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Prove that for any polynomial function $f$, and any number $a$, there is a polynomial function $g$,and a number $b$, such that $f(x)=(x-a)g(x) + b$ for all $x$.

The following is the answer I'm given:

If the degree of $f$ is $1$, then $f$ is of the form $f(x)= cx + d = c(x-a) + (d+ac)$, so we can let $g(x)=c$ and $b= d + ac$. Suppose that the result is true for polynomials of degree $\le k$. If $f$ has degree $k+1$, then $f$ has the form $$f(x)=a_{k+1}x^{k+1} + \cdots + a_{1}x + a_{0}.$$

Now the polynomial function $h(x)=f(x)-a_{k+1}(x-a)$ has degree $\le k$, so we can write $$f(x)-a_{k+1}(x-1)=(x-a)g(x) + b,$$

or $$f(x)=(x-a)[g(x)+a_{k+1}] + b,$$

which is the required form.

I wanted to know if that $(x-1)$ should be $(x-a)$ instead? I know that $a$ can be any number, but we add $a_{k+1}(x-1)$ to both sides to get $f(x) = (x-a)[g(x)+a_{k+1}]$.

Community
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Skm
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    Why $h(x)=f(x)-a_{k+1}(x-a)$ has degree $\leq k$? To get an $h$ whose degree $\leq k$, it is considerable to let $h(x)=f(x)-a_{k+1}(x-a)^{k+1}$ so that the term of order $k+1$ of $h$ will be 0. Then use the argument same with that in your question after defining $h$. – Feng Jul 13 '19 at 22:18
  • Special case of the Polynomial Division Algorithm. See my comment on the accepted answer regarding a possible circular proof. – Bill Dubuque Jul 13 '19 at 23:04

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Another way without using induction.

Let $h(x)=f(x)-f(a)$, then $h(a)=0$, so $(x-a)|h(x)$, i.e. there exists $g(x)$ such that $f(x)=(x-a)g(x)+f(a)$. Here $h$ and $g$ are polynomials.

Feng
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  • But it does use induction - which is hidden in the proof of the invoked Factor Theorem. Moreover, the Factor Theorem is usually proved use the OP case of the division algorithm, so using it here likely leads to a circular proof. DId you have some other way in mind to avoid that circularity? – Bill Dubuque Jul 13 '19 at 22:56
  • @Bill Dubuque Write $h(x)$ as the polynomial of $(x-a)$, then $h(a)=0$ implies the constant term is zero. Does this proof use induction? – Feng Jul 13 '19 at 23:13
  • Please give the full details of "Write $h(x)$ as the polynomial of $(x−a)$". It's not clear what you intend by that. – Bill Dubuque Jul 13 '19 at 23:17
  • @Bill Dubuque $h(x)=a_n(x-a)^n+\cdots+a_0$. $h(a)=0$ implies $a_0=0$. – Feng Jul 13 '19 at 23:22
  • How do you propose to prove that such a representation exists without using induction on $n$? – Bill Dubuque Jul 13 '19 at 23:24
  • Let $p(x)=h(x+a)$ then $p(x)$ is a polynomial, which can check by expanding $h(x+a)$. So $p(x)=a_nx^n+\cdots$. $h(x)=p(x-a)$. – Feng Jul 13 '19 at 23:31
  • Which again requires induction on the degree for a rigorous proof. – Bill Dubuque Jul 13 '19 at 23:32
  • I can't understand. $h(x)=b_nx^n+\cdots+b_0$ so $h(x+a)=b_n(x+a)^n+\cdots=b_n(x^n+ax^{n-1}+\cdots)+\cdots$. Dose this use induction? After all, I can write explicitly these "cdots" in the form of a summation, I mean, use the "\sum". – Feng Jul 13 '19 at 23:38
  • To define formal sums rigorously requires recursion, which uses induction. In any case I don't want to be overly pedantic.. My point was simply that one needs to be careful here about circularity and how to do so will depend on the context in the OP's case. In many informal presentations these details are glossed over with handwaving, but one should not gloss over circularity. – Bill Dubuque Jul 13 '19 at 23:44
  • Now I agree with you. I will watch out in my study in the future. Thanks Bill. – Feng Jul 13 '19 at 23:49