Given that $P(x)$ is a polynomial such that $P(x^2+1) = x^4+5x^2+3$, what is $P(x^2-1)$?

Question

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 $P(x^2-1)$.

Answer 1

Since$$x^4+5x^2+3=(x^2+1)^2+3(x^2+1)-1,$$ $P(x)=x^2+3x-1$ and therefore $P(x^2-1)=\cdots$

Answer 2

Hint:   let $y=x^2$, then the given condition writes as $P(y+1)=y^2+5y+3$ and it follows that $P(x^2-1)=P\big((y-2)+1\big)= (y-2)^2+5(y-2)+3 = \ldots$

Answer 3

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 $.

Answer 4

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 $P(x^2-1)=\ldots=x^4+x^2-3$.

Answer 5

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.