Question

The remainder theorem states that for any polynomial f(x) divided by (x-a) the remainder is always f(a).

How do I prove this?

Answer 1

It is advisable to use:

\[\color{green}{\textsf{Dividend = Divisor} \times \textsf{Quotient + Remainder}}\]

Answer 2

Here, f(x) is your Dividend, Suppose you divide this polynomial f(x) by (x-a), and suppose you get quotient as : q(x), some other polynomial with Remainder R, then:

\[f(x) = (x-a) \times q(x) + R\]

Answer 3

so now if we plug in x=a we get f(a)=R ok cool.

But now I notice something really interesting.

Answer 4

What is that??

Answer 5

\[\LARGE f(x)=q(x)(x-a)+f(a)\] if we rearrange this we can get: \[\LARGE \frac{f(x)-f(a)}{x-a}=q(x)\] which looks like the mean value theorem or something. So what's the connection going on here?

Answer 6

Remainder is f(a) only for x = a..

Answer 7

all it says is (x-a) divides f(x) evenly when we subtract f(a) from f(x)

Answer 8

limit definition :)