The remainder theorem states that when a polynomial $P(x)$ is divided by a linear polynomial $ax+b$, the remainder is equal to $P(\frac{-b}{a})$. It should be noted that this value $\frac{-b}{a}$ is not arbitrarily chosen. We get it by equating the same linear factor to $0$ and solving for $x$:
$$ \begin{equation}\begin{aligned} ax+b&=0\\ ax&=-b\\ \therefore x&=\frac{-b}{a}\\ \end{aligned}\end{equation} $$We then substitute this $x$ value into the polynomial to get the remainder: $$ \begin{equation}\begin{aligned} R=P(\frac{-b}{a})\\ \end{aligned}\end{equation} $$
Example
Consider the example we covered in the previous lesson, where we divided $x^2+5x+7$ by $x+3$ using long division. Here, we have a polynomial $P(x)=x^2+5x+7$ being divided by a linear polynomial $x+3$. First we find the $x$ to be substituted by equating the linear polynomial to $0$: $$ \begin{equation}\begin{aligned} x+3&=0\\ \therefore x&=-3\\ \end{aligned}\end{equation} $$
By the remainder theorem, we substitute this $x$ value into the polynomial to get the remainder: $$ \begin{equation}\begin{aligned} R&=P(-3)\\ &=(-3)^2+5(-3)+7\\ &=9-15+7\\ &=1\\ \end{aligned}\end{equation} $$
This $1$ is the answer we had gotten for remainder when at the last step of the long division involving $x^2+5x+7$ and $x-3$.
Example 2
When $x^3-2x^2+kx+3$ is divided by $x-1$, the remainder is $3$. Find the value of the constant $k$. First, we find what to substitute for $x$ in the polynomial: $$ \begin{equation}\begin{aligned} x-1&=0\\ \therefore x&=1\\ \end{aligned}\end{equation} $$
By the remainder theorem, we substitute this $x$ value into the polynomial to get the remainder: $$ \begin{equation}\begin{aligned} R&=P(1)\\ &=(1)^3-2(1)^2+k(1)+3\\ &=1-2+k+3\\ &=k+2\\ \end{aligned}\end{equation} $$
We are told that the remainder is $3$ thus our result from using the remainder theorem must the same as $3$: $$ \begin{equation}\begin{aligned} R&=k+2\\ \therefore 3&=k+2\\ 3-2&=k\\ \therefore k&=1\\ \end{aligned}\end{equation} $$
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