How to factor this algebraic identity used in Lagrange's theorem for polynomial congruences

Asked May 14 '21 at 17:53

Active May 14 '21 at 21:16

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The question is how to factor an identity. The identity is shown below but it uses $f(x)$ as defined in Theorem 5.21, so let me show $f$'s definition first.

In the proof, we eventually get to this identity:

We have two equations there. I'm good with the first. Just expanding it out makes it obvious. My problem is on the second equation. By taking the expansion and factoring $x$ on all non-$x_0$ terms, I get

$$x(c_n x^{n-1} + \cdots + c_2 x + c_1).$$

Similarly for $x_0$,

$$x_0(c_n x_0^{n-1} + \cdots + c_2 x_0 + c_1).$$

Question. How can I use these two expressions and get the second equation in the identity?

If I let $g(x) = (c_n x^{n-1} + \cdots + c_2 x + c_1)$, then I have $g(x)$ in the first expression above and $g(x_0)$ in the second. These are not the same, so I'm puzzled as how this is allowed.

edited May 14 '21 at 21:16

Bill Dubuque

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asked May 14 '21 at 17:53

user851954

1

Use that $a^k - b^k = (a-b)(a^{k-1}+a^{k-2}b+\cdots+ab^{k-2}+b^{k-1})$. – dxiv May 14 '21 at 18:00
1

By the linked Factor Theorem $,x-x_0\mid x^r-x_0^r$ so $,x-x_0,$ divides each summand hence it divides the entire sum, which yields the general Factor Theorem $,x-a\mid f(x)-f(a).\ $ Generally a nonzero polynomial over a domain has no more roots than its degree. – Bill Dubuque May 14 '21 at 21:14

How to factor this algebraic identity used in Lagrange's theorem for polynomial congruences

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