*Factorisation* can be thought of as reversing the process of expanding brackets.

The aim is to take an expression, usually a polynomial in one or more variables, and write it as the product of other expressions or *factors*.

We usually require that each factor of an expression is not equal to $1$ and cannot be further factorised.

Factorisation can always be checked by multiplying out the factors and seeing if this gives the original expression.

For example, in the expression $x^2+3x+2=(x+1)(x+2)$, $x+1$ and $x+2$ are the linear factors and both factors cannot be factorised further.

Sometimes we cannot find a factorisation; for example the quadratic $x^2+3x+3$ does not have any factors.

In the following examples we consider two types of mathematical expressions and consider their factorisation:

- The expression is a sum of terms where each term has the same or common factor.
- Quadratic expressions.

We also give examples of useful factorisations other than these.

We will use the following result in some of the examples. It can be a quick way of finding a linear factor.

If $p(x)$ is a polynomial in the variable $x$ and if $p(r)=0$, then \[p(x)=(x-r)q(x)\] where $q(x)$ is a polynomial of degree one less than the degree of $p(x)$.

**Note:** $x=r$ is called a *root* of the polynomial as it is a solution of $p(x)=0$.

For example, we could have factorised $x^2+3x+2$ using this result:

- Putting $x=-1$ gives $(-1)^2-3+2=1-3=0$, hence $x-(-1)=x+1$ is a factor.
- Putting $x=-2$ gives $(-2)^2+3\times (-2)+2=4-6+2=0$, hence $x-(-2)=x+2$ is a factor.

So $x^2+3x+2=(x+1)(x+2)$.

**Note:** You have to be careful in using this result. You may spend a long time in finding factors by this method - for example the polynomial may not have any roots! Best to use it to test for roots such as $-2,\;-1,\; 0,\; 1,\;2$.

If an expression is a sum of terms, each of which has the same factor, then we can factorise the whole expression:.

**Example**: In the expression $ax+a$, there is a factor of $a$ common to both terms, so it can be written as $ax+a = a(x+1)$.

Factorise $3xy+9x^2$.

The common factors in both terms are $3$ and $x$, because $3$ divides into both $3$ and $9$, and both terms have an $x$ in them.

\begin{align} 3xy+9x^2 &= 3(xy+3x^2)\\ &= 3x(y+3x) \end{align}

Factorise $16xy^2z + 8x^2y^3 - 4xy^4z^2$.

The same process applies when there are more than two terms; here the common factors are $4$, $x$ and $y^2$. \begin{align} 16xy^2z + 8x^2y^3 - 4xy^4z^2 &= 4(4xy^2z+2x^2y^3-xy^4z^2)\\ &= 4x(4y^2z + 2xy^3 - y^4z^2)\\ &=4xy^2(4z + 2xy - y^2z^2) \end{align}

A quadratic expression in one variable is of the form $ax^2+bx+c$ where $a \neq 0$.

Note that to factorise $q(x)=ax^2+bx+c$ we could write $q(x)=a(x^2+bx/a+c/a)$ with $a$ as the factor - but this is not considered to be a proper factorisation.

However if we can factorise $x^2+\frac{b}{a}x+\frac{c}{a}=(x+r)(x+s)$ then we have factorised the original quadratic.

So we look for factorising quadratics of the form $q(x)=x^2+px+q$.

This means we are looking for numbers $r$ and $s$ such that \[q(x)=x^2+px+q=(x+r)(x+s)=x^2+(r+s)x+rs.\] We can factorise the quadratic if we can find $r$ and $s$ such that

- They sum to the coefficient of $x$: $r+s=p$.
- Their product is the constant term: $rs=q$.

**Note:** It is not always possible to factorise a quadratic; for example the quadratic $x^2+2x+2$ cannot be factorised as there are no two numbers $r$ and $s$ such that $r+s=2, \;rs=2$. See the section on the discriminant in order to determine whether or not a quadratic expression can be factorised.

A useful factorisation to remember is that of a quadratic of the form $x^2-a^2$ . This is easily factorised into two brackets by: \[x^2-a^2 = (x-a)(x+a)\]

This works for any value of $a$ and can be checked by expanding the brackets.

\begin{align} (x-a)(x+a) &= x^2 -ax +ax -a^2\\ &= x^2-a^2 \end{align}

Factorise $x^2-7x+10$.

Start by thinking of two numbers that will multiply to give $10$; the options are \[10\times1, \;(-10)\times(-1), \;5\times2, \;(-5)\times(-2)\]

We can clearly tell that there is no combination of $10$ and $1$ that will give the answer $-7$. So the numbers must be $5$ and $2$, or $-5$ and $-2$. If we add together each of these pairs of numbers, we will find that some combination will give us $-7$. \begin{align} 5+2 &= 7 \\ (-5)+(-2) &= -7 \end{align}

A combination of $-5$ and $-2$ will give us the answer $-7$. So, \[x^2-7x+10 = (x-5)(x-2)\]

Always check the final answer by expanding the brackets to see that the expression matches the one you started with.

You could also have found this result by noting that $x^2-7x+10=0$ when $x=5$ or $x=2$ and using the Useful Result above.

Factorise $2x^2+x-15$.

First notice that this time the $x^2$ coefficient is $2$. This means the final factorised expression must take the form $(2x \,\dotso)(x \, \dotso)$ because this is the only way to obtain a $2x^2$ term. The next step is the same as in the previous example. Think of two numbers that will multiply to give $-15$; the options are \[(-5) \times 3, \;5 \times (-3), \;15 \times (-1), \;(-15) \times 1\]

Now, we have to take into account that if we were to expand these brackets, one of our chosen numbers would be multiplied by $2x$. This means that when trying to find the combination of numbers that add to give the coefficient of $x$, in this case $1$, one of our numbers must be multiplied by two. Just by looking at the numbers above, we can see that no combination of $15$ and $1$, with either multiplied by $2$, will give a value of $1$. So the numbers must be $-5$ and $3$, or $5$ and $-3$.

\begin{align} 2 \times (-5) + 3 &= -7 \\ (-5) + 2 \times 3 &= 1 \\ 2 \times 5 + (-3) &= 7 \\ 5 + 2 \times (-3) &= -1 \end{align}

A combination of $-5$ and $3$, with $3$ to be multiplied by the $2x$ term, works. \[2x^2+x-15 = (2x - 5)(x + 3)\]

**Note:** This time, it matters which way round the numbers go in the brackets. If it were the other way around, the $5$ would be multiplied by the $2$ and this would not give the right answer.

**Note:** In this case looking for values for $x$ such that $2x^2+x-15 =0$ and using the Useful Result above could be a long job! So it's not always a good idea to do this.

Prof. Robin Johnson factorises the expression $x^2+x-6$.

\[x^3-a^3=(x-a)(x^2+ax+a^2)\] \[x^3+a^3=(x+a)(x^2-ax+a^2)\]

Consider $p(x)=x^3-a^3$.

If we put $x=a$ we find that $p(a)=a^3-a^3=0$. Hence, using the Useful Result above, we know that $x-a$ is a factor and so $x^3-a^3=(x-a)q(x)$ for some polynomial $q(x)$ of degree 2.

It follows from inspection or by dividing $x^3-a^3$ by $x-a$ that $q(x)=x^2+ax+a^2$.

So we have \[x^3-a^3=(x-a)(x^2+ax+a^2)\]

Similarly for $p(x)=x^3+a^3$, if we put $x=-a$ then we find $p(-a)= (-a)^3+a^3 = -a^3+a^3=0$ and so $x+a$ is a factor.

It follows that \[x^3+a^3=(x+a)(x^2-ax+a^2)\]

Factorise $x^3+8$.

$x^3+8=x^3+2^3=(x+2)(x^2-2x+4)$.

We cannot factorise $x^2-2x+4$ any further.

Factorise $2y^3-54$.

\begin{align}2y^3-54&=2(y^3-27)\\&=2(y^3-3^3)\\ &=2(y-3)(y^2+3y+9)\end{align}

We cannot factorise $y^2+3y+9$ any further.

Factorise $p(x)=x^4-x^3-8x+8$.

Putting $x=1$ gives $p(1)=1-1-8+8=0$. Hence $x-1$ is a factor.

On dividing $p(x)$ by $x-1$, we obtain

\begin{align} x^4-x^3-8x+8&=(x-1)(x^3-8)\\&=(x-1)(x^3-2^3)\\&=(x-1)(x-2)(x^2+2x+4) \end{align}

We cannot factorise $x^2+2x+4$ any further.

In these two videos Professor Johnson finds the roots of two cubics by factorising the cubics. In each case one of the roots is given, which enables him to find a linear factor and then divide by that factor to find a quadratic which he can then solve. Thus all three roots are found.

This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.

Test yourself: Numbas test on factorising quadratics

Test yourself: Numbas test on Apply the factor and remainder theorems

- Factorising simple expressions resources at
**math**centre. - Factorising quadratics resources at
**math**centre.