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:

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