Factorising a spacial polynomial

Question

I have to factorise $2(x^2 + x^3 + x^4 + x^5 + x^6 + x^7)$. It seems that there a very simple expression to which simply this polynomial.

The only thing I found so far is $2 x^2 (1 + x + x^2 + x^3 + x^4 + x^5)$. Can you explain to me in details how to factorise this even more? If so, is there a general rule for such similar polynomial?

Answer 1

$ 2x^2(1+x+x^2+x^3+x^4+x^5) $

$ 2x^2(1(1+x)+x^2(1+x)+x^4(1+x)) $

$ 2x^2(1+x^2+x^4)(1+x) $

$ 2x^2(x^2+x+1)(x^2-x+1)(1+x) $

Hope this help you

Edit: Here's how $ (1+x^2+x^4) $ can be factorized to $ (x^2+x+1)(x^2-x+1) $

$ x^4+x^2+1 $

$ =(x^2)^2+2(x^2)(1)+1^2−x^2$

$ =(x^2+1)^2−x^2$

$ =(x^2+1−x)(x^2+1+x)$

Answer 2

$x^6 -1 = (x-1)(1+x+x^2+x^3+x^4+x^5)$

so all complex $6$-th roots of unity except $1$ are roots of $(1+x+x^2+x^3+x^4+x^5)$, group them by conjugate pair and you have a factorisation over $\Bbb R$.

Answer 3

$$2(x^2+x^3+x^4+x^5+x^6+x^7)$$

$$2x^2(1+x+x^2+x^3+x^4+x^5)$$

$$2x^2\frac{x^6-1}{x-1}$$

Answer 4

Hint:

$x^2+x^3=x^2(1+x)$

$x^4+x^5=x^4(1+x)$

You can continue...