Why is $x - a$ a factor of $p(x) - p(a)$?

Question

I'm reading a proof in a linear algebra book. It mentions $$p(x) -p(c)= (x - c) h(x),$$ where $c$ is a constant, and $p(x)$ and $h(x)$ are polynomials.

Can we always factor $p(x) - p(c)$ in this way?

Please give a proof.

Answer 1

If $R$ is a commutative ring, $f(x)$ is a polynomial in $R[x]$ and $c\in R$, then

$f(x)=(x-c)g(x)$ for some $g(x)\in R[x]$ if and only if $f(c)=0$

Indeed, long division of $f(x)$ by $x-c$ is possible because $x-c$ is monic; so $f(x)=(x-c)g(x)+r$, where $r\in R$. The conclusion is now easy.

What can you say about $f(x)=p(x)-p(c)$?

Answer 2

Let $ p(x) = \sum_{i=0}^n p_ix^i $ Now $$p(x) - p(a) = \sum_{i=o}^n p_i(x^i-a^i)$$ Use the formula $$x^i-a^i= (x-a)( \sum_{j=0}^{i-i}x^ja^{i-1-j} ) $$(This comes from a geometric series) Thus now you can see from each term a factor of $x-a$ comes out . Thus it is proved.