Question

Show that $x-1_F$divides $a_n{x}^n+.....+a_2{x}^2+a_{1}x+a_0$ in$F\left[x\right]$ if and only if$a_0+a_1+a_2+........+a_n=0_F$ .

Short answer

It is proved that $x-1_F$ divides $a_n{x}^n+...........+a_2{x}^2+a_{1}x+a_0\mathrm{in}F\left[x\right]$ if and only if $a_0+a_1+a_2+......+a_n=0_F$.

Step-by-step solution

Factor Theorem:

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

Proof:

Given function is:$a_n{x}^n+...........+a_2{x}^2+a_{1}x+a_0\mathrm{in}F\left[x\right]$

Rewritten as: $f\left(x\right)={\sum }_{i=0}^n{a}_i{x}^i\in F\left[x\right]$

At, $x=1$,we have:

$f\left(1\right)={\sum }_{i=0}^n{a}_i$

So, according to the factor theorem, we see, $x-1\left|f\left(x\right)\right.$ if and only if:

$f\left(1\right)={\sum }_{i=0}^n{a}_i=0$

Or, this can be written as:

$a_0+a_1+a_2+......+a_n=0_F$

Hence proved, $x-1_F$ divides $a_n{x}^n+...........+a_2{x}^2+a_{1}x+a_0\mathrm{in}F\left[x\right]$if and only if $a_0+a_1+a_2+......+a_n=0_F$ .