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 $\displaystyle \displaystyle \begin{align*} P(x) \end{align*}$ 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 $\displaystyle \displaystyle \begin{align*} -\frac{b}{a} \end{align*}$.