Question

$\text{Show that}(p-1)\text{is a factor of}{p}^{10}+p^8+p^6-p^4-p^2-1$

Answer 1

$\text{Let}f\left(p\right)=p^{10}+p^8+p^6-p^4-p^2-1$

$\text{Put}p=1,\text{we obtain}$

$f\left(1\right)=1^{10}+1^8+1^6-1^4-1^2-1$

$f\left(1\right)=1+1+1–1–1–1$

$f\left(1\right)=1+1+1–1–1–1=0$

$\text{Hence,}p-1\text{is a factor of}{p}^{10}+p^8+p^6-p^4-p^2-1$