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"In algebra, the polynomial remainder theorem or little Bézout's theorem is an application of Euclidean division of polynomials. It states that the remainder of the division of a polynomial f(x) by a linear polynomial x-a is equal to f(a). In particular, x-a is a divisor of f(x) if and only if f(a)=0." -wikipedia

The polynomial remainder theorem

Let $p(x)$ be any polynomial, and $d(x)=x-c$ for any $c.$ Then the remainder of the division $$R(\frac{p(x)}{d(x)})=p(c).$$

So I tried to prove this theorem before looking for the proof on the web. Looking through the web, I only found one traditional way to prove this, so I was wondering if my proof using polynomial division below is legitimate:

Let $p(x)= \sum_{i=0}^n a_n x^n , d(x)=x-c.$

Applying long division we have:

$a_n x^{n-1}+(a_{n-1}+a_nc)x^{n-2}+ \cdot \cdot \\ x-c| \overline{a_n x^n+a_{n-1} x^{n-1} +\cdot \cdot \cdot +a_0 } \\ \qquad a_n x^n-ca_nx^{n-1} \\ \qquad ------ \\ \qquad \quad (a_{n-1} + ca_n)x^{n-1} + a_{n-2} x^{n-2} \\ \qquad \quad (a_{n-1} + ca_n)x^{n-1}-c(a_{n-1} + ca_n)x^{n-2} \\ \qquad \quad --------------- \\ \qquad \qquad (a_{n-2}+ca_{n-1}+c^2a_n)x^{n-2}\\ \qquad \qquad \qquad \qquad . \\ \qquad \qquad \qquad \qquad . \\ \qquad \qquad \qquad \qquad . \\ \qquad \qquad \quad a_0+a_1c+a_2c^2+ \cdot \cdot \cdot + a_nc^n = p(c)$

Thus the remainder of the division is equal to $p(c)$ as the theorem states.

George
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    For the restricted case of division by $x-c$, it seems much simpler to note that $$p(x)-p(c)=\sum_{k=0}^na_kx^k-\sum_{k=0}^na_kc^k=\sum_{k=1}^na_k(x^k-c^k)$$ and that, for every $k\geqslant1$, $x-c$ divides $x^k-c^k$ since $$x^k-c^k=(x-c)(x^{k-1}+cx^{k-2}+\cdots+c^{k-1})$$ (This even gives an explicit formula for the quotient.) – Did Jan 14 '18 at 08:06

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Well first starters you would have to assume you have proved that the polynomial long division algorithm is correct (not your use of it but the algorithm itself). Assuming that then your proof looks correct but would not considered to be fully rigorous by many people since it's so visual.

Here is the proof I prefer:

Using the division algorithm for polynomials over a field:

$$\text{If } f,g \in F[x] \text{ and } g \neq 0, \text{ then there exists unique } q,r \in F[x] \text{ such that } f = qg+r \\ \text{ and deg } r < \text{ deg }g \text{ or } r = 0 $$

$\\$

Set $ g = x-c$ and note that deg $g = 1$. Choose $q,r$ such that $f = qg+r$ and $r = 0$

or deg $r < $ deg $g = 1$ Thus we must have either $r = 0$ or deg $r = 0$; in both cases, $r$ is just a constant.

Now use $f(x) = q(x)g(x) + r(x) = q(x)(x-c) + r(x) \implies f(c) = q(c) \times 0 + r(c) = r(c)$

In particular, $f(c) = r(c) = r \in F$ and hence $f(x) = q(x)(x-c) + f(c)$ and you're done.

David Reed
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