How to express $z^8 − 1$ as the product of two linear factors and three quadratic factors 
Verify
$$(z-e^{i \theta} ) (z - e^{-i \theta} ) ≡ z^2 - 2\cos \theta + 1$$
Hence express $z^8 − 1$ as the product of two linear factors and three quadratic factors, where all
coefficients are real and expressed in a non-trigonometric form.


For the first part I just expanded  the LHS and showed its equal to RHS.
Roots of $z^8-1$:
$$z= e^{i \frac{\pi k}{4}}$$

Where $k=0,1,2,3,4,5,6,7$

Ok since they want non-Trignometric so :
I know two roots, which are obvious:
$$z=1,-1$$
$$z^8-1=(z-1)(z+1)(z^6+z^5+z^4+z^3+z^2+z+1)$$
They want two linear factors which I believe I have found: $(z+1)$ and $(z-1)$
How do I make
$$(z^6+z^5+z^4+z^3+z^2+z+1)$$
to three quadratic factors? And they mentioned Hence , so I have to use the identity I verified. Please help.

Recall the two versions of the fundamental theorem of algebra:


*

*Any complex polynomial can be factored into a product of linear terms

*Any real polynomial can be factored into a product of linear and quadratic terms (and the quadratic terms have no real roots)


In this case, we can find the factorization as a complex polynomial into linear terms easily; the trick then to recover the factorization as a real polynomial is to multiply the linear terms together appropriately to convert pairs of complex linear terms into real quadratic terms.
In particular, you want to group the conjugate roots together, and multiply them as described by the identity you list.
