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

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- Aug 7th 2012, 04:54 AManthonyeremainder 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, 05:00 AMProve ItRe: remainder theorem issue
If you divide a polynomial $\displaystyle \displaystyle \begin{align*} P(x) \end{align*}$ by $\displaystyle \displaystyle \begin{align*} ax + b \end{align*}$, you will get a quotient of one less degree plus a remainder with a constant numerator. So

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

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

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

So yes, when you want to find the remainder when dividing a polynomial by $\displaystyle \displaystyle \begin{align*} ax + b \end{align*}$, you plug in $\displaystyle \displaystyle \begin{align*} -\frac{b}{a} \end{align*}$. - Aug 7th 2012, 05:03 AManthonyeRe: remainder theorem issue
Thank you.