Question

Use the Factor Theorem to show that ${\mathit{x}}^{\mathbf{7}}\mathbf{-}\mathit{x}$factors in${\mathbf{\mathbb{Z}}}_{\mathbf{7}}\left[x\right]$ as$\mathit{x}\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)\left(x-6\right)$ , without doing any polynomial multiplication.

Short answer

It is proved that $x\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)$can be expressed as factors$x^7-x$.

Step-by-step solution

Factor Theorem:

If F is any field such that$f\left(x\right)\in F\left[x\right]$and $\mathit{k}\mathbf{\in }\mathit{F}$. Then, kwill be the root of the polynomial f(x), if and only if (x - k) is a factor off(x) in $\mathit{F}\left[x\right]$.

Proof:

The given polynomial is: $x^7-x$.

Now, using Fermat's Little Theorem, for 7 being a prime number, we have:

$p^7=p .............\left\{\forall p\in {\mathrm{\mathbb{Z}}}_7\right\}$

This implies that every element in ${\mathrm{\mathbb{Z}}}_7$will be a root of $x^7-x$, so, we have:

$\prod _{i=0}^6\left(x-i\right)\left|x^7-x\right.$

Analyzing both the polynomials, we have:

$\text{deg}\left\{\prod _{i=0}^6\left(x-i\right)\right\}=\text{deg}\left\{x^7-x\right\}$

Clearly, both are associates.

Hence proved, $x\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)$ can be expressed as factors $x^7-x$.