I want to express $z^8-1$ as a product of two linear factors and three quadratic factors, where all coefficients are real and expressed in a non-trigonometric form.
I know that$z^8-1=(z^4-1)(z^4+1)=(z^2-1)(z^2+1)(z^4+1)=(z-1)(z+1)(z^2+1)(z^4+1)$ what should I do next?
Well, you have your two linear factor, and one quadratic factor. So it reamains to split $$z^4+1 = (z^2+\sqrt{2}z+1)(z^2-\sqrt{2}z+1)$$ and you are done.
How did I get this factorization? The roots of $z^4+1$ are $\pm \frac{\sqrt{2}}2 (1 \pm i)$ (using trigonometry). Then put together conjugate roots, and you will find the answer.
Decided to write this up as an answer because I find it useful to remember it like this, and it makes the factorisation a lot easier than using complex numbers as Crostul did.
To factorise $z^4 + 1$, we can write $z^4 + 1 = z^4 + 2z^2 + 1 - 2z^2 = (z^2 + 1)^2 - (\sqrt 2 z)^2 = (z^2 + \sqrt 2 z + 1)(z^2 - \sqrt 2 z + 1)$.
This method is neater and avoids complex numbers.
Your final factorisation (with real coefficients) is therefore:
$$z^8 - 1 = (z^4-1)(z^4+1) =(z^2-1)(z^2+1)(z^4+1)=(z-1)(z+1)(z^2+1)(z^2 + \sqrt 2 z + 1)(z^2 - \sqrt 2 z + 1)$$
