Expansion
Expansion is the opposite of factorisation. Use this resource if you need to review expansion of algebraic expressions.
Factorisation
While expansion is the process of removing brackets from algebraic expressions and equations, factorisation is the opposite. Both are important to make solving algebraic problems easier. Use this resource to learn the basics of factorisation.
Video tutorial – factorisation: common factors
Watch this video to learn about factorisation using common factors.
Hi, this is Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on finding common factors in algebra. Let’s start by recalling that expansion or removing brackets is all about multiplying all the terms inside the brackets by those outside. For instance A brackets B plus C is the same as A times B plus A times C, if A brackets two Y minus four A is equal to 10 A Y minus 20 A X.
Factorisation is the reverse of expansion or removing brackets. For example, here’s an example, 10 A Y minus 20 A X. At this stage just look at the right hand side which are the factors, five A and brackets two Y minus four X, they are the factors, so moving from right to left, multiplying five A by the brackets will give you 10 A Y minus 20 A X. Similarly the factors for this expression are three and X minus Y and multiplying those terms will give us three X minus three Y, so to factorise an algebraic expression we look for a common factor, if there is more than one common factor we multiply them to get the highest common factor, HCF, the highest common factor is then placed before the bracket and the terms inside the brackets are found by dividing each term by the highest common factor.
Let’s do some examples to explain these four important steps. Here’s an example, five Y plus 10. First of all let’s look for the common factor and here the common factor of five Y and 10 is five, in other words five will divide into five Y and five will divide into 10. In other words that’s the same as five times Y plus five times two, it’s the same as five brackets Y plus two, so note the four steps, we factorise, the common factor is five, we place the five before the brackets and what goes after that five will be the other factor, Y plus two, which is inside the brackets.
Here’s another example, but this time we’re using pro-numerals, P squared plus P. What is the common factor of P squared plus P, well it’s P, P will go into P squared and P will go into P, so that’s the same as P times P plus P times one which again is the same as P brackets P plus one. There are two factors P and P plus one, the common factor P, P is placed before the brackets, the P plus one is placed inside the brackets.
Here is a slightly more complicated one. We’re mixing numbers and pro-numerals here. What’s the common factor of two A B and 12 A, well two goes into both those terms but also A does as well so the common factor is two and A, so we call that the highest common factor here, we put the two together and call them one thing, two A, same as two A times B minus two A times six, which can be written in another way, two A brackets B minus six. So to repeat, factorising, highest common factor is two A, two A before the brackets, B minus six is placed inside the brackets, which is the other factor.
Now do some questions for yourself. The answers to these problems are on the next slide. Thanks for watching this short movie.
Factorising algebraic expressions
Factorisation is the process of writing a number of algebraic expression as a product of numbers or expressions. Something that can be helpful to grasp this concept is to remember that factors are numbers we can multiply together to get another number.
Factorisation involves putting terms into brackets. This is the opposite of expansion, which involves removing brackets.
The simplest way to factorise an algebraic expression is using common factors. A common factor is a number or pronumeral that is shared by every term in an algebraic expression. Being able to identify common factors is critical for factorisation.
Example 1 – identifying common factors
Identify the factors in \(42\) and \(-2xyz\).
\(42\) is \(6\times7\) and also \(2\times3\times7\), so the factors are: \(2\), \(3\), \(6\) and \(7\).
\(-2xyz\) is \(-1\times2\times x\times y\times z\), so the factors are: \(-1\), \(2\), \(x\), \(y\) and \(z\).
Remember that combinations are factors as well, i.e. \(-2\), \(xy\), \(xyz\), etc.
Identify the factors in \(6b+6\).
\(6b+6\) is \(6\times(b+1)\) and also \(2\times3\times(b+1)\), so the factors are: \(2\), \(3\), \(6\) and \((b+1)\).
Identify the factors in \(12x+15\).
\(12x+15\) is \(3\times(4x+5)\), so the factors are: \(3\) and \((4x+5)\).
Identify the factors in \(15ay-10ax\).
\(15ay-10ax\) is \(5a\times(3y-2x)\) and also \(5\times a\times(3y-2x)\), so the factors are: \(5\), \(a\) and \((3y-2x)\).
Identify the factors in \(4prs+16pr+2ps+8p\).
\(4prs+16pr+2ps+8p\) is \(2p\times(2rs+8r+s+4)\) and also \(2\times p\times(2rs+8r+s+4)\), so the factors are: \(2\), \(p\) and \((2rs+8r+s+4)\).
Factorising using common factors
To factorise using common factors, you can use the following steps:
- Look at each term in the expression and find all of their factors.
- Identify the factors that every expression has in common.
- If there is more than one common factor, multiply them to give highest common factor (HCF).
- Divide each term by the HCF and write them in brackets. Make sure to keep the original operations that connect the terms.
- Place the HCF before the bracket.
Let's look at some examples where these steps are applied.
Example 1 – factorising using common factors
Factorise \(5y+10\).
The factors of \(5y\) are \(5\) and \(y\).
The factors of \(10\) are \(2\), \(5\) and \(10\).
The HCF is \(5\).
Dividing each term by the HCF:
\[\begin{align*} 5y \div 5 & =y\\
10 \div 5 & =2
\end{align*}\]
When we keep the original operations and place the HCF before the bracket, we get:
\[5(y+2)\]
Factorise \(3x+3y\).
The factors of \(3x\) are \(3\) and \(x\).
The factors of \(3y\) are \(3\) and \(y\).
The HCF is \(3\).
Dividing each term by the HCF:
\[\begin{align*} 3x \div 3 & =x\\
3y \div 3 & =y
\end{align*}\]
When we keep the original operations and place the HCF before the bracket, we get:
\[3(x+y)\]
Factorise \(p^{2}+p\).
\(p^{2}\) only has one factor, \(p\).
\(p\) also has one factor, \(p\).
The HCF is \(p\).
Dividing each term by the HCF:
\[\begin{align*} p^{2} \div p & =p\\
p \div p & =1
\end{align*}\]
When we keep the original operations and place the HCF before the bracket, we get:
\[p(p+1)\]
Factorise \(7y^{2}+7y\).
The factors of \(7y^{2}\) are \(7\) and \(y\).
The factors of \(7y\) are \(7\) and \(y\).
There is more than one common factor, so we multiply them to get the HCF, \(7\times y=7y\).
Dividing each term by the HCF:
\[\begin{align*} 7y^{2} \div 7y & =y\\
7y \div 7y & =1
\end{align*}\]
When we keep the original operations and place the HCF before the bracket, we get:
\[7y(y+1)\]
Factorise \(2abc-12ac\).
The factors of \(2abc\) are \(2\),
\(a\), \(b\) and \(c\).
The factors of \(-12ac\) are \(-1\), \(2\), \(3\), \(4\), \(6\), \(12\), \(a\) and \(c\).
There is more than one common factor, so we multiply them to get the HCF, \(2\times a\times c\).
Dividing each term by the HCF:
\[\begin{align*} 2abc \div 2ac & =b\\
12ac \div 2ac & =6
\end{align*}\]
When we keep the original operations and place the HCF before the bracket, we get:
\[2ac(b-6)\]
Factorise \(-2a-2b\).
The factors of \(-2a\) are \(-1\), \(2\) and \(a\).
The factors of \(-2b\) are \(-1\), \(2\) and \(b\).
There is more than one common factor, so we multiply them to get the HCF, \(-1\times2=-2\).
Dividing each term by the HCF:
\[\begin{align*} -2a \div -2 & =a\\
-2b \div -2 & =b
\end{align*}\]
When we keep the original operations and place the HCF before the bracket, we get:
\[-2(a+b)\]
Factorise \(-3x+6xy\).
The factors of \(-3x\) are \(-1\), \(3\) and \(x\).
The factors of \(6xy\) are \(2\), \(3\), \(6\),
\(x\) and \(y\).
There is more than one common factor, so we multiply them to get the HCF, \(3\times x=3x\).
Dividing each term by the HCF:
\[\begin{align*} -3x \div 3x & =-1\\
6xy \div 3x & =2y
\end{align*}\]
When we keep the original operations and place the HCF before the bracket, we get:
\[3x(-1+2y)=3x(2y-1)\]
Exercise – factorising using common factors
- Factorise the following expressions.
- \(3x+3y\)
- \(2a-2b\)
- \(8a-8b+8c\)
- \(xy-5x\)
- \(x^{2}-x\)
- \(7x+21y\)
- \(5m-2n\)
- \(c^{2}-2bc-3c\)
- \(5mn-10n\)
- \(3m^{2}-3mnp\)
- \(7x+21x^{2}\)
- \(12m^{2}-18mn\)
- \(5xy-10xz\)
- \(5pq-pq^{2}-3pqr\)
- \(2ab^{2}c+6abc^{2}\)
- \(rst+5rst^{2}-2rs\)
- \(5mn+10m-pqr\)
- \(5xyz-x^{2}yz^{2}+10x\)
- Factorise the following expressions by removing a negative factor.
- \(-3x-6y\)
- \(-15xy+25xz\)
- \(-2xy+4xyz\)
- \(14xyz-7xy\)
- \(-6xyz-15yz-3xy^{2}z\)
- \(7x-21y\)
-
- \(3(x+y)\)
- \(2(a-b)\)
- \(8(a-b+c)\)
- \(x(y-5)\)
- \(x(x-1)\)
- \(7(x+3y)\)
- Cannot be factorised further
- \(c(c-2b-3)\)
- \(5n(m-2)\)
- \(3m(m-np)\)
- \(7x(1+3x)\)
- \(6m(2m-3n)\)
- \(5x(y-2z)\)
- \(pq(5-q-3r)\)
- \(2abc(b+3c)\)
- \(rs(t+5t^{2}-2)\)
- Cannot be factorised further
- \(x(5yz-xyz^{2}+10)\)
-
- \(-3(x+2y)\)
- \(-5x(3y-5z)\)
- \(-2xy(1-2z)\)
- \(-7xy(-2z+1)\) or \(-7xy(1-2z)\)
- \(-3yz(2x+5+xy)\)
- \(-7(x+3y)\)
