Proving $(x-c)|(p(x)-p(c))$

Solution 1:

If you divide $p(x)$ with $x-a$ you get such polynomials $k(x)$ and $r(x)$ (remainder theorem) that $$ p(x)=k(x)(x-a)+r(x)$$ Since $r(x)$ must be constant and previous formula is true for all $x$ it is also for $x=a$. Thus you get $r(x)=r(a)=p(a)$. So we have $$p(x)-p(a)=k(x)(x-a)$$ Thus $x-a$ divide $p(x)-p(a)$ if $p\in F[x]$.

Solution 2:

Consider $(p(x)-p(c))=q(x)$ then $q(c)=0$ then because $F$ is a field so $x-c$ should devide $q(x)$

Solution 3:

Use the constant remainder theorem: Let $f(x)$ be a polynomial in $F[x]$, then the remainder of the division of any polynomial $g(x)\in F[x]$ by $x-c$ is $g(c)$. This implies that:

$$ x-c\,\big\vert\, f(x) \quad \text{if and only if} \quad f(c)=0 $$ then you can apply it to the polynomial $f(x)$ defined as: $$f(x)=p(x)-p(c)$$