how to write product of real quadratic polynomials?

Question

could anyone tell me how to write $x^{2n}+x^{2n-1}+\dots+x+1$ as product of real quadratic polynomial. Thanks for your help.

Answer 1

$$1+x+x^2+\cdots+x^{2n}=0 \\ \implies (1-x)(1+x+x^2\cdots x^{2n})=1-x^{2n+1}=0$$ The roots of the second equation are $$\exp\left(i\frac{k\pi}{2n+1}\right)$$ for $k=0,1,\cdots,2n$ so the quadratic factors are of form $$\left(x+\exp\left(i\frac{t\pi}{2n+1}\right)\right)\left(x+\exp\left(i\frac{(2n-t+1)\pi}{2n+1}\right)\right)$$ For $t=1,2,\cdots,n$

Answer 2

The form is a geometric progression, hence is equal to $\dfrac{x^{2n+1} - 1}{x - 1}$, so the roots of the form are the roots of $x^{2n+1} = 1$, except $x = 1$. The pairs of conjugated roots will give you quadratic polynomials. I guess you can continue.