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The theorem I've been tasked with proving is that for any polynomial function $f:\mathbb{R} \to \mathbb{R}$ and any $a \in \mathbb{R}$ there exists some polynomial $g:\mathbb{R} \to \mathbb{R}$ and some $b \in \mathbb{R}$ such that $$f(x)=(x-a)g(x)+b.$$

Using induction on the degree of $f$, I came up with the following proof.

For the case $n = 1$, we have a first degree polynomial, i.e. $f(x) = cx+d$. So if we let $g(x)=c$ and $b=ca+d$, then we have $$(x-a)g(x)+b=c(x-a)+ca+d=cx-ca+ca+d=cx+d=f(x)$$ proving it for the base case. Now we assume that for any polynomial of degree $n$ the theorem is true, so we can represent $f$ in the form $f(x) = (x-a)g(x) + b$. Now, for a polynomial of degree $n+1$ we have $$\begin{align} f(x) &= a_{n+1}x^{n+1}+a_nx^n+\cdots+a_0 \\ &= x^n(a_{n+1}x+a_n)+a_{n-1}x^{n-1}+\cdots+a_0 \\ &= x^n[(x-a)g_1(x)+b_1]+(x-a)g_2(x)+b_2 \tag{By hypothesis} \\ &= x^n(x-a)g_1(x)+(x-a)g_2(x)+(b_1x^n+b_2) \\ &= (x-a)x^ng_1(x)g_2(x)+(x-a)g_3(x)+b_3 \tag{By hypothesis}\\ &= (x-a)x^ng_1(x)g_2(x)g_3(x)+b_3 \end{align}$$ so we let $g(x) = x^ng_1(x)g_2(x)g_3(x)$ and $b = b_3$, proving that it also holds in the $n+1$ case. $\square$

Does anyone know of any neater/cleaner/more enlightening proofs of this theorem? I'm particularly interested in seeing ones that don't use induction (I've seen one simpler proof here: Proof of Remainder Theorem for polynomials).

Community
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2 Answers2

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Note that $x^k-a^k$ is divisible by $x-a$. For we have the familiar identity $$x^k-a^k=(x-a)(x^{k-1}+x^{k-2}a +\cdots +xa^{k-2}+a^{k-1}).$$

It follows that $$\sum_{k=0}^n b_kx^k -\sum_{k=0}^n b_k a^k$$ is divisible by $x-a$.

Thus $P(x)=(x-a)Q(x)+P(a)$ for some polynomial $Q$.

André Nicolas
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Whatever way you turn the proof, your proof will be pointing at the quotient $g$ of Euclidean division of $f$ by $x-a$, whose coefficients depend on those of$~f$ in the usual long-division manner. This means some iteration or induction will be applied at some level anyway. There are different way to hide this though; the answer by André Nicholas explicitly gives the contribution to the quotient for each monomial, which can then be combined by linearity. Another way, essentially the answer referred to in the question, is to put $y=x-a$, expand $f(x)=f(y+a)$ as a polynomial in$~y$, and let $b$ be its constant term.

Marc van Leeuwen
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