Determining the roots of polynomial $R(x)$

Question

$$P(x) = x^2 -1 $$

$$R(x) = P(P(x))$$

• Determine the roots of the polynomial $R(x)$. ($1$, $0$ or $-1$)

My Attempt:

$$x^2-1 = 0$$

Rewriting the equation as

$$(x+1)(x-1)=0$$

Which yields two solutions

$$x = -1 \space \lor \space x = 1 $$

However, the correct root seems to be $0$. Can you explain why?

Regards!

Answer 1

$R(x) =P(P(x))$

$R(x) =(x^2-1)^2-1$

$R(x) = x^4+1-2x^2-1$

$R(x) = x^4-2x^2$

now:

$R(x)=0\implies x^2(x^2-2)=0$

$\implies x=0,\pm\sqrt2$

Answer 2

$$R(x)=0\iff P(P(x))=0\iff P(x)=\pm1\iff x^2-1=\pm1$$ so we get $x=0$, $\sqrt2$ or $-\sqrt2$.