Tuesday, 26 November 2013

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.


No comments:

Post a Comment

Our Latest Post

Introduction of Circle

All of my lessons and teaching videos are in English and most of them are for students of Logistics Management. But many of mys students an...

Popular Post