2} - 1\right)$

#### Explanation:

${x}^{2} \left({x}^{2} + 1\right) = \frac{P \left({x}^{2}\right)}{P \left(x\right)} {\cong}_{x \to \infty} {x}^{2 n} / \left({x}^{n}\right) = {x}^{n} \setminus \implies n = 4$

{(P(x)=P(-x)=(P(x^2))/(x^2(x^2+1))), (P(0)=0):}\ \ => \ P(x)=ax^4+bx^2

$P \left({x}^{2}\right) = a {x}^{8} + b {x}^{4} = {x}^{2} \left({x}^{2} + 1\right) \left(a {x}^{4} + b {x}^{2}\right) = \left(b + a\right) {x}^{6} + \ldots$

$\implies b = - a$

$P \left(x\right) = a \left({x}^{4} - {x}^{2}\right)$

$P \left(2\right) = a \left(16 - 4\right) = 12 \setminus \setminus \implies a = 1$

$P \left(x\right) = {x}^{2} \left({x}^{2} - 1\right)$