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# Chapter 4: Factorising algebraic expressions

In previous years you learnt that we can use the distributive law to expand an algebraic expression. This is when we multiply the number or variable before a bracket with all the terms inside the bracket. We can represent the distributive law as follows:

expand We expand an algebraic expression by using the distributive law to remove brackets and then collecting like terms.

distributive law The distributive law states that the multiplying factor or coefficient of a bracket can be distributed across each of the terms inside the bracket, for example $a(b+c)=ab+bc$.

Last year you learnt that we can write an algebraic expression as a product of factors. Suppose we start with the expression $ab+ac$. To write the expression as a product of factors, we have to factorise it. This means we have to identify the factors that can be multiplied to give the expression. The variable $a$ is a common factor of the two terms. We therefore know that it is one of the factors of the expression. If we divide each term by $a$, we are left with $b+c$. Therefore $ab+ac=a(b+c)$.

factorise We factorise an algebraic expression by writing it as a product of factors that can be multiplied to give the original expression.

Factorising an algebraic expression is the reverse of expanding it. We can represent the two processes as follows:

## 4.1 Revision of common factors

Last year you learnt how to factorise an algebraic expression using a common factor. The terms in an algebraic expression have a common factor if the same number or variable is a factor of each term. For example, the number 5 divides into both $5x$ and $10y$ without a remainder. Therefore, 5 is a common factor of the terms in the expression $5x+10y$.

common factor When an algebraic expression has the same factor in two or more terms, this is called a common factor.

### Worked example 4.1: Factorising an algebraic expression using a common factor

Factorise the following expression.

1. Step 1: Find the HCF of the numbers.

Write the two numbers as a product of their prime factors in index form.

$20=2^2\times5$ and $15=3\times5$

Identify the prime factors that are contained in both numbers, to the lowest index.

2. Step 2: Find the HCF of the variables.

Apply the same principles you used in Step 1. Identify the variables that are contained in both terms, to the lowest index.

We have $x^2y$ and $x^3$.

3. Step 3: Combine the answers from Step 1 and 2 to find the HCF for the full expression. This is the first factor.

4. Step 4: To find the second factor, divide each term by the first factor.

\begin{align} \require{cancel} &\frac{20x^2y}{5x^2}-\frac{15x^3}{5x^2} \\ =&\frac{20 \times x \times x \times y}{5\times x \times x}-\frac{15\times x \times x \times x}{5\times x \times x} \\ =&\frac{\cancel{20}^4 \times \cancel x \times \cancel x \times y}{\cancel{5}^1 \times \cancel x \times \cancel x}-\frac{\cancel{15}^3 \times \cancel x \times \cancel x \times x}{\cancel{5}^1 \times \cancel x \times \cancel x} \\ =&4y-3x \end{align}
5. Step 5: Write down the two factors from Steps 3 and 4.

### Exercise 4.1: Factorise algebraic expressions using a common factor

Factorise each of the following algebraic expressions.

1. \begin{align} & \, 10p+20q^2-30rs \\ =& \, 10(p+2q^2-3rs) \end{align}
2. \begin{align} & \, 4x-20y+16z \\ =& \, 4(x-5y+4z) \end{align}
3. \begin{align} & \, b^2-3ab^2 \\ =& \, b^2(1-3a) \end{align}
4. \begin{align} & \, 4ab-8abc+2a^2 \\ =& \, 2a(2b-4bc+a) \end{align}
5. \begin{align} & \, 18mn^2+24m^3n \\ =& \, 6mn(3n+4m^2) \end{align}
6. \begin{align} & \, -3a^3+15a^2-6a \\ =& \, -3a(a^2-5a+2) \end{align}

If the first term is negative, your common factor should also be negative. Be careful with the signs in the bracket. Remember:
$(+)\times(+)=(+)$
$(-)\times(-)=(+)$
$(+)\times(-)=(-)$
$(-)\times(+)=(-)$

7. \begin{align} & \, -b^3-3b^4+2ab^3 \\ =& \, -b^3(1+3b-2a) \end{align}
8. \begin{align} & \, -16x^2yz-40x^2y+32x^3y^2 \\ =& \, -8x^2y(2z+5-4xy) \end{align}
9. \begin{align} & \, 2p(q-4)+3q(q-4) \\ =& \, (q-4)(2p+3q) \end{align}

A bracket can also be a common factor.

10. \begin{align} & \, 2y(x-y)-x^2(x-y) \\ =& \, (x-y)(2y-x^2) \end{align}

## 4.2 Grouping terms

For some algebraic expressions, there is no common factor for all the terms. We can, however, factorise such expressions by grouping terms together and then identifying a common factor for each group. Algebraic expressions where grouping can be used normally have four terms.

### Worked example 4.2: Factorising an algebraic expression using grouping

Factorise:

1. Step 1: Always try to factorise using a common factor first.

There is no number or variable that is a factor of all four terms.

2. Step 2: Group the terms in pairs so that there is a common factor in each pair. Rewrite the expression.

Number the terms from 1 to 4:

\begin{array}{cccc} \;^1 & \;^2 & \;^3 & \;^4 \\ 2a^2 & +\;ab & -\;2a & -\;b \end{array}

Try grouping terms 1 and 2 in one pair: $2a^2+ab$
This pair has a common factor.

Check that terms 3 and 4 also have a common factor: $-2a-b$
This pair does not have a common factor, so this grouping does not work.

Now try grouping terms 1 and 3 in one pair: $2a^2-2a$
This pair has a common factor.

Check that terms 2 and 4 also have a common factor: $ab-b$
This pair also has a common factor, so this grouping works.

Rewrite the expression so that the terms in a pair are next to each other:
$2a^2-2a+ab-b$

3. Step 3: Factorise each group.

\begin{align} & \, 2a^2-2a+ab-b \\ =& \, 2a(a-1)+b(a-1) \\ \end{align}

Remember, $\frac{2a}{2a}=1$ and $\frac{b}{b}=1$.

4. Step 4: If the resulting terms have a bracket that is a common factor, factorise further.

\begin{align} & \, 2a^2-2a+ab-b \\ =& \, 2a(a-1)+b(a-1) \\ =& \, (a-1)(2a+b) \end{align}

### Exercise 4.2: Factorise algebraic expressions using grouping

Factorise each of the following algebraic expressions.

1. \begin{align} & \, ac+bc+ad+bd \\ =& \, c(a+b)+d(a+b) \\ =& \, (a+b)(c+d) \end{align}
2. \begin{align} & \, 3ax+3bx+by+ay \\ =& \, 3x(a+b)+y(b+a) \\ =& \, 3x(a+b)+y(a+b) \\ =& \, (a+b)(3x+y) \end{align}

Remember that $a+b=b+a$.

3. \begin{align} & \, 2n+6+3m+mn \\ =& \, 2(n+3)+m(3+n) \\ =& \, 2(n+3)+m(n+3) \\ =& \, (n+3)(2+m) \end{align}
4. \begin{align} & \, 2xz+yz+2x^2+yx \\ =& \, z(2x+y)+x(2x+y) \\ =& \, (2x+y)(z+x) \\ \end{align}
5. \begin{align} & \, ab-b-2a+2 \\ =& \, b(a-1)-2(a-1) \\ =& \, (a-1)(b-2) \\ \end{align}
6. \begin{align} & \, x^2-3x+2xy-6y \\ =& \, x(x-3)+2y(x-3) \\ =& \, (x-3)(x+2) \\ \end{align}
7. \begin{align} & \, a^2-3a-ba+3b \\ =& \, a(a-3)-b(a-3) \\ =& \, (a-3)(a-b) \\ \end{align}
8. \begin{align} & \, -ps-qs-pt-qt \\ =& \, -s(p+q)-t(p+q) \\ =& \, (p+q)(-s-t) \\ =& \, (p+q)(-1)(s+t) \\ =& \, -(p+q)(s+t) \end{align}

If you end with a bracket where both terms inside the bracket are negative, take -1 out as a common factor.

9. \begin{align} & \, 9xy+9yz+3ax+3az \\ =& \, 3(3xy+3yz+ax+az) \\ =& \, 3\Big( 3y(x+z)+a(x+z) \Big) \\ =& \, 3\Big((x+z)(3y+a) \Big) \\ =& \, 3(x+z)(3y+a) \\ \end{align}

Remember that you must factorise using a common factor first.

10. \begin{align} & \, -20x^3+5x^2+20xy^3-5y^3 \\ =& \, -5(4x^3-x^2-4xy^3+y^3) \\ =& \, -5\Big( x^2(4x-1)-y^3(4x-1) \Big) \\ =& \, -5\Big((4x-1)(x^2-y^3) \Big) \\ =& \, -5(4x-1)(x^2-y^3) \\ \end{align}

Remember that you must factorise using a common factor first.

## 4.3 Difference of two squares

Last year you learnt a short method to expand two similar brackets that have opposite signs:

• Square the term that stands first in each bracket.
• Square the term that stands last in each bracket.
• Put a minus sign between the two squares.

Here are some examples from last year:

Because you have squared both terms, and then put a minus sign between the two squares, we can call this expression the difference of two squares.

Factorising an algebraic expression that is the difference of two squares is the reverse of expanding two similar brackets that have opposite signs.

### Worked example 4.3: Factorising the difference of two squares

Factorise:

1. Step 1: Write down two empty brackets.

2. Step 2: Determine the square root of the first term and put it in the first position of each bracket.

3. Step 3: Determine the square root of the last term and put it in the last position of each bracket.

### Exercise 4.3: Factorise the difference of two squares

Factorise each of the following algebraic expressions.

1. \begin{align} & \, 45x^2-5y^2 \\ =& \, 5(9x^2-y^2) \\ =& \, 5(3x+y)(3x-y) \end{align}

If an expression looks like the difference of two squares, but one or both of the terms do not have a square root, try to factorise using a common factor first.

A polynomial is an algebraic expression with two or more terms. In a polynomial, one term can be a constant, and the other terms each consist of coefficients with variables raised to positive powers. A binomial has two terms, for example $2x+y$. A trinomial has three terms, for example $3x+2y+5$.

A quadratic trinomial is a trinomial of which the highest power of any variable is two. Here are some examples of quadratic trinomials:

trinomial A trinomial is an algebraic expression with three terms of which at least two terms consist of coefficients with variables raised to positive powers.

quadratic trinomial A quadratic trinomial is a trinomial of which the highest power of any variable is two.

There are different methods to factorise quadratic trinomials. We will only consider one such method. Keep in mind that not all quadratic trinomials can be factorised.

### Worked example 4.4: Factorising quadratic trinomials using the box method

Factorise:

1. Step 1: Draw a $2\times2$ grid.

\begin{array}{|c|c|} \hline \phantom{0}&\phantom{0} \\ \hline \phantom{0}&\phantom{0} \\ \hline \end{array}
2. Step 2: Put the first term of the trinomial in the top left corner. Then put the last term of the trinomial in the bottom right corner.

The minus sign is part of the last term.

\begin{array}{|c|c|} \hline 6x^2&\phantom{0} \\ \hline \phantom{0}&-2y^2 \\ \hline \end{array}
3. Step 3: Multiply the coefficients of the terms in the corners of the box.

4. Step 4: Write down all the factors of 12 that could give an answer of $-12$.

\begin{align} 1 & \times -12 \\ 2 & \times -6 \\ 3 & \times -4 \\ 4 & \times -3 \\ 6 & \times -2 \\ 12 & \times -1 \end{align}
5. Step 5: Look at the middle term of the trinomial. Determine which combination in Step 4 will give this term when added. Include the variables of the middle term.

The middle term is $-xy$.

From the combinations in Step 4: $3xy+(-4xy)=-xy$

6. Step 6: Put the two terms identified in Step 5 in the empty corners of the box. It does not matter which term goes in what corner.

\begin{array}{|c|c|} \hline 6x^2&3xy \\ \hline -4xy&-2y^2 \\ \hline \end{array}
7. Step 7: Add another column and another row to the box.

\begin{array}{|c|c|c} \hline 6x^2&3xy & \phantom{000} \\ \hline -4xy&-2y^2 & \phantom{000} \\ \hline \phantom{0}&\phantom{000} \\ \end{array}
8. Step 8: Find the HCF of each row and each column and enter it in the added spaces.

\begin{array}{|c|c|c} \hline 6x^2&3xy & \color{red}{3x} \\ \hline -4xy&-2y^2 & \color{red}{-2y} \\ \hline \color{green}{2x} & \color{green}{y} \\ \end{array}

Only include a minus sign in the common factor of a row or column if both terms in that row or column are negative.

9. Step 9: Write down the column and row you completed in Step 8 as two brackets. Include the signs of the terms. The brackets are the factors of the trinomial.

The column forms the bracket $(3x-2y)$.
The row forms the bracket $(2x+y)$.

Therefore:

You learnt the distributive law last year: $(a+b)(c+d)=ac+ad+bc+bd$

\begin{align} & \, (3x-2y)(2x+y) \\ =& \, 6x^2+3xy-4xy-2y^2 \\ =& \, 6x^2-xy-2y^2 \end{align}

The factors in Step 9 are correct.

### Exercise 4.4: Factorise quadratic trinomials

Factorise each of the following algebraic expressions.

1. If you multiply the coefficients of the first two terms you put in the corners, you get -6.

The common factors of -6 that you can add to get the middle term, $x$, which is the same as $1x$, are $3$ and -2, so the terms in the other two corners are $3x$ and $-2x$.

Then you need to find the HCF of each row and each column.

\begin{array}{|c|c|c} \hline x^2 & 3x & x \\ \hline -2x & -6 & -2 \\ \hline x & 3 \\ \end{array}
2. \begin{array}{|c|c|c} \hline a^2 & -5a & a \\ \hline 3a & -15 & 3 \\ \hline a & -5 \\ \end{array}
3. \begin{array}{|c|c|c} \hline 2p^2 & 4p & 2p \\ \hline 3p & 6 & 3 \\ \hline p & 2 \\ \end{array}
4. \begin{array}{|c|c|c} \hline x^2 & -11x & x \\ \hline 2x & -22 & 2 \\ \hline x & -11 \\ \end{array}
5. \begin{array}{|c|c|c} \hline b^2 & 4bc & b \\ \hline 2bc & 8c^2 & 2c \\ \hline b & 4c \\ \end{array}
6. \begin{array}{|c|c|c} \hline x^2 & -7x & x \\ \hline -4x & 28 & 4 \\ \hline x & 7 \\ \end{array}

Note that we have an exception here: $(x+4)(x+7)=x^2+11x+28$. In the $2 \times 2$ grid, look at the diagonals. Both top right and bottom left are negative, while both top left and bottom right are positive. In this case the signs in the brackets change from plus to minus.

7. \begin{array}{|c|c|c} \hline 4y^2 & 12y & 4y \\ \hline y & 3 & 1 \\ \hline y & 3 \\ \end{array}

The only HCF of $y$ and $3$ in the bottom row is 1.

8. \begin{array}{|c|c|c} \hline a^2 & -3ab & a \\ \hline 2ab & -6b^2 & 2b \\ \hline a & -3b \\ \end{array}
9. \begin{array}{|c|c|c} \hline x^2 & -5xy & x \\ \hline -4xy & 20y^2 & 4y \\ \hline x & 5y \\ \end{array}

Look at the diagonals of the $2 \times 2$ grid. In one, both blocks are positive. In the other, both blocks are negative. So the signs in the brackets change from plus to minus.

10. \begin{array}{|c|c|c} \hline 6a^2 & -9ab & 3a \\ \hline 4ab & -6b^2 & 2b \\ \hline 2a & -3b \\ \end{array}

## 4.5 Practical applications

In previous years you learnt that algebraic expressions can be very useful to solve problems. Now you can also use factorisation for everyday problems.

### Exercise 4.5: Use factorisation to solve problems

1. The area of a rectangle is $x^2+7x+10$. If the breadth of the rectangle is $x+2$, determine its length.

For a rectangle:

\begin{align} l \times b&=x^2+7x+10 \\ \text{But }x^2+7x+10 &=(x+2)(x+5) \\ \therefore l \times b&=(x+2)(x+5)\\ l \times (x+2)&=(x+2)(x+5) \\ \therefore l&=x+5 \end{align}
2. The product of your age $t$ years ago and your age in $t$ years' time is $x^2-9$. Determine the value of $t$.

Suppose your current age is $x$.
Then your age $t$ years ago was $x-t$.
Your age in $t$ years' time is $x+t$.

\begin{align} (x+t)(x-t)&=x^2-9 \\ \text{But }x^2-9 &=(x+3)(x-3) \\ \therefore (x+t)(x-t)&=(x+3)(x-3)\\ \therefore t&=3 \end{align}
3. The perimeter of a rectangle is $4x+20$. If the breadth of the rectangle is $x$, determine its length.

For a rectangle:

\begin{align} 2(l+b)&=4x+20 \\ \text{But }4x+20 &=2(2x+10) \\ \therefore 2(l+b)&=2(2x+10)\\ 2(l+x)&=2(x+10+x) \\ \therefore l&=x+10 \end{align}
4. The area of a square is $4x^2+12x+9$. Determine the perimeter of the square.

For a square:

\begin{align} s^2&=4x^2+12x+9 \\ \text{But }4x^2+12x+9 &=(2x+3)(2x+3) \\ \therefore s^2&=(2x+3)^2\\ \therefore s&=2x+3 \end{align}

For a square:

\begin{align} P &= 4s \\ P&=4(2x+3) \\ &=8x+12 \end{align}
5. Write the expression $2x+ax+ay+2y$ as the product of two factors.

\begin{align} & \,2x+ax+ay+2y \\ =& \,x(2+a)+y(a+2)\\ =& \,(2+a)(x+y) \end{align}

## 4.6 Summary

• We factorise an algebraic expression by writing it as a product of factors that can be multiplied to give the original expression.
• A common factor is a factor of two or more terms in an algebraic expression. For example, $ab+ac+ad=a(b+c+d)$.
• For some algebraic expressions, there is no common factor for all the terms. We can, however, factorise such expressions by grouping terms together and then identifying a common factor for each group. For example, $ay+az+xy+xz=a(y+z)+x(y+z)=(y+z)(a+x)$.
• The factors of the difference of two squares are two similar brackets that have opposite signs. For example, $a^2-b^2=(a+b)(a-b)$.
• A trinomial is an algebraic expression with three terms of which at least two terms consist of coefficients with variables raised to positive powers.
• A quadratic trinomial is a trinomial of which the highest power of any variable is two.
• We can use the box method to factorise a quadratic trinomial. For example, the box for $x^2+5x+6$ is:
\begin{array}{|c|c|c} \hline x^2 & 3x & x \\ \hline 2x & 6 & 2 \\ \hline x & 3 \\ \end{array}
Therefore $x^2+5x+6=(x+2)(x+3)$
• We can use factorisation to solve problems.