Factoring the polynomial $ P(X)= X^6-X^5-X+1 $ over real and complex numbers

Question

I'm trying to solve this exercise, I'm starting with polynomials and I'm wondering how to answer the 2 and 3 with the help of 1).

We have the polynomial $$ P(X)= X^6-X^5-X+1 $$

1. Prove that 1 is a root of P(X) and factor it by $(X-1)$

I calculated $P(1)=0$ and found that $$ P(X)=(X-1)(X^5-1)$$

2. Factor $P(X)$ on $\mathbb C$.
3. Factor $P(X)$ on $\mathbb R$.

How to do 2) and 3), is 1) useful here?

Answer 1

You already found one factor in part 1. So what could be a factor of $X^5-1$? Can you see/guess another real root? Then you will be left with a degree 4 polynomial. Try to factor this (over the reals) into two quadratics. All roots of $X^5-1$ have absolute value $|x|=1$, so what could be the constant term of such a quadratic factor (recall that it is the product of all roots of the factor!)

Answer 2

Hints:

b) Solutions to an equation of the form $x^n-1$ are called the $n$th roots of unity. They have a general form.

c)

$$(a^5-b^5)=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)$$