Prove that if $c$ is any element in a field $\mathbb{F}$ and $f(x) \in \mathbb{F}$, then $f(x) \equiv f(c)$ mod $(x-c)$

Question

I believe I'm supposed to produce something of the from $f(x)-f(c) = (x-c)p(x)$ where $p(x) \in \mathbb{F}[x]$, however I'm not sure where to start.

You meant $f(x)\in F[x]$. Then $f(x+c)-f(c)$ has $0$ in its roots it implies that it is $=x g(x)$. No need that $F$ is a field. — reuns, Dec 06 '20 at 02:49
$f(x) \equiv f(c) \mod (x-c)$ just means that $x-c$ divides $f(x) - f(c)$. Since c is a root of $f(x)-f(c)$, by the factor theorem $(x-c)$ divides $f(x)-f(c)$. So I think that's all you need to do. — Ameet Sharma, Dec 06 '20 at 02:51

score 2 · Accepted Answer · answered Dec 06 '20 at 03:10

2

Start by using the division algorithm to show that there exists polynomials $g(x), h(x)\in \mathbb{F}[x]$ such that $f(x) = (x-c)g(x) + h(x)$ and $\deg{h(x)} < \deg{(x-c)} = 1$.

Thus, $h(x)$ must be a constant, lets call it $a$.

Then, plug in the value $c$ for $x$ to get, $f(c) = (c-c)g(x) + a = 0\cdot g(x) + a = a$

$\Rightarrow f(c) = a$

$\Rightarrow f(x) = (x-c)g(x) + h(x) = (x-c)g(x) + a = (x-c)g(x) + f(c)$

Thus, you have proved your result.

answered Dec 06 '20 at 03:10

aefrrs

146

Please strive not to add more dupe answers to FAQs. – Bill Dubuque Dec 06 '20 at 03:48
Again, this isn't a dupe answer. This is one way of proving the required result. – aefrrs Dec 06 '20 at 03:57
Again, it is a dupe of the linked answer(s). – Bill Dubuque Dec 06 '20 at 03:59

Prove that if $c$ is any element in a field $\mathbb{F}$ and $f(x) \in \mathbb{F}$, then $f(x) \equiv f(c)$ mod $(x-c)$

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