### Remainder Theorem

If a polynomials   f(x)   is divided by    (x-a)   i.e. a polynomial of degree  1   then the remainder is    f(a)   .

We know that
f(x) = g(x) q(x) + r(x)

where degree    r(x) <  degree   g(x)

choose     g(x) = (x-a)

there fore    f(x) =  (x-a)  q(x)  +  r(x)

where degree  r(x) <  degree g(x)  ,   i.e.  <1  ,  or   degree  r(x) =0   or   say   r(x) = r  .

therefore ,   f(x) = (x-a) q(x) + r

therefore ,   f(a) = (a-a) q(a) + r

or   f(a) = r =  remainder when the polynomial  f(x)   is divided by   x-a     .

therefore  ,     f(x) = (x-a) q(x) +f(a)

Here is an example of Reminder Theorem

Show that the polynomial remainder theorem holds for an arbitrary second degree polynomial $f(x) = ax^2 + bx + c$ by using algebraic manipulation:
\begin{align} \frac{f(x)}{{x - r}} &= \frac{{a{x^2} + bx + c}}{{x - r}} \\ &= \frac{{ax(x - r) + (b + ar)x + c}}{{x - r}} \\ &= ax + \frac{{(b + ar)(x - r) + c + r(b + ar)}}{{x - r}} \\ &= ax + b + ar + \frac{{c + r(b + ar)}}{{x - r}} \\ &= ax + b + ar + \frac{{a{r^2} + br + c}}{{x - r}} \end{align}
Multiplying both sides by (x − r) gives
$f(x) = ax^2 + bx + c = (ax + b + ar)(x - r) + {a{r^2} + br + c}$.
Since $R = ar^2 + br + c$ is our remainder, we have indeed shown that $f(r) = R$.