Cubics

Method 1: Factorising Cubics given 1 Factor

When factorising a cubic expression, you will be able to put it into (up to) 3 brackets. You may be given 1 factor that you can work with, which will make the factorisation easier.

Example: (x-2) is a factor of f(x) = 2x^3 + x^2 - 13x + 6. Hence, express f(x) as a product of three linear factors.

Step 1: Write down the factor we know, \textcolor{orange}{(x-2)}, next to another set of brackets:

\textcolor{orange}{(x-2)}( \quad \quad \quad \quad \quad \quad ) = 2x^3 + x^2 - 13x + 6

Step 2: Put an x^2 term at the start of the second set of brackets – this will need to multiply together with x from the first bracket to make 2x^3:

\textcolor{orange}{(x-2)}(\textcolor{red}{2x^2} \quad \quad \quad \quad ) = 2x^3 + x^2 - 13x + 6

Step 3: Find the constant term for the end of the second bracket – this will need to be multiplied together with -2 to make 6:

\textcolor{orange}{(x-2)}(\textcolor{red}{2x^2} \quad \quad \quad \textcolor{limegreen}{-3}) = 2x^3 + x^2 - 13x + 6

Step 4: The x and -3 above multiply to make -3x, but we need -13x. So we need an extra -10x from somewhere – we need to multiply -2 by 5x to make -10x:

\textcolor{orange}{(x-2)}(\textcolor{red}{2x^2} + \textcolor{blue}{5x} \textcolor{limegreen}{-3}) = 2x^3 + x^2 - 13x + 6

Step 5: Check that this works, by multiplying out the brackets and seeing if this matches f(x):

\begin{aligned} &\, \textcolor{orange}{(x-2)}(\textcolor{red}{2x^2} + \textcolor{blue}{5x} \textcolor{limegreen}{-3}) \\ &= 2x^3 + 5x^2 - 3x - 4x^2 - 10x + 6 \\ &= 2x^3 + x^2 - 13x + 6 \end{aligned}

So, this works so far.

Step 6: Factorise the quadratic 2x^2 + 5x -3 into 2 linear factors:

\textcolor{red}{2x^2} + \textcolor{blue}{5x} \textcolor{limegreen}{-3} = \textcolor{orange}{(2x-1)(x+3)}

Step 7: Put it all together:

2x^3 + x^2 - 13x + 6 = \textcolor{orange}{(x-2)(2x-1)(x+3)}

Method 2: Factorising Cubics using the Factor Theorem

If you are given no factors, then you can use the Factor Theorem to find one factor, and then use Method 1 from above.

Reminder: The Factor Theorem is defined as:

“If f(x) is a polynomial, and f(k)=0, then (x-k) is a factor of f(x)”

or

“If f \left(\dfrac{b}{a} \right)=0, then (ax-b) is a factor of f(x)”

When using the factor theorem here, you will need to try small values of k, e.g. f(1), f(-1), f(2), etc., until you find f(k)=0.

Cubic Graphs

You can sketch a cubic if you know its factors. You have to find where the function is 0.

All cubic graphs have a general shape:

• If the coefficient of x^3 is positive, then the graph goes from ‘the bottom left to the top right’
• If the coefficient of x^3 is negative, then the graph goes from ‘the top left to the bottom right’

The y-intercept of a cubic graph y=ax^{3}+bx^{2}+cx+d is always d, so by expanding brackets we can find where the graph crosses the y-axis.

Example: Sketch the graphs of:

\begin{aligned} f(x) &= x(x-1)(x+2) \\ g(x) &= (x+1)(x^2 + x + 1) \\ h(x) &= (1-x)(x+2)^2 \\ m(x) &= (1-2x)^3 \end{aligned}