On re-writing polynomials of the form, $P(x) = x^n - b^n$

Question

This is a follow up question to this answer, in which it says:

Let $b$ be fixed, and consider the polynomial $P(x) = x^n - b^n$. Then there is a polynomial $Q(x)$, and a constant $r$, such that

$P(x) = (x - b)Q(x) + r$

Could someone please tell me why this is true (without using the factorization formula for $a^n - b^n$ or the fact that $b$ is a root, since that's what the answer is trying to prove)? (Does it involve any theorems or facts from congruence-class arithmetic or am I just thinking too hard about this?)

It is not assuming what it is trying to prove because of the $r$ term. Once you allow $r$ this is a consequence of polynomial division. The remainder is of lower degree than the polynomial you divide out. Since we are dividing out a linear polynomial, the remainder is a constant. The point of the proof is that $r=0$, so this is a factorization of the polynomial. — Ross Millikan, Dec 29 '20 at 03:55

score 1 · Answer 1 · answered Dec 29 '20 at 03:16

Given polynomials $P(x)$ and $x-b$,

by Euclidean division of polynomials there exist polynomials $Q(x)$ and $r(x)$

such that $P(x)=Q(x)(x-b)+r(x)$, where the degree of $r(x)$ is less than the degree of $x-b$.

Since $x-b$ is linear (first degree), $r(x)$ must be a constant.

score 1 · Accepted Answer · answered Dec 29 '20 at 03:16

Polynomial remainder theorem:
If a polynomial $f(x)$ is divided by $x-a$, the remainder is $f(a)$.
Proof. Suppose $f(x)$ is divided by $x-a$. Then we can write $$f(x)=(x-a)q(x)+r(x)$$ where $q(x)$ is the quotient and $r(x)$ is the remainder. Since the remainder is at most one degree less than the divisor, we can conclude that $r(x)$ is a constant and rewrite the above as $$f(x)=(x-a)q(x)+R.$$ By substitution $x=a$, we have $$f(a)=(a-a)q(a)+R\implies f(a)=R.$$

See here.

On re-writing polynomials of the form, $P(x) = x^n - b^n$

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