# A quick way to factorise cubics.

There are many ways of factorising a cubic expression.

The method you are about to see, once learnt, becomes the quickest method ever to factorise cubics.

As if with any polynomial, factorisation commences with finding one factor. Use the Factor Theorem to find this factor. This means, find the values(roots) that will make your expression to equal zero.

Cubics have the form:

$${f(x)=ax^3+bx^2+cx+d}$$

Using the factor theorem, you will first need to find $$f(x) = 0$$. Then draw the table of four rows and four columns.

To help you see this, let us try a few examples:

1. Factorise $$f(x)=x^3-5x^2-8x+12$$

The steps:

Find the factor, using factor theorem. This means try different values of $$x$$ that will make $$f (\,x)\, =0$$. In this case, $$x = 1$$ will make $$f (\,x)\, =0$$. The factor is therefore, $$(\,x-1)\,$$.

1 1 -5 -8 12
0 1 -4 -12
1 -4 -12 0

Procedure

1. In the first column, put a zero in the middle row.

2. To get the third row elements, add the first two rows. For example, $$1+0 =1$$

3. To get the remaining row two elements, multiply the preceding row three elements with the factor, in this case, $$1$$. For example, $$1 \times 1 = 1$$. Continue in this manner.

The final answer is the factor times the first three values of the third row. These values arre coefficients of a quadratic expression. We now have:

\begin{gather} f(\,x)\, = x^3-5x^2-8x+12\\ f(\,x)\ = (\,x-1)\,(\,x^2-4x-12)\,\\ f(\,x)\ = (\,x-1)\,(\,x-6)\,(\,x+2)\, \end{gather}
2. Factorise $$f(x)=-x^3+10x^2-17x-28$$

The factor is $$(\,x+1)\,$$.

By following the steps in (1), we get:

-1 -1 10 -17 -28
0 1 -11 -28
-1 11 -28 0

The expression becomes:

\begin{gather*} f(\,x)\,= -x^3+10x^2-17x-28\\ f(\,x)\, = (\,x+1)\,(\,-x^2+11x-28)\, \\ f(\,x)\, = (\,x+1)\,[\,-(\,x^2-11x+28)\,]\, \\ f(\,x)\, = (\,x+1)\,[\,-(\,x-7)\,(\,x-4)\,]\, \\ f(\,x)\, = -(\,x+1)\,(\,x-6)\,(\,x-4)\, \end{gather*}
3. Now you have factorised the expression.