Wednesday, September 2, 2026

2026/073) How do you find the remainder when $x^10$ is divided by $x^4+x^3+x^2+x+1$

This we can do it by long division,

Here we realize that  

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

And $x^{10}-1= (x^5+1)(x^5-1)$

Or   $x^{10}-1= (x^5+1)(x^5-1)= (x^5+1)(x-1)(x^4+x^3+x^2+x+1)$

Or  $x^{10}= (x^5+1)(x-1)(x^4+x^3+x^2+x+1) + 1$

Hence remainder is 1