## Lecture

\[\frac{f(x)}{g(x)} \equiv q(x) + \frac{r(x)}{g(x)} \]

There's an algorithm for doing this and I estimated just over half of my students could remember how to find the right-hand side of the equation.

\[ \frac{2x^2-7x-6}{x-1} \]

This was the layout I was taught at school:

\[ \frac{2x^2-7x-6}{x-1} \equiv 2x-5 - \frac{11}{x-1} \]

\[ 2x^2 -7x -6 \equiv (2x-5)(x-1)-11 \]

## Seminar 1

## Seminar 2

\[ x^3 -x^2 -14x+24 = 0 \]

In this case we set \( f(x)=x^3-x^2-14x+24 \) and start searching for factors:

\begin{align}

f(1)=10 \\

f(2) = 0

\end{align}

\[ f(x) = (x-2)(x^2+x-12) \]

and this can be factorised using methods seen in Week 2 to get

\[ f(x) = (x-2)(x-3)(x+4) \]

And the solution to the original cubic is found easily:

\[ \begin{align}

x^3 -x^2 -14x+24 = 0 \\

(x-2)(x-3)(x+4) = 0

\end{align} \]

Therefore \( x = 2, 3 -4 \).