# remainder theorem issue

• Aug 7th 2012, 05:54 AM
anthonye
remainder theorem issue
Hi;
When using the remainder theorem do I always plug the root in?
If yes why?

ie dividing by (x - 1) plud in a 1 and (x + 1) plug in -1
• Aug 7th 2012, 06:00 AM
Prove It
Re: remainder theorem issue
Quote:

Originally Posted by anthonye
Hi;
When using the remainder theorem do I always plug the root in?
If yes why?

ie dividing by (x - 1) plud in a 1 and (x + 1) plug in -1

If you divide a polynomial \displaystyle \begin{align*} P(x) \end{align*} by \displaystyle \begin{align*} ax + b \end{align*}, you will get a quotient of one less degree plus a remainder with a constant numerator. So

\displaystyle \begin{align*} \frac{P(x)}{ax + b} &= Q(x) + \frac{R}{ax + b} \\ P(x) &= (ax + b)Q(x) + R \end{align*}

Now if you let \displaystyle \begin{align*} x = -\frac{b}{a} \end{align*}, you have \displaystyle \begin{align*} ax + b = 0 \end{align*}, so

\displaystyle \begin{align*} P\left(-\frac{b}{a}\right) &= 0\,Q\left(-\frac{b}{a}\right) + R \\ P\left(-\frac{b}{a}\right) &= R \end{align*}

So yes, when you want to find the remainder when dividing a polynomial by \displaystyle \begin{align*} ax + b \end{align*}, you plug in \displaystyle \begin{align*} -\frac{b}{a} \end{align*}.
• Aug 7th 2012, 06:03 AM
anthonye
Re: remainder theorem issue
Thank you.