Surds
Use this resource if you need a refresher on how to simplify surds.
So far, you will have come across a few situations where a quadratic cannot be further factorised into two linear factors using common factors, perfect squares, or the difference of two squares. In these cases, you can use a different method where you complete the square.
Before you begin, make sure you are confident with factorising using common factors, the difference of two squares rule, and the difference of two squares rule.
Watch this video to learn about completing the square.
Completing the square is a method that involves making the quadratic a perfect square, so that you can then factorise using \((x+b)^{2}=x^{2}+2bx+b^{2}\) or \((x-b)^{2}=x^{2}-2bx+b^{2}\). It allows any quadratic to be factorised.
You can complete the square using the following steps:
- Take \(\dfrac{1}{2}\) the coefficient in front of \(x\) and square it.
- Add the term to the expression. Remember that if you add a term, you must also subtract it to keep the expression balanced.
- Group the terms so that you can use the DOTS rule.
Note that DOTS can only be used if you are looking at the DIFFERENCE between squares. If the expression is the SUM of two squares, there will be no real linear factors.
See how this is applied to expressions in the form \(ax^{2}+bx+c\) where \(a=1\) with the following examples.
Factorise \(x^2+8x-5\) by completing the square.
Find the term by calculating \(\dfrac{1}{2}\) of the coefficient in front of the \(x\), squared:
\[\left(\frac{8}{2}\right)^{2} = (4)^{2}\]
Add and subtract the term:
Find the term by calculating \(\dfrac{1}{2}\) of the coefficient in front of the \(x\), squared:
\[\left(\frac{-6}{2}\right)^{2} = (-3)^{2}\]
Add and subtract the term:
Find the term by calculating \(\dfrac{1}{2}\) of the coefficient in front of the \(x\), squared:
\[\left(\frac{5}{2}\right)^{2}\quad\textrm{keep as fraction}\]
Add and subtract the term:
We are in trouble here. We cannot apply the DOTS rule because we have the SUM of two squares not the DIFFERENCE. So in this case, the quadratic \(x^{2}+5x+9\) does not have any real factors and solutions. Complex factors and solutions do exist but are not considered here.
Geometrically, the lack of real factors means that the graph of \(x^{2}+5x+9\) does not intersect the \(x\)-axis.
When \(a\neq1\), we need to first take the \(a\) out by dividing each term by \(a\) before factorising the quadratic. That is:
\[ax^{2}+bx+c=a\left[x^{2}+\frac{b}{a}x+\frac{c}{a}\right]\]
We complete the square on the term in the square brackets. Don’t forget to include the \(a\) in your final answer!
Factorise \(2x^{2}-10x+2\).
Divide all terms in the quadratic by \(a\). Here, \(a=2\).
\[\begin{align*} 2x^{2}-10x+2 & = 2\left[x^{2}-\frac{10}{2}x+\frac{2}{2}\right]\\
& = 2\left[x^{2}-5x+1\right]
\end{align*}\]
Complete the square. Find the term by calculating \(\dfrac{1}{2}\) of the coefficient in front of the \(x\), squared:
Add and subtract the term:
\[\begin{align*} 2x^{2}-10x+2 & = 2\left[ x^{2}-5x+1+\left(\frac{-5}{2}\right)^{2}-\left(\frac{-5}{2}\right)^{2} \right]\quad\textrm{group terms}\\
& = 2\left[ x^{2}-5x+\left(\frac{-5}{2}\right)^{2}+1-\left(\frac{-5}{2}\right)^{2} \right]\\
& = 2\left[ x^{2}-5x+\left(\frac{25}{4}\right)+1-\frac{25}{4}\right]\quad\textrm{factorise quadratic using perfect squares}\\
& = 2\left[ \left(x-\frac{5}{2}\right)^{2}-\frac{21}{4}\right]\quad\textrm{take square root of \(\frac{21}{4}\) and square it}\\
& = 2\left[ \left(x-\frac{5}{2}\right)^{2}-\left(\frac{\sqrt{21}}{2}\right)^{2}\right]\quad\textrm{use DOTS and take square root of both sides}\\
& = 2 \left(x-\frac{5}{2}+\frac{\sqrt{21}}{2}\right)\left(x-\frac{5}{2}-\frac{\sqrt{21}}{2}\right)
\end{align*}\]
Surds
Use this resource if you need a refresher on how to simplify surds.
Factorisation
Use this resource if you need a refresher on how to identify common factors.
Quadratic factorisation
Use this resource if you need a refresher on how to factorise quadratic expressions using common factors.
Difference of two squares
Use this resource if you need a refresher on the difference of two squares (DOTS) rule.