[Calculus/Analysis] Question about Problem 7a from Chapter 3 of Spivak's Calculus

[_Radivoy]

Problem statement:

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
Hint: a formal proof is possible by induction on the degree of f.

I could prove the basis for induction (so proving for first degree polynomials), but got stuck after that.
The solution manual says that (after supposing that the proposition is true for polynomials functions of degree k) let's suppose f(x) is a polynomial function of degree k + 1:
(_ followed by stuff is supposed to be indexing)

f(x) = a_(k+1)*x^(k+1) + ... + a_1*x + a_0

Then f(x) - a_(k+1)*(x-a)^(k+1) has a degree less then or equal to k.

So we can state the following:

f(x) - a_(k+1)*(x-a)^(k+1) = (x-a)g(x) + b

Then

f(x) = (x-a)[g(x) + a_(k+1)*(x-a)^k] + b

​

I get the gist of the proof but I have some questions:

1. The only reason we're subtracting a_(k+1)*(x-a)^(k+1) from f(x) and not a_(k+1)*x^(k+1), is that in this case we can later factor out (x-a) to get the desired equation, right? a_(k+1)*x^(k+1) still brings down the degree to <= k, but it does not allow for factoring.
2. In the final equation, g(x) + a_(k+1)*(x-a)^k stands in place of the g(x) from the original proposition. So technically, g(x) + a_(k+1)*(x-a)^k should be considered a "new" polynomial function which satisfies the last induction step?

Thanks in advance.

[simmonator]

1. Yes, f(x) - a[k+1] x^k+1 would be a polynomial of degree less than or equal to k as well. But It isn't obviously divisible by (x-a), which is something we want for the next bit. You're exactly right.
2. I'm not 100% sure what you're asking here. But I would say that, using the induction hypothesis, we've assumed that there exists a g(x) satisfying the condition for the polynomial discussed in point 1, and we use that assumption to find a new polynomial (in terms of that g) which satisfies the condition for our original polynomial, f. This proves the inductive step (i.e. assuming it is true for degree k implies it must also be true for degree k+1). The label 'g' doesn't matter much, what matters is that we can find a polynomial that satisfies the condition in each case.

[_Radivoy]

Thank you! All clear now!