Given that $P(x)$ is a polynomial such that $P(x^2+1) = x^4+5x^2+3$, what is $P(x^2-1)$? How would I go about solving this? I can't find a clear relation between $x^2+1$ and $x^4+5x^2+3$ to solve $P(x^2-1)$.
 A: Since$$x^4+5x^2+3=(x^2+1)^2+3(x^2+1)-1,$$ $P(x)=x^2+3x-1$ and therefore $P(x^2-1)=\cdots$
A: Hint:   let $y=x^2$, then the given condition writes as $P(y+1)=y^2+5y+3$ and it follows that $P(x^2-1)=P\big((y-2)+1\big)= (y-2)^2+5(y-2)+3 = \ldots$
A: Here is essentially
division by $x^2+1$
that is just
plug and grind.
$\begin{array}\\
x^4+5x^2+3
&=x^4+x^2+4x^2+3\\
&=x^2(x^2+1)+4x^2+4-1\\
&=x^2(x^2+1)+4(x^2+1)-1\\
&=(x^2+4)(x^2+1)-1\\
&=(x^2+1+3)(x^2+1)-1\\
&=(x^2+1)^2+3(x^2+1)-1\\
\end{array}
$
so,
as 
José Carlos Santos got,
$P(x)=x^2+3x-1$.
Then
$P(x^2-1)
=(x^2-1)^2+3(x^2-1)-1
=x^4-2x^2+1+3x^2-3-1
=x^4+x^2-3
$.
A: Here is another way. Let $t :=\sqrt{x-1}$. Then note that 
$$\begin{align}
t&=\sqrt{x-1};\\
t^2&=x-1;\\
t^2+1&=x;\\
t^4&=x^2-2x+1.
\end{align}$$
So that $$\begin{align}
P(x)=P(t^2+1)&=t^4+5t^2+3\\
&=(x^2-2x+1)+5(x-1)+3\\
&=x^2+3x-1.
\end{align}.$$
Lastly, evaluate $P(x^2-1)=\ldots=x^4+x^2-3$.
A: One more alternative is to start by deducing that $P(x)$ is a quadratic based on the fact that its degree has to be $2$ in order to output a quartic when the argument is a quadratic expression. So let $P(x) = ax^2 + bx + c$.
When $x^2 + 1 = 0, x = \pm i$, so we get From $P(0) = 1 - 5 + 3 = -1$, and also $P(0) = c$ we can immediately get $c=-1$.
When $x=0, x^2 + 1  = 1$ so we have that $P(1) = 3$ and $P(1) = a+b-1$, so $a+b = 4$.
When $x = 1, x^2 + 1 = 2$, so we have that $P(2) = 9$ and $P(2) = 4a + 2b - 1$, so $4a + 2b = 10$.
Solving the latter two for $a$ and $b$, we get $a=1, b=3$.
Hence $P(x) = x^2 + 3x - 1$ and $P(x^2-1)$ can be worked out.
