Functions and Algebra: Manipulate and simplify algebraic expressions

# Unit 2: Factorisation

Natashia Bearam-Edmunds

### Unit outcomes

By the end of this unit you will be able to:

• Factorise by identifying and taking out the common factor.
• Factorise by grouping in pairs.
• Factorise the difference of two squares.
• Factorise trinomials.

## What you should know

Before you start this unit, make sure you can:

## Introduction

When we learn about fractions we are shown that the highest common factor (HCF) of two numbers is the largest number that divides evenly into both numbers. For example, $\scriptsize 3$ is the (HCF) of $\scriptsize 9$ and $\scriptsize 15$ because it is the largest number that divides exactly into both numbers. The HCF of works the same way.

Many polynomial expressions can be written in simpler forms by factorising. Finding the HCF is just one of the ways that we factorise expressions. In this unit, we will look at different methods to factorise polynomials.

## Find the highest common factor

Factorisation reverses the process of multiplying and expanding brackets that we covered in Unit 1. For example, if we expand $\scriptsize 3(x+2)$ we get $\scriptsize 3x+6$. When we factorise, we start with $\scriptsize 3x+6$ and end up with $\scriptsize 3(x+2)$. The two expressions are exactly the same in value no matter what values we substitute for $\scriptsize x$.

Each part of a product is called a factor of the expression. If $\scriptsize c=ab$, then $\scriptsize a$ and $\scriptsize b$ are factors of $\scriptsize c$. Similarly, $\scriptsize 3$ and $\scriptsize x+2$ are factors of $\scriptsize 3(x+2)$ since $\scriptsize 3x+6=3(x+2)$. Factorising based on common factors relies on there being factors common to all the terms.

Let’s factorise the expression $\scriptsize 10a+5$. Both $\scriptsize 10a$ and $\scriptsize 5$ have a common factor of $\scriptsize 5$:

$\scriptsize 10a=2\times \underline{5}\times a$
$\scriptsize 5=1\times \underline{5}$

So we can write $\scriptsize 10a+5$ as $\scriptsize 2\times 5\times a+1\times 5$. We can factorise by taking out the HCF of $\scriptsize 5$ from both terms and we will get:

$\scriptsize \begin{array}{l}10a+5&=5(2\times a+1)\\&=5(2a+1)\end{array}$

We have divided each term by the HCF of $\scriptsize 5$; this is called factorising by taking out the highest common factor.

When factorising a polynomial, our first step should always be to check for a HCF. Look for the HCF of the coefficients, and then look for the HCF of the variables and combine them to find the HCF of the expression. This is shown in the next example.

### Example 2.1

Factorise:

1. $\scriptsize 4{{x}^{3}}+2{{x}^{2}}+6x$
2. $\scriptsize 3(a-b)-x(a-b)$
3. $\scriptsize 2(x-y)-a(y-x)$

Solutions

1. First, find the HCF of the coefficients.
The HCF of $\scriptsize 4,2$ and $\scriptsize 6$ is $\scriptsize 2$.
$\scriptsize 4=\underline{2}\times 2$
$\scriptsize 2=\underline{2}\times 1$
$\scriptsize 6=\underline{2}\times 3$
Next, find the HCF of the variables.
$\scriptsize {{x}^{3}}=\underline{x}\times x\times x$
$\scriptsize {{x}^{2}}=\underline{x}\times x$
$\scriptsize x=\underline{x}\times 1$
.
The HCF of $\scriptsize {{x}^{3}},{{x}^{2}}$ and $\scriptsize x$ is $\scriptsize x$. Note: the HCF of a set of expressions of the form $\scriptsize {{x}^{n}}$ will always be the power with the smallest exponent.
.
We combine the HCF’s of both the coefficients and variables to get that the HCF of the entire expression is $\scriptsize 2x$.
.
Lastly, we divide each term by the HCF to factorise.
$\scriptsize \begin{array}{l}4{{x}^{3}}+2{{x}^{2}}+6x&=2x\left( {\displaystyle \frac{{4{{x}^{3}}}}{{2x}}+\displaystyle \frac{{2{{x}^{2}}}}{{2x}}+\displaystyle \frac{{6x}}{{2x}}} \right)\\&=2x(2{{x}^{2}}+x+3)\end{array}$
.
Note: you can check your answer after factorising by finding the product or expanding the brackets. If you have factorised correctly you will end up with the original expression. Try and confirm this for yourself.
2. A common bracket can be taken out as the HCF provided the bracket is identical in both terms. There is a common factor bracket of $\scriptsize a-b$ in$\scriptsize 3(a-b)-x(a-b)$. To take out the HCF, simply divide each term by the common bracket.
$\scriptsize \begin{array}{l}3(a-b)-x(a-b)&=(a-b)\left[ {\displaystyle \frac{{3(a-b)}}{{(a-b)}}-\displaystyle \frac{{x(a-b)}}{{(a-b)}}} \right]\\&=(a-b)[3-x]\end{array}$
.
Factorising by finding a HCF which is a common bracket is called grouping.
3. In this example, the brackets look similar but the terms within have different signs. Use a ‘switch around’ method to find the common factor.
$\scriptsize (y-x)=-1(x-y)$ Divide each term by negative one to switch around the terms.
$\scriptsize \begin{array}{l}2(x-y)-a(y-x)&=2(x-y)-a(-1)(x-y)\\&=2(x-y)+a(x-y)\\&=(x-y)(2+a)\\&=(x-y)(a+2)\end{array}$
.
Note:$\scriptsize (2+a)=(a+2)$. There is no need to change signs outside the bracket when both terms are positive.

### Exercise 2.1

Factorise:

1. $\scriptsize -a{{x}^{2}}+bx+{{x}^{3}}$
2. $\scriptsize \left( {a-b} \right)x+a-b$
3. $\scriptsize \left( {a-b} \right)x-a+b$
4. $\scriptsize -12a+24{{a}^{2}}-3$

The full solutions are at the end of the unit.

## Factorising by grouping

Finding and taking out the HCF is the starting point of all factorising. We know that the factors of $\scriptsize 2x+2$ are $\scriptsize 2$ and$\scriptsize (x+1)$. Similarly the factors of $\scriptsize 3{{x}^{2}}+3x$ are $\scriptsize 3x$ and $\scriptsize (x+1)$. If you have one expression that combines all of these terms, $\scriptsize 2x+2+3{{x}^{2}}+3x$, there is no common factor to all four terms, but we can factorise as follows:

$\scriptsize \begin{array}{l}2x+2+3{{x}^{2}}+3x&=(2x+2)+(3{{x}^{2}}+3x)\\&=2(x+1)+3x(x+1)\end{array}$

Now there is a common factor of $\scriptsize (x+1)$.

By taking out the common bracket of $\scriptsize (x+1)$ we get $\scriptsize 2(x+1)+3x(x+1)=(x+1)(2+3x)$. As you saw in Example 2.1, this is called factorising by grouping.

### Example 2.2

Find the factors of: $\scriptsize 6x+3y+2ax+ay$

Solution

Step 1: There are no factors common to all terms.

Step 2: Group terms with common factors together.

$\scriptsize 3$ is a common factor of the first two terms and $\scriptsize a$ is a common factor of the second two terms.

$\scriptsize 6x+3y+2ax+ay=3(2x+y)+a(2x+y)$

Step 3: Take out the common bracket as the HCF.

$\scriptsize \begin{array}{l}3(2x+y)+a(2x+y)&=\displaystyle \frac{{3(2x+y)}}{{(2x+y)}}+\displaystyle \frac{{a(2x+y)}}{{(2x+y)}}\\&=(2x+y)(3+a)\end{array}$

Step 4: Write the final answer.

The factors of $\scriptsize 6x+3y+2ax+ay$ are $\scriptsize (2x+y)$ and $\scriptsize (3+a)$.

Note: you can get to the same answer by grouping $\scriptsize 6x$ with $\scriptsize 2ax$ and $\scriptsize 3y$ with $\scriptsize ay$, try it yourself to confirm this.

### Exercise 2.2

Factorise:

1. $\scriptsize {{x}^{2}}-6x+5x-30$
2. $\scriptsize 6t-15s+2yt-5ys$
3. $\scriptsize ab-{{a}^{2}}-a+b$
4. $\scriptsize 3ax+bx-3ay-by-9a-3b$

The full solutions are at the end of the unit.

## Difference of two squares

A difference of squares is a perfect square subtracted from another perfect square. A difference of squares can be rewritten as a product of binomials containing the same terms but opposite signs because the middle terms will cancel each other out if the two factors are multiplied.

Given a difference of squares, you can factorise it into binomials by:

• confirming that the first and last term are perfect squares and that the expression has one negative sign and one positive sign.
• write the factorised form as $\scriptsize (\sqrt{{{{a}^{2}}}}+\sqrt{{{{b}^{2}}}})(\sqrt{{{{a}^{2}}}}-\sqrt{{{{b}^{2}}}})$.

Note: We cannot factorise the sum of two squares.

### Activity 2.1: Find factors for the difference of two squares

Time required: 10 minutes

What you need:

• pen and paper

What to do:

1. Complete the following table and see if you can find a pattern (rule), which you can use to predict the answers to the first column’s calculations without needing to square the numbers.
 $\scriptsize {{3}^{2}}-{{2}^{2}}=$ $\scriptsize 3+2=$ $\scriptsize 3-2=$ $\scriptsize (3+2)(3-2)=$ $\scriptsize {{4}^{2}}-{{3}^{2}}=$ $\scriptsize 4+3=$ $\scriptsize 4-3=$ $\scriptsize (4+3)(4-3)=$ $\scriptsize {{5}^{2}}-{{4}^{2}}=$ $\scriptsize 5+4=$ $\scriptsize 5-4=$ $\scriptsize (5+4)(5-4)=$ $\scriptsize {{6}^{2}}-{{5}^{2}}=$
2. Now predict the answers to each of the following without squaring. Check your answers. Does the rule that you discovered in Question 1 also hold for the following cases?
1. $\scriptsize {{16}^{2}}-{{12}^{2}}$
2. $\scriptsize {{27}^{2}}-{{18}^{2}}$
3. Write the rule you discovered in a formula using $\scriptsize {{a}^{2}}-{{b}^{2}}$

What did you find?

1. .
 $\scriptsize {{3}^{2}}-{{2}^{2}}=5$ $\scriptsize 3+2=5$ $\scriptsize 3-2=1$ $\scriptsize (3+2)(3-2)=5\times 1=5$ $\scriptsize {{4}^{2}}-{{3}^{2}}=7$ $\scriptsize 4+3=7$ $\scriptsize 4-3=1$ $\scriptsize (4+3)(4-3)=7\times 1=7$ $\scriptsize {{5}^{2}}-{{4}^{2}}=9$ $\scriptsize 5+4=9$ $\scriptsize 5-4=1$ $\scriptsize (5+4)(5-4)=9\times 1=9$ $\scriptsize {{6}^{2}}-{{3}^{2}}=27$ $\scriptsize 6+3=9$ $\scriptsize 6-3=3$ $\scriptsize (6+3)(6-3)=9\times 3=27$

If you multiply the sum of the square roots by the difference of the square roots you end up with the same answer as the difference of the squared numbers in column 1. The sum and difference of the squares are the factors of the expressions in column 1.

2. .
1. $\scriptsize {{16}^{2}}-{{12}^{2}}=(16+12)(16-12)=(28)(4)=112$
2. $\scriptsize {{27}^{2}}-{{18}^{2}}=(27+18)(27-18)=45\times 9=405$
Yes, the rule is the same in the above cases too.
3. A difference of two squares can be rewritten as two factors containing the same terms but opposite signs. Perfect squares $\scriptsize {{a}^{2}}-{{b}^{2}}$ have factors of $\scriptsize (a+b)(a-b)$.

We can apply the rule for factorising a difference of squares to algebraic expressions too.

$\scriptsize 16{{x}^{2}}-25$ is made up of two perfect squares because $\scriptsize 16{{x}^{2}}={{(4x)}^{2}}$ and $\scriptsize 25={{(5)}^{2}}$ and the terms are separated by a minus sign. So we can factorise and rewrite $\scriptsize 16{{x}^{2}}-25$ as $\scriptsize (4x+5)(4x-5)$.

### Exercise 2.3

Factorise:

1. $\scriptsize 1-{{x}^{2}}$
2. $\scriptsize 8{{x}^{2}}-18{{y}^{2}}$
3. $\scriptsize -36+{{t}^{2}}$
4. $\scriptsize 2a({{a}^{2}}-9)-7({{a}^{2}}-9)$

The full solutions are at the end of the unit.

### Note

When you have access to an internet connection watch the video called “Factor a Sum or Difference of Cubes” which explains how to factorise the sum and difference of cubes.

## Factorising a trinomial

A quadratic expression is any expression where the variable has a highest power (or degree) of two. $\scriptsize a{{x}^{2}}+bx,\text{ }{{x}^{2}},\text{ }{{a}^{2}}-{{b}^{2}}$ and $\scriptsize a{{x}^{2}}+bx+c$ are all examples of quadratic expressions. We use the expression $\scriptsize a{{x}^{2}}+bx+c$ so often that it has a special name. $\scriptsize a{{x}^{2}}+bx+c$ is called a quadratic trinomial in standard form. Remember that a trinomial has three terms. Factorising using grouping or a difference of two squares was relatively simple. However, quadratic trinomials require a little more work to arrive at a general method.

### Activity 2.2: Find factors of a quadratic trinomial

Time required: 10 minutes

What you need:

• pen and paper

What to do:

1. Complete the table by finding the product.
 Product First term in the product Two middle terms of product Sum of the two middle terms Product of the constant terms Factors of the constant term $\scriptsize (x+2)(x+3)$ $\scriptsize (x+1)(x+6)$ $\scriptsize (x-2)(x-3)$ $\scriptsize (x-1)(x-6)$
2. Which terms do you multiply together to find the $\scriptsize {{x}^{2}}$ in the product?
3. How do you find the co-efficient of the $\scriptsize x$ term in the product?
4. Which terms do you multiply to find the constant term in the product?
5. What are the factors of $\scriptsize {{x}^{2}}+5x+6$?
6. What are the factors of $\scriptsize {{x}^{2}}-5x+6$?
7. What are the factors of $\scriptsize {{x}^{2}}+7x+6$?
8. What are the factors of $\scriptsize {{x}^{2}}-7x+6$?

What did you find?

1. .
 Product First term in the product Two middle terms of product Sum of the two middle terms Product of the constant terms Factors of the constant term $\scriptsize (x+2)(x+3)$ $\scriptsize {{x}^{2}}+5x+6$ $\scriptsize {{x}^{2}}$ $\scriptsize 2x\text{ and }3x$ $\scriptsize 5x$ $\scriptsize 6$ $\scriptsize 1\text{ and }6$, $\scriptsize 2\text{ and }3$, $\scriptsize -1\text{ and }-6$, $\scriptsize -2\text{ and }-3$ $\scriptsize (x+1)(x+6)$ $\scriptsize {{x}^{2}}+7x+6$ $\scriptsize {{x}^{2}}$ $\scriptsize x\text{ and 6}x$ $\scriptsize 7x$ $\scriptsize 6$ $\scriptsize 1\text{ and }6$, $\scriptsize 2\text{ and }3$, $\scriptsize -1\text{ and }-6$, $\scriptsize -2\text{ and }-3$ $\scriptsize (x-2)(x-3)$ $\scriptsize {{x}^{2}}-5x+6$ $\scriptsize {{x}^{2}}$ $\scriptsize -2x\text{ and }-3x$ $\scriptsize -5x$ $\scriptsize 6$ $\scriptsize 1\text{ and }6$, $\scriptsize 2\text{ and }3$, $\scriptsize -1\text{ and }-6$, $\scriptsize -2\text{ and }-3$ $\scriptsize (x-1)(x-6)$ $\scriptsize {{x}^{2}}-7x+6$ $\scriptsize {{x}^{2}}$ $\scriptsize -x\text{ and }-\text{6}x$ $\scriptsize -7x$ $\scriptsize 6$ $\scriptsize 1\text{ and }6$, $\scriptsize 2\text{ and }3$, $\scriptsize -1\text{ and }-6$, $\scriptsize -2\text{ and }-3$

2. We see that the $\scriptsize {{x}^{2}}$ term in the quadratic is the product of the x-terms in each bracket.
3. The middle term is the sum of two middle terms when the brackets are expanded.
4. The $\scriptsize 6$ in the quadratic is the product of the constant terms in the brackets.
5. Factors of $\scriptsize {{x}^{2}}+5x+6$ are $\scriptsize (x+2)(x+3)$.
6. Factors of $\scriptsize {{x}^{2}}+7x+6$ are $\scriptsize (x+1)(x+6)$.
7. Factors of $\scriptsize {{x}^{2}}-5x+6$ are $\scriptsize (x-2)(x-3)$.
8. Factors of $\scriptsize {{x}^{2}}-7x+6$ are $\scriptsize (x-1)(x-6)$.

The first method we will look at to factorise quadratic trinomials involves trial and error. As you saw in Activity 2.2 the signs of the terms in the brackets make a difference to the answers of the product. For example, even though both $\scriptsize 2$ and $\scriptsize 3$ and $\scriptsize -2$ and $\scriptsize -3$ are factors of $\scriptsize 6$, they give us different values for the middle term when the brackets are expanded. Hence $\scriptsize (x-2)(x-3)$ are factors of $\scriptsize {{x}^{2}}-5x+6$ but not factors of $\scriptsize {{x}^{2}}+5x+6$.

### Note

For any quadratic trinomial $\scriptsize a{{x}^{2}}+bx+c$, if $\scriptsize c$ is positive, then the factors of $\scriptsize c$ must either both be positive or both be negative. If $\scriptsize c$ is negative, it means only one of the factors of $\scriptsize c$ is negative and the other one must be positive.

Once you have factorised, you can multiply out your brackets again just to make sure it really works. The following example uses the method of trial and error to factorise a quadratic trinomial.

### Example 2.3

Factorise: $\scriptsize 2x-1+3{{x}^{2}}$

Solution

Step 1: Check that the quadratic is in the standard form of $\scriptsize a{{x}^{2}}+bx+c$. If it is not, then rewrite it in standard form.

$\scriptsize 2x-1+3{{x}^{2}}=3{{x}^{2}}+2x-1$

Step 2: Write down a set of factors for $\scriptsize a$ and $\scriptsize c$.

The possible factors of $\scriptsize 3{{x}^{2}}$ are $\scriptsize 1x$ and $\scriptsize 3x$.

The possible factors of $\scriptsize -1$ are $\scriptsize 1$ and $\scriptsize -1$.

Step 3: Set up the brackets with terms for $\scriptsize a$ and $\scriptsize c$ arranged vertically to see which factors would add up to the middle term by cross multiplying. The factors are read horizontally.

Let’s try factors of $\scriptsize (3x+1)$ and $\scriptsize (x-1)$. By cross multiplying we are working out what the middle term will be when the brackets are expanded.

We see that the middle term would work out to be $\scriptsize -2x$ instead of $\scriptsize 2x$. So these are not the correct factors of $\scriptsize 3{{x}^{2}}+2x-1$.

Let’s try $\scriptsize (3x-1)$ and $\scriptsize (x+1)$.

We see that the factors of $\scriptsize (3x-1)$ and $\scriptsize (x+1)$ will give us a middle term of $\scriptsize 2x$ when expanded and a constant term of $\scriptsize -1$. So these are the correct factors.

Step 4: Write the final answer.

$\scriptsize 3{{x}^{2}}+2x-1=(3x-1)(x+1)$.

### Exercise 2.4

Factorise by trial and error:

1. $\scriptsize {{x}^{2}}+7x-8$
2. $\scriptsize 2{{x}^{2}}+4x-6$
3. $\scriptsize 2{{y}^{2}}+5y-3$

The full solutions are at the end of the unit.

### Factorising quadratic trinomials by grouping

The method of factorising a quadratic trinomial by trial and error can take a while to master. Now, we will look at an alternative method to factorise trinomials. You may choose whichever method you find the easiest to use to factorise quadratic trinomials.

### Example 2.4

Factorise: $\scriptsize 2{{x}^{2}}+x-6$

Solution

Step 1: Write down the values of $\scriptsize a,b$ and $\scriptsize c$.

$\scriptsize a=2,\text{ }b=1,\text{ }c=-6$

Step 2: Determine $\scriptsize a\cdot c$ and list the factors of $\scriptsize a\cdot c$. In this case $\scriptsize ac=-12$.

 Factors of $\scriptsize -12$ $\scriptsize 1,-12$ $\scriptsize -12,1$ $\scriptsize 2,-6$ $\scriptsize -6,2$ $\scriptsize 3,-4$ $\scriptsize -3,4$

Step 3: Find $\scriptsize p,q$ a pair of factors of $\scriptsize ac$ with a sum of $\scriptsize b$.

 Factors of $\scriptsize -12$ Sum of factors $\scriptsize 1,-12$ $\scriptsize 1-12=-11$ $\scriptsize -1,12$ $\scriptsize +11$ $\scriptsize 2,-6$ $\scriptsize -4$ $\scriptsize -2,6$ $\scriptsize +4$ $\scriptsize 3,-4$ $\scriptsize -1$ $\scriptsize -3,4$ $\scriptsize +1$

For $\scriptsize b=1$, $\scriptsize p=-3$ and $\scriptsize q=4$.

Step 4: Rewrite the original expression as $\scriptsize a{{x}^{2}}+px+qx+c$ and factorise by grouping in pairs.

$\scriptsize \begin{array}{l}2{{x}^{2}}-3x+4x-6=(2{{x}^{2}}-3x)+(4x-6)\\=x(2x-3)+2(2x-3)\\=(2x-3)(x+2)\end{array}$

Step 5: Write the final answer.

$\scriptsize 2{{x}^{2}}+x-6=(2x-3)(x+2)$

Check your understanding of this method by trying the exercise below.

### Exercise 2.5

Factorise the following by using the method of grouping:

1. $\scriptsize {{x}^{2}}-7x+6$
2. $\scriptsize 2{{x}^{2}}+9x+9$
3. $\scriptsize 5{{x}^{2}}+7x-6$

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

• How to find the highest common factor of a polynomial.
• How to find the factors of a polynomial by grouping in pairs.
• How to factorise a difference of two squares.
• How to factorise a quadratic trinomial in two different ways.

# Assessment

#### Suggested time to complete: 30 minutes

1. Factorise the following fully:
1. $\scriptsize {{q}^{2}}y-2{{q}^{3}}y$
2. $\scriptsize 6{{x}^{2}}yz+78{{x}^{2}}{{y}^{2}}z-36{{x}^{2}}y{{z}^{2}}$
3. $\scriptsize 3x-6y-3xy+6{{y}^{2}}$
4. $\scriptsize {{x}^{3}}-3x-4{{x}^{2}}+12$
5. $\scriptsize {{a}^{4}}-1$
6. $\scriptsize {{(x+y)}^{2}}-16{{y}^{2}}$
7. $\scriptsize 2{{y}^{2}}-y-21$
8. $\scriptsize 6{{x}^{2}}+xy-12{{y}^{2}}$
9. $\scriptsize 2{{m}^{2}}-40m+200$
2. A school is creating a pitch in the centre of the school grounds. The school ground is a square with side length $\scriptsize 20\text{ m}$ as shown below. The pitch will be a square plot with an area of $\scriptsize {{x}^{2}}-6x+9\text{ }{{\text{m}}^{2}}$.

Find the length of the pitch by factorising.

The full solutions are at the end of the unit.

# Unit 2: Solutions

### Exercise 2.1

1. .
$\scriptsize \begin{array}{l}-a{{x}^{2}}+bx+{{x}^{3}}&=x(-ax+b+{{x}^{2}})\text{ We can write with descending powers of }x.\\&=x({{x}^{2}}-ax+b)\end{array}$
2. $\scriptsize a-b=(a-b)$
3. $\scriptsize -a+b=-1(a-b)$
$\scriptsize \begin{array}{l}\left( {a-b} \right)x-a+b&=\left( {a-b} \right)x-1(a-b)\\&=(a-b)(x-1)\end{array}$
4. .
$\scriptsize \begin{array}{l}-12a+24{{a}^{2}}-3&=3(-4a+8{{a}^{2}}-1)\\&=3(8{{a}^{2}}-4a-1)\end{array}$

Back to Exercise 2.1

### Exercise 2.2

1. .
$\scriptsize \begin{array}{l}{{x}^{2}}-6x+5x-30&=x(x-6)+5(x-6)\\&=(x-6)(x+5)\end{array}$
2. .
$\scriptsize \begin{array}{l}6t-15s+2yt-5ys&=3(2t-5s)+y(2t-5s)\\&=(2t-5s)(3+y)\end{array}$
3. .
$\scriptsize \begin{array}{l}ab-{{a}^{2}}-a+b&=a(b-a)+1(b-a)\\&=(b-a)(a+1)\end{array}$
.
OR
.
$\scriptsize \begin{array}{l}ab-{{a}^{2}}-a+b&=-{{a}^{2}}-a+ab+b\\&=-a(a+1)+b(a+1)\\&=(a+1)(b-a)\end{array}$
4. Here we need to group three sets of binomial together to find a common bracket.
$\scriptsize \begin{array}{l}3ax+bx-3ay-by-9a-3b&=(3ax+bx)+(-3ay-by)+(-9a-3b)\\&=x(3a+b)-y(3a+b)-3(3a+b)\\&=(a+b)(x-y-3)\end{array}$

Back to Exercise 2.2

### Exercise 2.3

1. $\scriptsize 1-{{x}^{2}}=(1-x)(1+x)$
2. Remember to always check for the HCF first when factorising.
$\scriptsize \begin{array}{l}8{{x}^{2}}-18{{y}^{2}}&=2(4{{x}^{2}}-9{{y}^{2}})\\&=2(2x-3y)(2x+3y)\end{array}$
3. We can rearrange $\scriptsize -36+{{t}^{2}}$ as $\scriptsize {{t}^{2}}-36$.
$\scriptsize {{t}^{2}}-36=(t-6)(t+6)$
4. Take out the common factor$\scriptsize ({{a}^{2}}-9)$.
$\scriptsize ({{a}^{2}}-9)(2a-7)$
You must go one step further and fully factorise the expression. Factorise the difference of squares in the first bracket.
$\scriptsize ({{a}^{2}}-9)(2a-7)=(a+3)(a-3)(2a-7)$

Back to Exercise 2.3

### Exercise 2.4

1. $\scriptsize {{x}^{2}}+7x-8=(x+8)(x-1)$
2. Take out the HCF before factorising the trinomial.
$\scriptsize \begin{array}{l}2{{x}^{2}}+4x-6&=2({{x}^{2}}+2x-3)\\&=2(x+3)(x-1)\end{array}$
3. $\scriptsize 2{{y}^{2}}+5y-3=(2y-1)(y+3)$

Back to Exercise 2.4

### Exercise 2.5

1. .
$\scriptsize \begin{array}{l}a=1,b=-7,c=6\\ac=6\end{array}$
 Factors of $\scriptsize 6$ Sum of factors $\scriptsize 1,6$ $\scriptsize 7$ $\scriptsize -1,-6$ $\scriptsize -7$ $\scriptsize 2,3$ $\scriptsize 5$ $\scriptsize -2,-3$ $\scriptsize -5$

$\scriptsize p=-1,q=-6$
$\scriptsize \begin{array}{l}{{x}^{2}}-7x+6&={{x}^{2}}-x-6x+6\\&=x(x-1)-6(x-1)\\&=(x-1)(x-6)\end{array}$

2. .
$\scriptsize \begin{array}{l}2{{x}^{2}}+9x+9&=2{{x}^{2}}+6x+3x+9\\&=2x(x+3)+3(x+3)\\&=(2x+3)(x+3)\end{array}$
3. .
$\scriptsize \begin{array}{l}5{{x}^{2}}+7x-6&=5{{x}^{2}}-3x+10x-6\\&=x(5x-3)+2(5x-3)\\&=(5x-3)(x+2)\end{array}$

Back to Exercise 2.5

### Unit 2: Assessment

1. .
1. $\scriptsize {{q}^{2}}y-2{{q}^{3}}y={{q}^{2}}y(1-2q)$
2. $\scriptsize 6{{x}^{2}}yz+78{{x}^{2}}{{y}^{2}}z-36{{x}^{2}}y{{z}^{2}}=6{{x}^{2}}yz(1+13y-6z)$
3. .
$\scriptsize \begin{array}{l}3x-6y-3xy+6{{y}^{2}}&=3(x-xy-2y+2{{y}^{2}})\\&=3\left[ {x(1-y)-2y(1-y)} \right]\\&=3(1-y)(x-2y)\end{array}$
4. .
$\scriptsize \begin{array}{l}{{x}^{3}}-3x-4{{x}^{2}}+12&=x({{x}^{2}}-3)-4({{x}^{2}}-3)\\&=({{x}^{2}}-3)(x-4)\end{array}$
5. .
$\scriptsize \begin{array}{l}{{a}^{4}}-1&=({{a}^{2}}-1)({{a}^{2}}+1)\\&=(a-1)(a+1)({{a}^{2}}+1)\end{array}$
6. .
$\scriptsize \begin{array}{l}{{(x+y)}^{2}}-16{{y}^{2}}&=\left( {\sqrt{{{{{(x+y)}}^{2}}}}-\sqrt{{16{{y}^{2}}}}} \right)\left( {\sqrt{{{{{(x+y)}}^{2}}}}+\sqrt{{16{{y}^{2}}}}} \right)\\&=(x+y-4y)(x+y+4y)\\&=(x-3y)(x+5y)\end{array}$
7. $\scriptsize 2{{y}^{2}}-y-21=(2y-7)(y+3)$
8. $\scriptsize 6{{x}^{2}}+xy-12{{y}^{2}}=(2x+3y)(3x-4y)$
9. .
$\scriptsize \begin{array}{l}2{{m}^{2}}-40m+200&=2({{m}^{2}}-20m+100)\\&=2(m-10)(m-10)\end{array}$
2. The pitch is a square therefore its lengths are equal.
$\scriptsize \begin{array}{l}\text{A=}l\times l\\{{l}^{2}}&={{x}^{2}}-6x+9\\&=(x-3)(x-3)\\&={{(x-3)}^{2}}\\\therefore l&=x-3\end{array}$

Back to Unit 2: Assessment