Insanely Huge Symmetric Polynomial

The given equation is a $\color{#3D99F6}{\text{Reciprocal Equation}}$ (also called $\color{#3D99F6}{\text{Palindromic Equation}}$ ) as $\color{#D61F06}{\text{Co - efficient of} \, z^{n} \, = \, \text{Co - efficient of}\,z^{11-i}}$ , where $\color{#D61F06}{i \,\,\epsilon \,\,(0,1,2,..,10,11)}$ . Generally, a palindromic equation of degree $\color{#D61F06}{n}$ can be reduced to an equation of degree $\color{#D61F06}{\frac{n}{2}}$ , where $\color{#D61F06}{n}$ is even, by making a substitution $\color{#20A900}{z+\frac{1}{z} \,=\, t}$ . But here $\color{#D61F06}{n}$ is odd, so we need to perform an additional step of manipulation which is as follows :- $4z^{11}+4z^{10}-21z^9-21z^8+17z^7+17z^6+17z^5+17z^4-21z^3-21z^2+4z+4=0 \implies (z+1)\underbrace{(4z^{10}-21z^{8}+17z^{6}+17z^{4}-21z^{2}+4)}_\text{This can be reduced !}=0$ . $\color{#20A900}{(z+1)}$ gives $\color{magenta}{z=-1}$ . Now, considering, $\color{#20A900}{4z^{10}-21z^{8}+17z^{6}+17z^{4}-21z^{2}+4=0}$ . Dividing the entire equation by $\color{#20A900}{z^{5}}$ and then replacing $\color{#20A900}{z+\frac{1}{z}}$ by $\color{#20A900}{t}$ , followed by simplification yields :- $4t^{5}-41t^{3}+100t=0 \implies t(t^{2}-4)(4t^{2}-25)=0$ $\implies t = 0, \,\pm\, 2, \,\pm \,\frac{5}{2} \implies z+\frac{1}{z} = 0,\, \pm \,2, \,\pm \,\frac{5}{2} \implies \color{magenta}{z = \underbrace{\pm\, i}_\text{-1 occurs here as well}, \pm \,1, \pm \,\frac{1}{2}, \pm\, 2 }$ . So, total number of distinct roots is $\color{#3D99F6}{8}$ . Furthermore, $\boxed {\color{magenta}{4z^{11}+4z^{10}-21z^9-21z^8+17z^7+17z^6+17z^5+17z^4-21z^3-21z^2+4z+4=(x-2)(x-1)^{2}(x-i)(x+i)(x+1)^{3}(x+2)(2x-1)(2x+1)}}$