Symmetric Polynomial

$\begin{aligned} z^8 + 4z^6-10z^4+4z^2+1 & = 0 \quad \quad \small \color{#3D99F6}{\text{Dividing both sides with } z^4} \\ z^4 + 4z^2-10+4\frac{1}{z^2}+\frac{1}{z^4} & = 0 \\ \left(z^2 + \frac{1}{z^2}\right)^2 + 4 \left(z^2 + \frac{1}{z^2}\right) - 12 & = 0 \\ \left(z^2 + \frac{1}{z^2} - 2\right) \left(z^2 + \frac{1}{z^2} + 6 \right) & = 0 \end{aligned}$

$\Rightarrow \begin{cases} z^4 - 2z^2 + 1 = 0 & \Rightarrow (z^2 - 1)^2 = 0 & \Rightarrow z = \begin{cases} -1 \\ 1 \end{cases} \\ z^4 + 6z^2 + 1 = 0 & \Rightarrow z^2 = -3 \pm 2\sqrt{2} = - (1 \mp \sqrt{2})^2 & \Rightarrow z = \begin{cases} -(1 + \sqrt{2})i \\ -(1 - \sqrt{2})i \\ (1 - \sqrt{2})i \\ (1 + \sqrt{2})i \end{cases} \end{cases}$

Therefore, there are $\boxed{6}$ distinct roots.

Actually, replace all z's with x, 'cause why not?

$x^8+4x^6-10x^4+4x^2+1=0$

$(x^8+4x^6+6x^4+4x^2+1)-16x^4 = 0$

Now, do you notice something?

$(x^2+1)^4-16x^4 = 0$

$((x^2+1)^2-4x^2)((x^2+1)^2+4x^2)=0$

Equating each factor to 0, we get

$(x^2+1)^2-4x^2=0$

$(x^2-2x+1)(x^2+2x+1) = 0$

$(x-1)^2(x+1)^2=0$

$x= {-1,1}$ (Two solutions)

Now, for the other factor:

$(x^2+1)^2+4x^2 = 0$

$x^4+6x^2+1 = 0$

Now, since this is not a perfect square trinomial, we will have 2 distinct solutions for $x^2$ , and so, we will have four distinct solutions for $x$ . So, the number of distinct solutions we will have is $4+2 = \boxed{6}$

That's the oldest way to solve. Note that $1$ and $-1$ are both roots of the polynomial so $z^2 -1$ divides it. So rearranging (or by using the division by polynomial algorithm) we get

$P(z) = z^8 - 4z^6 - 10z^4 + 4z^2 + 1 = (z^6 + 5z^4 - 5z^2 -1)(z^2 -1)$

The first factor has still $1$ and $-1$ as roots so we can divide it again by $z^2 -1$ so we get

$P(z) = (z^4+6z^2-1)(z^2-1)(z^2-1)$

The first factor can be easily solved placing $t = z^2$ and we get two distinct values for $t$ namely $3 + \sqrt{10}$ and $3 - \sqrt{10}$ .

In the REAL field (since the latter is negative) we get only two distinct values for $z$ from $z^2 = 3 + \sqrt{10}$ and we only get $\boxed{4}$ solutions (the last two, and $1$ and $-1$ each of multeplicity $2$ ).

In the COMPLEX field we also get two distinct solutions from $z^2 = 3 - \sqrt{10}$ , so we have a total of $\boxed{6}$ solutions, which is the right answer to the problem.

Roots are : $1,-1,i(\sqrt{2}-1),i(\sqrt{2}+1),-i(\sqrt{2}+1),-i(\sqrt{2}-1)$

Since the powers are even and coefficients are symmetric , we can first take $X=Z^2,\ and\ divide\ the\ equation\ by\ X^2.\\ and\ take\ t=(X+\frac 1 X)$
On solving and going back to Z we get $Z=\pm\ 1,\ \pm 1,\ (two\ distict) \ \ \ \pm\ (1\ \pm\ \sqrt2) ...(four).\ \ total=2+4+6.$

Not a level 5 qs Replace z^2 with t to get (t-1)^2(t^2+6t+1)