1. ## Polynomial long division.

Can someone help me to simplify this expression:
(x^8+x^6+x^4+x^2+1)/(x^4+x^3+x^2+x+1).
The solution is x^4-x^3+x^2-x+1, but I don't know how to get it.

2. Just use long division.

3. In the instruction it is said that I should use this formula: 1+q^2+q^3+...+q^n=(1-q^n+1)/(1-q). And it is ok for denominator, but I don't know what I should to do with numerator.

4. $\displaystyle x^8+x^6+x^4+x^2+1 = (x^2)^4 + (x^2)^3 + (x^2)^2 + (x^2) + 1$

let $\displaystyle q = x^2$ and use your given formula

5. Hello, Frannny!

$\displaystyle \text{Simplify: }\:\frac{x^8+x^6+x^4+x^2+1}{x^4+x^3+x^2+x+1}$

$\displaystyle \text{The solution is: }\:x^4-x^3+x^2-x+1$

We have: .$\displaystyle \frac{x^8+x^6+x^4+x^2+1}{x^4+x^3+x^2+x+1}$

$\displaystyle \text{Multiply by }\frac{x^2-1}{x^2-1}\!:\;\;\frac{x^2-1}{x^2-1}\cdot \frac{x^8+x^6+x^4+x^2+1}{x^4+x^3+x^2+x+1}$

. . . . . . . . . . . . . $\displaystyle =\;\frac{x^{10}-1}{(x^2-1)(x^4+x^3+x^2+x+1)}$

$\displaystyle \text{Multiply by }\frac{x-1}{x-1}\!:\;\;\frac{x-1}{x-1}\cdot \frac{x^{10}-1}{(x^2-1)(x^4+x^3+x^2+x+1)}$

. . . . . . . . . . . . . $\displaystyle =\;\frac{(x-1)(x^{10}-1)}{(x^2-1)(x^5-1)}$

$\displaystyle \text{Factor: }\;\frac{(x-1)(x^5-1)(x^5+1)}{(x-1)(x+1)(x^5-1)}$

$\displaystyle \text{Reduce: }\;\frac{x^5+1}{x+1}$

$\displaystyle \text{Factor: }\;\frac{(x+1)(x^4-x^3+x^2-x+1)}{x+1}$

$\displaystyle \text{Reduce: }\;x^4-x^3+x^2-x+1$