The method you are about to see, once learnt, becomes the quickest method ever to factorise cubics.
As 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/equation to equal zero.
Cubics have the form:
$${\normalsize f(x)=ax^3+bx^2+cx+d}$$Using the factor theorem, you will first need to find \(f(\,x)\,=0\). Then draw the table of three rows and five columns.
To help you see this, let us try a few examples:
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Factorise \(\normalsize f(x)=x^3-5x^2-8x+12\)
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 The procedure to filling this table:
- The first value of the row is the value of \(\normalsize x\) that makes the expression zero (a root). The other values of the first row are the coefficients of the cubic expression.
- Fill the second row, column two with a zero on the second column and move rightwards
- To get the third row elements, add the elements of the first and second rows column-wise. For example, \(1+0 =1\)
- To get the remaining row elements, multiply each row-three element by the root and place the result in the next column of row two. Continue in this manner.
NB: all column 1 elements except row 1 are empty
The final answer is the factor \(x-1\) multiplpied by the quadratic formed by the first three values of the third row:
$$ \normalsize \begin{align*} f(\,x)\, &= x^3-5x^2-8x+12\\ f(\,x)\, &= (\,x-1)\,(\,x^2-4x-12)\,\\ f(\,x)\, &= (\,x-1)\,(\,x-6)\,(\,x+2)\, \end{align*} $$ -
Factorise \( \normalsize f(x)=-x^3+10x^2-17x-28\)
The factor is \(\normalsize (\,x+1)\,\).
By following the steps in (1), we get:
-1 -1 10 -17 -28 0 1 -11 -28 -1 11 -28 0 The expression becomes:
$$ \normalsize \begin{alignat}{1} f(\,x)\,&= -x^3+10x^2-17x-28\\ f(\,x)\, &= (\,x+1)\,(\,-x^2+11x-28)\, \\ f(\,x)\, &= (\,x+1)\,[\,-(\,x^2-11x+28)\,]\, \\ f(\,x)\, &= (\,x+1)\,[\,-(\,x-7)\,(\,x-4)\,]\, \\ f(\,x)\, &= -(\,x+1)\,(\,x-6)\,(\,x-4)\, \end{alignat} $$Now you have factorised the expression.