Cubic factorisation
Being able to do long division of polynomials is critical for solving cubic equations. Go back to this resource to review your understanding.
Solving cubic equations
Learn how to solve cubic equations, where the highest power of the variable is three. These types of equations can model various real-world scenarios, from physics to engineering.
The factor theorem
The long division of polynomials is an important tool for solving cubic equations. However, the factor theorem is equally important. It is a trial-and-error approach to solving polynomial equations.
The factor theorem states that \((x-a)\) is a factor of the equation \(f(x)\) if \(f(a)=0\).
In other words, if we substitute \(a\) in place of the variable (e.g. \(x\)) in the polynomial and the answer is \(0\), then \((x-a)\) is a factor.
We can use the factor theorem to find one factor of a polynomial, and then use polynomial long division to find the remaining factor(s).
This will all make more sense once we look at some examples.
Example – using the factor theorem
Solve the equation \(x^{3}-5x^{2}-2x+24=0\).
Use the factor theorem to find a linear factor of \(x^{3}-5x^{2}-2x+24\). We let:
\[ f(x)=x^{3}-5x^{2}-2x+24 \]
We can try \(x=1\) first.
\[\begin{align*} f(1) & = (1)^{3}-5(1)^{2}-2(1)+24\\
& = 1-5-2+24\\
& = 18\\
& \neq 0
\end{align*}\]
Since \(f(1)\neq0\), \((x-1)\) is not a factor of the polynomial.
We can try \(x=-1\), too.
\[\begin{align*} f(-1) & = (-1)^{3}-5(-1)^{2}-2(-1)+24\\
& = -1-5+2+24\\
& = 20\\
& \neq 0
\end{align*}\]
Thus, \((x+1)\) is also not a factor of the polynomial.
Let's try \(x=2\).
\[\begin{align*} f(2) & = (2)^{3}-5(2)^{2}-2(2)+24\\
& = 8-20-4+24\\
& = 8\\
& \neq 0
\end{align*}\]
\((x-2)\) is not a factor of the polynomial.
How about \(x=-2\)?
\[\begin{align*} f(-2) & = (-2)^{3}-5(-2)^{2}-2(-2)+24\\
& = -8-20+4+24\\
& = 0
\end{align*}\]
Since \(f(-2)=0\), this means that \((x+2)\) is a factor. We can then use polynomial long division to find the other factor. We divide \(x^{3}-5x^{2}-2x+24\) by \(x+2\).
\[\begin{array}{r} x^2-7x+12\phantom{)} \\
x+2{\overline{\smash{\big)}\,x^3-5x^2-2x+24\phantom{)}}} \\
\underline{-(x^3+ 2x^2)\phantom{bbbbbbbbbi)}} \\
0- 7x^2 \,-2x\phantom{bbbbi)} \\
\underline{-~\phantom{()}(-7x^2-\!14x)}\phantom{bbbi)} \\
12x+24 \phantom{)} \\ \underline{-~\phantom{()}(12x+24)} \\
0+0\phantom{)}
\end{array}\]
This means that:
\[x^{3}-5x^{2}-2x+24=(x+2)(x^{2}-7x+12)\]
The quadratic factor can also be further factorised:
\[x^{3}-5x^{2}-2x+24=(x+2)(x-3)(x-4)\]
As \(x^{3}-5x^{2}-2x+24=0\), we can solve the cubic equation using the null factor law.
\[\begin{align*} x+2=0 \textrm{, } x-3 & = 0 \textrm{ or } x-4=0\\
\therefore x=-2 \textrm{, } x & = 3 \textrm{ or } x=4
\end{align*}\]
Exercise – using the factor theorem
Solve the following cubic equations.
- \(x^{3}+7x^{2}+11x+5=0\)
- \(4x^{3}+2x^{2}-2x=0\)
- \(-x^{3}-3x^{2}+x+3=0\)
- \(x^{3}-7x-6=0\)
- \(x=-1\) and \(x=-5\)
- \(x=-1\), \(x=0\) and \(x=\dfrac{1}{2}\)
- \(x=1\), \(x=-1\) and \(x=-3\)
- \(x=-1\), \(x=-2\) and \(x=3\)
