The most basic cubics questions might ask you to factorise a simple cubic where a factor of x can be taken out first. For instance, the terms in the expression $3x^3-2x^2+5x$ have a common factor of x and so factor x out to give $x\left(3x^2-2x+5\right)$. You might think that this can be factorised further, however, in this case the quadratic cannot be factorised. This can be seen by noting that the discriminant of $3x^2-2x+5$ is $(-2)^2-4\times 3\times 5=-56$ which is negative and so $3x^2-2x+5$ has no roots. – see Discriminants. It follows that the cubic cannot be factorised further. In most cases, however, there will be some more factoring required. See Example 1.
In addition to the above, other cubics questions might involve factorising a more general cubic and may require knowledge of the factor theorem. See Example 2.
- Firstly, identify whether the cubic is positive or negative.
- Then, substitute $x=0$ into the cubic expression to identify the $y$-intercept.
- Next, factorise if possible and set $y=0$ to identify the roots. Note that, in $y=x(x+1)^2$ for example, $x=-1$ is a repeated root and the curve must touch the x-axis at $x=-1$.
- Finally, place the graph on the axes so that all the above criteria are satisfied.
Factorise the following:
- Given that $x=-2$ is a root of the cubic $x^3+x^2-x+2$, factorise it completely.
- In addition, factorise $f(x)=x^3-x^2-x+1$ completely.
- Since $x=-2$ is a root, $(x+2)$ is a factor and factoring it out gives $(x+2)(x^2-x+1)$ which can’t be factorised any further.
- By inspection, we can see that $x=1$ is a root of $f(x)$ and so $(x-1)$ is a factor. Using polynomial division or inspection we have $f(x)=(x-1)(x^2-1)$ which factorises completely to $f(x)=(x-1)(x-1)(x+1)=(x-1)^2(x+1)$.