## 3 Algebra

### Big ideas

Simplifying, expanding, and factorising are interconnected processes that provide different ways to understand and manipulate algebraic expressions.

### Curriculum Outcomes

VC2M9A02

simplify algebraic expressions, apply the distributive law to expand algebraic expressions including binomial products, and factorise monic quadratic expressions

VC2M9A06

use mathematical modelling to solve applied problems involving change, including financial contexts involving simple interest; formulate problems, choosing to use either linear or quadratic functions or other simple variations; interpret solutions in terms of the context; evaluate the model and report methods and findings

## 3.01 Algebra review

After this lesson, you will be able to…

• review adding and subtracting algebraic terms, focusing on collecting like terms.

• review multiplying and dividing algebraic terms, highlighting how indices may change.

• apply the order of operations in the context of simplifying algebraic expressions.

### Algebra review

When working with algebraic terms, it can be helpful to remember some vocabulary.

3xy, and 2xy

Previously we have seen how to:

• Add and subtract algebraic terms

• When adding and subtracting algebraic terms, we collect like terms by adding or subtracting their coeﬃcients, for example 2a + 3a = 5a

• Multiply algebraic terms

• Divide algebraic terms

• When multiplying and dividing algebraic terms, terms do not have to be like terms. We multiply or divide the coeﬃcients and may use index laws, for example 8a 4 ÷ 2a = 4a 3

When simplifying algebraic expressions we may need to use the order of operations.

Example 1

Simplify: a 9a + 3b + 2a 4b

Create a strategy

There are two pairs of like terms, 9a and 2a, and 3b and 4b. Like terms can be collected together.

Apply the idea

Group like terms = 11a – b

Combine like terms

Reﬂect and check

We can identify two types of terms, terms in a and terms in b. Our solution has two terms, one in a and one in b

Create a strategy

wy 2 = w × y × y and w 2 y = w × w × y, so terms in wy 2 and terms in w 2 y are unlike.

Here:

• 5wy 2 and wy 2 are like

• 7w 2 y and 2w 2 y are like

• wy does not have a like term, so it will not change

Apply the idea 5

Reﬂect and check

The original statement had three different types of terms: wy 2 , w 2 y, and wy

Our solution has three terms, one in wy2, one in w 2 y, and one in wy

### Example 2

Simplify: a 5r × ( 7)

Create a strategy

Group like terms

Combine like terms

Multiply the numeric components and then simplify the algebraic components.

Apply the idea

5r × ( 7) = 5 × ( 7) × r Group the numeric and algebraic components together

= 35 × r Multiply the numbers

= 35r Simplify

b 3c 2 × 5d3

Create a strategy

Multiply the numeric components and then simplify the algebraic components.

Apply the idea

3c 2 × 5d3 = 3 × 5 × c 2 × d3

= 15 × c 2 × d3

= 15c 2 d3

Reﬂect and check

Group the numeric and algebraic components together

Multiply the numbers

Simplify

Since c 2 and d3 have different bases, we cannot use the index laws to simplify them.

c 8ab2 × 3abc

Create a strategy

Multiply the numeric components and then simplify the algebraic components.

Apply the idea

8ab2 × 3abc = 8 × 3 × a × a × b × b × b × c

= 24 × a 2 × b3 × c

= 24a 2 b3 c

Reﬂect and check

Looking at the original statement:

• The coeﬃcients are 8 and 3 and 8 × 3 = 24

Group the numeric and algebraic components together

Multiply the numbers, and then the variables in turn

Simplify

• The ﬁrst term has an a and the second term has an a, and a × a = a 2

• The ﬁrst term has a b2 and the second term has a b, and b2 × b = b3

• There is only one c

### Example 3

Simplify:

a 30m ÷ 5m

Create a strategy

Remove common factors from the numeric components and then simplify the algebraic components. Often it is easier to work with algebraic division if it is written as a fraction.

Apply the idea

Write as a fraction

Remove the common factor of 5

Simplify the algebraic terms

Reﬂect and check

• Considering the coeﬃcients, 30 ÷ 5 = 6

• Considering the algebraic part, m ÷ m = 1

b

Create a strategy

Remove common factors from the numeric components and then simplify the algebraic components. Often it is easier to work with algebraic division if it is written as a fraction.

Apply the idea

Introduce multiplication signs and bring out highest common factor (HCF) of numbers

Remove the common factor, 5

Remove the common factor, w

Remove the common factor, x

Remove the unnecessary multiplication signs

Reﬂect and check

We can think about working on each factor separately: , then , and then

c

Create a strategy

Remove common factors from the numeric components and then simplify the algebraic components. Often it is easier to work with algebraic division if it is written as a fraction.

Apply the idea

Reintroduce the multiplication signs

Remove the common factors of 3, a and b

Group the numeric and algebraic components together

Simplify

Reﬂect and check

We could use the approach we use with index laws:

Use multiplication law in numerator

Divide coeﬃcients and use division law

Simplify

Simplify:

Create a strategy

Use the order of operations and what we know about algebraic terms.

Apply the idea

Perform the multiplication = 38a 2

Add the like terms

Reﬂect and check

The multiplication produces like terms that can be collected together, otherwise, we would not be able to simplify further.

b (7a 2a) × 3ab

Create a strategy

Use the order of operations and what we know about algebraic terms.

*Example 4*

Apply the idea

(7a 2a) × 3ab = 5a × 3ab

Collect the like terms in the brackets = 15a 2 b

Perform the multiplication

c

Create a strategy

Use the order of operations and what we know about algebraic terms.

Apply the idea

Collect the like terms in the numerator and perform the multiplication in the denominator

Divide out the common factor of 15

Introduce multiplication signs

Divide out the common x and simplify.

Reﬂect and check

We could also use the division and negative index laws:

Collect the like terms in the numerator and perform the multiplication in the denominator

Divide out the common factor of 15

Division index law

Negative index law

### Idea summary

Simplifying algebraic expressions makes use of:

• Adding and subtracting algebraic terms, which may use like terms

• Multiplying and dividing algebraic terms, which may use index laws

• Applying the order of operations

## Practice questions

### Choose your question path

Explorer

1, 2, 3a-j, 4a-j, 5a-j, 6a-c, gh, 7a-g, 8, 9, 13–15, 17

### Understanding

1 For the expression:

a How many terms are there?

c What is the coeﬃcient of x?

e What is the constant term?

Adventurer Trailblazer

1, 2, 3acegikmoqsuw, 4acegikmo, 5acegikmo, 6acegik, 7acegikmo, 9, 10, 11, 14–16, 18

3x 2 xy + 5x 3y 9 + 5y

2 Are these statements true or false?

a 9xyz = 9zxy

b 4 × 5 × a × b = 9ab

c p 5 = p × p × p × p × p

1, 2, 3m-x, 4h-p, 5h-p, 6dg, i-l, 7dfik, m-p, 10, 11, 12, 15–19

b What are the variables?

d What is the coeﬃcient of xy?

f Identify any like terms.

d 2x 2 × 5xy cannot be simpliﬁed because they are not like terms.

e c 8 ÷ c 2 = c 4

f x ÷ 2 =

### Fluency

3 Simplify:

a 7a + 3a

c 6b + 10b

e 7c + 3c c

g 5a + 5b + 2a + b

b 5x + 8x

d 10y 7y

f 3x 4x x

h 6c + 3d + 3c d

i 7x + 4y 2x + y j 5p 10q + 3p 3q

k 2x + 7y + 3x + z l 3x + 7 2y + x

m x 2 + 6x + 5x + 10 n x 2 7x + 3x 21

o x 2 4x + 4x 16 p 4a + 7b + c a + b + 5c

q 5ab + 2bc 2ab + 3cb

s 5x ( 3x) + 7 + ( 2)

u xy + 2x 2 y 7 2xy 2 + 3xy

w x ( 3x) 8x + x 2

r 5a 2 b + 7ab2 + 2a 2 b 3ab2

t 3ab + 9cd ( 2ab) ( cd)

v 7c 2 + 3c + 7 2 c + 7cd

x 5cd + 3c 2 d cd2 + 4c 2 d 2cd

4 Simplify:

a 9 × 5u

c 2 × x × y × 3

b 7r × 4

d 4r × 6s

e 3u 3 × 5v 5 f 12p × 5q 2

g ( 3c) × ( 5d)

i ( 3x) × ( 10y)

k 9a × 3a × ( 2a)

m 5a 3 b2 × 4a 4 b3

h 5e × ( 5f )

j 6x 2 × 2y 2 × 3z 2

l 5a × 2b × a × 2b

n 3w 3 x 2 y × ( 3)wx 2 y 3 × 2xy

o 7c 5 d3 × 3c 2 × 2d5 p 3e 7 f 2 × 5e 6 f 3 × e × f 2

5 Simplify:

a 6m ÷ 36 b 10ab ÷ 50ab

c 15x 6 ÷ 5x 6 d ( 32p7) ÷ 8p 7

e (9cd) ÷ 12de f ( 12fg) ÷ ( 3g2) g h

i j k l

m n o p

6 Simplify:

a b

c d e f g 5u × 8v ÷ 4w

h ( 3a) × ( 5b) ÷ 10a 2

i 11xy × 5z ÷ x 2 y 2 j 25g ÷ 5 × 7f

k (8y ÷ 4x) × (12y ÷ 3x)

7 Simplify:

a (7c 3c) × 2c

c (12a 4a) ÷ (8 × 4a2)

l (9a × 4b) ÷ (5 × 2c)

b (19t + 5t) ÷ 4u

d 5f 3 × 4f + 7f

e 2 (4x 2 + x × 2x) ÷ 16x f 5 (3a × 2b) + 4a × ( 3b)

g h i j k l m n o 3x 5 × 7y 4 2x 3 y 2 × 3x 2 y 2 p

*Ex*

### Reasoning

8 a Explain why we cannot combine 3a + 4b as 7ab

b Is it ever possible that 3a + 4b = 7ab? Explain your reasoning.

9 Explain why x 2 and x are not like terms.

10 Is there any value of a for which 3a × 2a = 5a2? If so, give an example. If not, explain why.

11

Simplify the expression 7a + 5a × 2 in two different ways:

• Perform the multiplication ﬁrst and simplify.

• Repeat the problem and do the addition ﬁrst.

Explain why an order of operations is needed when simplifying algebraic expressions.

12 Consider the problem

Identify and correct the error in each student’s work:

Lawrence: Marika:

### Problem solving

13 An isosceles triangle has 2 sides of length 3b cm. If the third side has a length of 2c cm, write a simpliﬁed expression for the perimeter of the triangle.

14 A rectangle has a length of (3x + y) cm and a width of (2x 2y) cm. Find the perimeter of the rectangle in terms of x and y

15 Addition pyramids work by the value in a box being equal to the sum of the two boxes below. The example shows how this works with numbers:

Complete the addition pyramids. a b

16 A rectangle has a length 4 times its width. The width of the rectangle is 3a cm. Find an expression for the area of the rectangle.

17 A purse contains x $10 notes, y $5 notes, and z $1 coins. Write an expression for the number of dollars in the purse.

18 Emma bakes x batches of chocolate chip cookies, with y cookies in each batch.

a How many chocolate chip cookies does Emma bake in total?

b If there are an average of y chocolate chips in each cookie, and she bakes the same number of cookies on 3 consecutive days, how many chocolate chips does she use?

19 A piece of paper with dimensions 8.5 centimetres by 11 centimetres will have a square with length x cut out from each corner to form a box.

Write and simplify an expression that can be used to represent the perimeter of the shape formed once the corners have been removed.

### 3.02 Distributive law

After this lesson, you will be able to…

• review and apply the distributive law for both numerical and algebraic expressions.

• use algebra tiles to illustrate the distributive law and create equivalent mathematical statements.

• expand and simplify real-world algebraic expressions using the distributive law.

• apply the distributive law to expand algebraic terms with brackets.

### The distributive law

### Interactive exploration

Explore online to answer the questions

Use the interactive exploration in 3.02 to answer these questions.

1. Show that 3 (x + 2) is equivalent to 3x + 6 using algebra tiles.

2. Create your own distribution problem and have your partner create the expression using algebra tiles to show the simplified expansion.

3. Find two different expressions that are equivalent to 4x 8 using algebra tiles.

4. Using the pattern from the applet, what do you think x(x 6) is equivalent to?

We use the distributive law to expand algebraic terms with brackets. The term on the outside of the brackets is multiplied by each term inside the brackets.

A(B + C) = AB + AC

When multiplying variables, we may need to use the multiplication index law: a m × a n = a m+ n

### Example 1

Expand:

a 6s (9t + 8)

Create a strategy

Use the distributive law A (B + C) = AB + AC to expand the expression.

Apply the idea

6s (9t + 8) = 6s × 9t + 6s × 8

Expand the expression = 54st + 48s Simplify

Reﬂect and check

We can use arcs to help us remember to multiply each term inside the brackets by the factor outside of the brackets.

b 4u 7 (4u 7 + 2u6)

Create a strategy

Use the distributive law A (B + C) = AB + AC to expand the expression. We will also use the multiplication index law a m × a n = a m + n

Apply the idea

4

Reﬂect and check

Expand the expression

Simplify

Remember that we multiply the coeﬃcients and add the indices for each product.

### Example 2

Expand and simplify: x (x 6) 8 (x + 7).

Create a strategy

Use the distributive law A (B + C) = AB + AC to expand the expression.

Apply the idea

x (x 6) 8 (x + 7) = x 2 6x 8x – 56

= x 2 14x – 56

Expand both brackets

Simplify

Reflect and check

When the factor outside of the brackets is negative, we need to multiply each term in the brackets by the negative.

### Idea summary

Expand means to write an algebraic expression without brackets.

We can use the distributive law to expand an algebraic expression with brackets:

A (B + C) = AB + AC

and if the second term in the brackets is negative:

A (B C) = AB AC

### Practice questions

Choose your question path

Explorer

1, 2, 3a, 4, 5acegikm, 6a-c, e, 7ac, 8b-d, ghkl, 9, 11, 13–15

Adventurer

1–3, 4bc, 5cdghklop, 6abef, 7bc, 8bdfhjln, 9, 11–15

Trailblazer

1–3, 4cd, 5k-p, 6e-h, 7cd, 8adeijmn, 9, 10, 13–17

Understanding

1 Ludivina states that (a3) (a3) = a9. Dortha states that (a3) (a3) = a6. Who is correct?

2 Explain why 2(a 3) is the same as (a 3) × 2.

3 Insert (=) or (≠) in each statement.

a 3a(2a + b) ⬚ 3a × 2a + 3a × b

b 6m 2 v(3 2m 2 + v) ⬚ 6

c 4ab(3b + 2

### Fluency

4 Complete the statements:

a 3 (2x + 4) = ⬚x + 12

b 2 ( 4s + 5) = ⬚s + ⬚

c 5 (⬚p + ⬚) = 15p 30 d ⬚ (3

⬚) = 6g + 14

5 Expand:

a a(b + 7) b m(11n + 19p)

c 4 (m 2 + 10) d m(m + 7)

e y(1 + y) f m(m 3) g w(w + 1) h y(4y 10)

i m(8 m) j 5t(t + 14) k 4t(4t + 3) l 7s 3 (8t5 9)

m 2b(2b + 16) n 5a(3a 4) o p 6wy( y + w)

6 Expand:

a 7y( y 2 + 4) b (5y 6) ( 3y2)

c 6a 4 (3a 6 + 6a4) d 3u 4 (8u + 10u3)

e 3x 2 (5x 2x4) f g h

7 Expand:

a 2f ( f 8g 7h) b 6a(a 5b + 3c)

8 Expand and simplify:

c 8 ( y + 6) + 3 ( y + 9)

(5x + 9y 2) (4x) d 2a 2 (a 2 + 4a 7)

d 5 (9x 6y) 3(8y + 2x)

e 2 (4 (x 3) + 5) f 3 (2 (x 4) + 5)

g 3c(9c 2 + 2c + 7) h h(h + 5) + 9 (h + 9)

i 4y 2 9y + 6y( y 8)

k 4c(8c 2 3c 5)

m 8y 2 7y + 9y( y 5)

### Reasoning

j 6u(5u + 8v) 7u(4v + 2u)

l x(x + 4) + 2 (x + 5)

n 9x(6x + 8y) 7x(2y + 4x)

9 Four students gave incorrect answers when expanding 6(4x 25).

a Explain the errors made by each student.

i Leon: 24x 25

iii Claire: 24x 150

b What is the correct answer?

10 There are three consecutive integers.

ii Jill: 2x 25

iv Beryl: 24x 150

a If x is the ﬁrst and smallest of the integers, write an expression for the: i Second integer ii Third integer

b The sum of the ﬁrst, and two times the second is 15 more than two times the third. Write an equation for this.

c What are the three integers?

d Are there any other sets of three consecutive integers that satisfy the equation in part (b)? Explain how you know.

11 The perimeter of a circle is 3π + 2.

Find and expand an expression that represents the perimeter of 5 of these circles.

### Problem-solving

12 Complete the equation so that x = 7 is the only solution: 2 (x 12) + 9 = 3x ⬚

13 a Write two or more expressions for the area:

b By expanding each of the expressions given in part (a), show that they are equivalent and hence come up with a single expression for the area.

c Explain why x cannot be negative.

14 A rectangular garden has a length that is one metre less than twice the width (w).

a Write and expand an expression for the area of the rectangle.

b When a border 2 metres wide is added around the garden, the total area (garden and border) is 2w 2 + 7w + 12. What is the area of the border only?

c The landscaper designing the garden has 100 square metres of pavers to use for the border. Determine the largest garden the landscaper can border before running out of pavers.

15 Expand and simplify 2a(a 3 4a + 2) + 5a

16 The rule for the distributive law is that A(B + C) = AB + AC

a Write A(B + C) if A = x + 1, B = 2x and C = 3.

b What is the expansion for A(B + C)?

17 A prism has dimensions of width = x, length = 2x 1 and height = 6x

a Write an expanded expression for the surface area.

b If materials needed to make the prism dictate that the width could be between 1 cm and 10 cm, ﬁnd the range of the possible surface area.

### 3.03 Binomial expansion

After this lesson, you will be able to…

• apply the distributive law for multiplying binomials, even when variables are involved.

• recognise the pattern for squaring a binomial.

### Binomial expansion

A binomial is an expression with two non-like terms that are being added or subtracted. For example:

• 3x + 2

• x 2 + 9

• 10xy 2 140xyz

## Exploration

Complete the area models for multiplication shown:

+ 9)

Complete the new area models for multiplication:

1. What do the area models have in common?

2. What’s different about the area models?

3. How do you think multiplying binomials can be related to multiplying integers?

Consider the expression ( A + B) (C + D). We can represent it visually using this area model.

This diagram shows that ( A + B) (C + D) = AC + AD + BC + BD. In other words, the area of the whole shape is the sum of the smaller areas.

Let’s look at using this model to expand (x + 3) (x + 5).

Here we can see that: (x + 3) (x + 5) = x 2 + 3x + 5x + 15

We can simplify this to: x 2 + 8x + 15

### Example 1

Expand and simplify (x + 1) (x + 3)

Create a strategy

Use the rule ( A + B) (C + D) = AC + AD + BC + BD.

Apply the idea

(x + 1) (x + 3) = x

= x 2 + 3x + x + 3

= x 2 + 4x + 3

### Idea summary

Apply the rule

Simplify the products

Combine like terms

We can expand the product of two binomial expressions using the rule: ( A + B) (C + D) = AC + AD + BC + BD

### Square of a binomial

Sometimes binomials such as (a + b) (a + b) are written as (a + b)2

If we need to expand binomials like this, we can still follow the same structure.

(a + b)2 = (a + b) (a + b)

= a 2 + ab + ab + b2

= a 2 + 2ab + b2

Write as binomial product

Binomial expansion

Simplify

Notice the middle term. It is 2ab, which is twice the product of a and b.

### Example 2

Expand (x + 9)2

Create a strategy

Use the pattern for squaring a binomial.

Apply the idea

(x + 9)2 = (x + 9) (x + 9) Write as a binomial product

= x 2 + 9x + 9x + 81

= x 2 + 18x + 81

Binomial expansion

Simplify

### Example 3

Expand (5x + 2y)2

Create a strategy

Use the pattern for squaring a binomial.

Apply the idea

(5x + 2y)2 = (5x + 2y) (5x + 2y)

= 25x 2 + 10xy + 10xy + 4y 2

= 25x 2 + 20xy + 4y 2

### Idea summary

Write as a binomial product

Binomial expansion

Simplify

Squaring a binomial means multiplying a binomial by itself.

(a + b)2 = (a + b) (a + b)

= a 2 + ab + ab + b2

= a 2 + 2ab + b2

### Practice questions

### Choose your question path

Explorer

1–5, 6acegi, 7a-e, 8a-c, f, 9abd, 10ace, 11–13, 15aeg, 16a-c, e, 17ab, 19, 20

Adventurer

1cd, 2–5, 6acegi, 7a-e, 8bc, 9bdf, 10bdf, 11–13, 15bdf, 16bdf, 17bc, 18, 20, 21

Understanding

1 Simplify:

a (8)2 b ( 5)2

c (3x)2 d ( 12y)2

2 Which are binomial products?

a 5 (2x 3)

c 2x(x 3) + 4 (7 x)

b (3x 4) (2x 7)

d (2x + 5)2

3 Are the expressions equivalent to ( y 5) ( y 8)?

a y( y 5) 8 ( y 5)

b 5 ( y 8) ( y 5)

Trailblazer

1cd, 2–5, 6bdfhj, 7bdfg, 8eh, 9ach, 10f-h, 11–14, 15h, 16deh, 17bd, 18, 20, 21

c y( y 5) + y 8 d y( y 8) 5 ( y 8)

4 The area of this ﬁgure is represented by the expression ( y + 3) ( y + 6).

a Find the area of:

i Rectangle A

ii Rectangle B

iii Rectangle C

iv Rectangle D

b Write an equivalent, simpliﬁed expression for the area of the ﬁgure.

5 Is each statement true or false?

a (3x + 5)2 = 9x 2 + 25

b The simpliﬁed product of any two binomials will be a trinomial.

c After completing a binomial expansion, adding products in different orders will always give the same answer.

d The product (3x 8) (3x + 8) will have two terms.

### Fluency

6 Expand and simplify:

a (x + 6) (x 12) b (x + 2) (x + 6) c (v + 11) (v + 10) d (x 4) (x 7)

e (m 5) (m + 6) f (10m + 9) (m 7) g (3w + 5) (7w + 6) h (3y 3) (10y + 9)

i (7y + 9) ( 4y 4) j

7 Expand and simplify:

a (a + 3) (a 3) b (x + 2) (x 2) c ( y + 5) ( y 5) d (2x + 5) (2x 5)

e (4x + 6) (4x 6) f (5x + 2) (5x 2) g (3x + 7) (3x 7) h (3y + 4) (3y 4)

i (9m + 8) (9m 8) j (12x + 1) (12x 1)

8 Expand and simplify:

a (x + 7)2 b (x + 4)2

c (x 3)2 d (x 5)2

e (x + 6)2 f (x 2)2 g h

9 Expand and simplify:

a (3a + 4b)2 b (2a 3)2

c (2x 7y)2 d (6p + 3q)2

e (4m 5n)2 f (x 2y)2 g h

10 Expand:

a (a + 6) (b 12)

c (q 2 7) (t + 9)

b (k 2r) (k + 5)

d (x + 5y) (x 2 + y)

e (3m 2 4p) (m p) f (4g 2 + 3h) (6g + 5h3)

g (3a 3 b2) (7a 6b2)

h (5x 2 y + 4x) (7y 3 12x5)

*Ex 1*

*Ex 4, 5*

*Ex 2*

*Ex 3*

### Reasoning

11 Marvin has a square bedroom. His sister Jane has a room that is 1 m shorter in length than Marvin’s room, but 1 m wider.

a Show that Marvin has the larger bedroom.

b How much bigger is Marvin’s bedroom compared to Jane’s bedroom?

12 Are (x 4)2 and (4 x)2 equivalent expressions? Explain your answer.

13 Rahul attempts to expand the expression (x + 9)2 and gets a result of x 2 + 81.

Is his answer correct? Explain where you think Rahul has gone wrong and identify the correct expansion.

14 Anette decides to use the area model to expand and simplify (10x + 2) (10x + 3). To warm up, she decides to try a simple integer example of (10 + 2) (10 + 3):

She then tried this method on

Find the error in Anette’s answer. Correct her work and solution.

Problem-solving

15 Expand and simplify: a 3 ( y + 3) ( y + 5) b 8 (x 2) (x 6) c (x + 5) (x + 6) d 4 ( y + 1) ( y + 9) e 9 ( y 7) ( y + 6) f 3 ( 4x + 3) (4x 3) g 4 (6x 2) (x + 2) h

16 Complete:

a (t + 8) ⬚ = t2 + 10t + 16 b (t + 6) ⬚ = t2 + 17t + 66

c (t 2) ⬚ = t2 7t + 10 d (t 2) ⬚ = t2 + 4t 12

e (t + 2) (t + 9) = t(t + 9) + ⬚ (⬚) f (v 2) (v 3) = v(v 3) ⬚ (⬚)

g (t 2) (t + 8) = t(t + 8) ⬚ h (t 5) ⬚ = t2 + 7t 60

17 Expand:

a (x + 3)3

c (2x + 3) (x + 5y 4z)

b (2x 4)3

d (t + 3) (2t 1) (3t + 5)

18 A square has a perimeter of 8x + 20. What is its area?

19 Find the area of a rectangle which has a length of (x + 4) cm and a width of (x + 3) cm.

20 What expression, when multiplied by n + 6, gives n 2 + 14n + 48?

21 Consider the rectangular prism shown with its dimensions labelled. Write an expression for the volume, and then expand and fully simplify the result.

(x + 7) m (x + 5) m 4 m

### 3.04 Algebraic factors

After this lesson, you will be able to…

• identify common factors in algebraic expressions.

• rewrite algebraic expressions in their factorised form.

• apply factorization techniques to simplify algebraic expressions.

### Factorise algebraic expressions

Factorising can be thought of as being the reverse process to expanding. Expanding would be:

6(x + 3) = 6x + 18

Factorising would be putting the expanded form into brackets by taking out the highest common factor (HCF):

6x + 18 = 6(x + 3)

We can view this as the opposite of the distribution law:

AB + AC = A(B + C)

A is the highest common factor, which is taken out. This extends to three or more terms:

AB + AC + AD = A(B + C + D)

In any algebraic term, such as 12ab or 9ac, there can be numeric and algebraic components. When factorising we need to ﬁnd the highest common numeric factor and the highest common algebraic factor.

In the case of 12ab and 9ac:

• the highest common numeric factor is 3

• the highest common algebraic factor is a

We can see this when we write the terms out in an expanded way:

12ab = 2 × 2 × 3 × a × b

9ac = 3 × 3 × a × c

By identifying all the factors that are common to both terms we can see that the highest common factor of 12ab and 9ac is 3a.

### Exploration

The expanded form for 5x

The expanded form for 2y 3

1. How many y’s are common in both expressions?

2. How many z’s are common in both expressions?

3. What is 5x 3 y 4 z + 2y 3 z5 in factorised form?

4. What can we do to check our answer?

The expression 5x 3 y 4 z + 2y 3 z 5 = y 3 z(5x 3 y + 2z4) when factorised. We can always expand our answer to check if we have factorised correctly.

### Example 1

Find the highest common factor for the following pairs of terms: a 10xy and 15yz

Create a strategy

Break down the terms into factors.

Apply the idea

10xy = 2 × 5 × x × y Factorise 10xy 15yz = 3 × 5 × y × z Factorise 15yz

HCF = 5 × y Select the common factors = 5y Simplify

b 6a 2 b and 21ab2

Create a strategy

Break down the terms into factors.

Apply the idea

6a 2 b = 2 × 3 × a × a × b Factorise 6a 2 b 21ab2 = 3 × 7 × a × b × b Factorise 21ab2

HCF = 3 × a × b Select the common factors = 3ab Simplify

### Example 2

Factorise 2s 10

Create a strategy

Check all the terms for common numeric and algebraic factors. Factorise using AB + AC = A (B + C).

Apply the idea

There are no common algebraic factors in 2s and 10, but there is a common numeric factor, 2 2s 10 = 2 × s + ( 2) × 5

= 2 (s + 5)

Reﬂect and check

Write the terms as products of 2

Factorise

It is important to be careful about factorising a negative because the sign of each term will change. If the leading term in an expression is negative, then we should factorise the negative out.

### Example 3

Factorise 42x x 2

Create a strategy

Check all the terms for common numeric and algebraic factors. Factorise using AB + AC = A (B + C).

Apply the idea

There are no common numeric factors in 42x and x 2 but there is a common algebraic factor, x 42x x 2 = x × 42 x × x

= x (42 x)

Reﬂect and check

Write the terms as products of x

Factorise

Let’s check our solution of x(42 x), by expanding it.

x(42 x) = x × 42 x × x

x(42 x) = 42x x 2

This is the same as the original expression, so we know that we have factorised this correctly.

## Example 4

Factorise 2ab2 + 16ab

Create a strategy

Check all the terms for common numeric and algebraic factors. Factorise using AB + AC = A (B + C).

Apply the idea

2ab2 and 16ab have a common numeric factor of 2 and a common algebraic factor of ab.

2ab2 + 16ab = 2ab × b + 2ab × 8

Write the terms as products of 2ab = 2ab (b + 8)

Factorise

### Idea summary

We can factorise expressions by writing the highest common factor of the terms outside of the brackets:

AB + AC = A(B + C)

Factorising is the reverse of expanding.

### Practice questions

### Choose your question path

Explorer

1aceg, 2, 3b-d, 4ac, 5, 6a-e, 7aceg, 8aceg, 9ac, 10, 11, 13–15

### Understanding

Adventurer

1bdfh, 2, 3bc, 4ace, 5bdf, 6acegi, 7aceg, 8f, h-j, 9df, 10, 11, 13, 14, 16, 18

1 Find the highest common numeric factor of each pair of terms:

a 8ab and 6cd

c 12j2 k and 16x 2 y

e 24a 2 n 3 y 6 and 8b4 m 2 h6

g 27ax 2 z and 18pr 2 k

Trailblazer

1bdfh, 2, 3ef, 4bdf, 5cef, 6bdfhj, 7bdfh, 8ijln, 9f-h, 10–12, 14, 16–18

b 10x 2 and 25yz

d 18xyz 2 and 6ab2 c

f 36x 7 y and 6a 7 z

h 100z 10 a 8 k5 and 20j10 b8 n 5

2 Expand each expression to see which ones are a factorised form of 12x + 20x 2 :

a 4 × 3x + 5 b 2 (6x 10)

c 4x (5 3x)

d x( 12 + 20x)

3 All of these terms have a common factor of 5b. Write out the terms as a product of their factors, cross out the 5b that is common and determine what is remaining for each.

a 10ab b 15a 2 b

e 5ab2 f 120a 2 b2

### Fluency

c 5ab d 5a 2 b

4 Find the highest common algebraic factor of each term:

a 9x and 8x

c 18b2 n and 7b2 m

e 8my and 12am

5 Complete:

a y 2 + 5y = y (⬚ + ⬚)

c 3y 2 + 6y = ⬚ ( y + 2)

e y 2 2y = ⬚ ( y + 2)

6 Factorise:

b 9xn and 4ax

d 13y 2 an and 17my 2 b

f 9bxm and 5bxa

b 2t2 + 2t = 2t (⬚ + ⬚)

d m 2 + 19m = ⬚ (m 19)

f 8v v 2 = v (⬚ ⬚)

a 4a + 12 b 5a 10 c 15 + 20a d 18a + 16

e 4ab + 6a f 11 km mn g 12s + 10 h 8hj 9gh

i 16jkt 5kt j 35m + 49n

7 Factorise:

a 3a 6 b 4 6a

c 5c 30 d 8 12d

e 4b 5ab f abc 5a 2 g w 2 6w h 45mnq 72mq

8 Factorise:

a 44uv 8u 2 v b 3x x 2

c k2 9k d y 2 + 4y

e 5u 2 15u f 8w 2 + 3w 2 y g 5k2 t + 40k2 t2 h 49p 2 q 28pq 2

i 16a 2 18a 2 b j 30w 2 25w 2 y k 10u 2 v + 9uv 2 l z 2 + 8z 6

m z 2 + 6z 4 n 63r 2 + 28

9 Factorise:

a 4x + 12 + 16yx b 3x + 9 + 12yx

c 2x + 10x 2 y + 8yx d 4x 2 y + 8xy 2 + 12xz 2

e 5ab2 c + 25bc3 + 100abc f 30b2 c + 10abc + 20c 2

g 5a 2 b2 + 2ab 3a 2 b2 4ab h 5x 2 y 2 + 15yz 2 + 25y 2 + 60y

Reasoning

10 A student incorrectly factorised and wrote 5x 2 15x = 5x (x 3).

Explain how to correct the error, and provide the correct answer.

11 Alex and Beth are both asked to factorise 5x + 10y. Alex wrote down 5(x 2y) and Beth wrote down 5(2y x). Who is correct? Explain.

12 Louis was asked to fully factorise 2x 3 + 8x 2 4x and gave as his answer 2(x 3 + 4x 2 2x).

What has he done wrong?

### Problem-solving

13 The rectangle shown has an area of 4x 2 16x square units. What expression describes the length of the rectangle?

14 A property management company has an area of land, 20n + 5n 2 available for parking. When factorised, this represents the length and width of the block of land.

a Factorise the expression to determine the dimensions of the block.

b Given the the length is longer than the width for all n > 1, which term is the length and which is the width?

c The length is further divided into 5 equal sections, and one of these sections is fenced off for storage. What is the length of fencing needed?

15 You are to design a photo collage made of two large square photos with a side length of x and four smaller rectangular photos that have a height of x and a width of 4 units.

a Find an algebraic expression for the area of the rectangle formed if the photos are all placed in a single row. Draw an example of what this arrangement would look like.

b Fully factorise your answer from part (a) and then draw a photo arrangement that would match these dimensions.

16 A rectangle has a width of 2xy cm and a length of 2(x y) cm.

a Write an expression for the area of the rectangle, and expand it.

b Write an expression for the length of a rectangle with the same area if its width is 4xy cm.

c Write a factorised expression for the length of a rectangle with the same area if its width is: i 4 cm ii xy cm

17 A container in the shape of a rectangular prism has a length of (2a b) metres, a width of 6ab metres, and a height of 5ab metres.

a Write an expression for the volume of the prism, and expand it.

b Write a factorised expression for the volume of a container that has twice the volume of the rectangular prism.

c What percentage of the rectangular prism has been ﬁlled if water is poured in to a depth of 2ab m?

d Write an expression for the percentage of the rectangular prism that has been ﬁlled if water is poured in to a depth of 3a metres.

18 Factorise 3t (t 5) + 4 (t 5)

## 3.05 Factorise by grouping

After this lesson, you will be able to…

• apply factorising by grouping in algebraic expressions.

• identify the characteristics of a monic quadratic trinomial and the steps to factorise it.

• apply factorising techniques to solve problems involving algebraic expressions with three or more terms.

### Factorise by grouping

Common factors can be more than 1 term. For example, 5 (a + 7) and b (a + 7) have a common factor of (a + 7).

So,

For expressions with 4 terms, sometimes separating them into 2 groups can help fully factorise the expression.

### Example 1

Factorise: 3x + 6 + 2x 2 + 4x

Create a strategy

Using inspection, determine the 2 groups that will likely end up with a common factor.

Apply the idea

3x + 6 + 2x 2 + 4x = (3x + 6) + (2x 2 + 4x)

= 3 (x + 2) + 2x (x + 2)

= (x + 2) (3 + 2x)

Separate into two groups

Factorise each group

Identify (x + 2) as the common factor and ﬁnish the factorisation

Reﬂect and check

There may be multiple ways to separate the original expression into two groups that share a common factor. For example, we could have factorised the expression this way: 3x + 6 + 2x 2 + 4x = 2x 2 + 3x + 4x + 6

= (2x 2 + 3x) + (4x + 6)

= x (2x + 3) + 2 (2x + 3)

= (2x + 3) (x + 2)

Rearrange the terms

Separate into two groups

Factorise each group

Identify (2x + 3) as the common factor and ﬁnish the factorisation

Notice that this is the same result we got before, the terms are simply rearranged.

### Example 2

Fully factorise: 5(a + b) + v(a + b)

Create a strategy

Factorise using the law, AB + AC = A(B + C).

Apply the idea

Both terms have a common factor of (a + b) so we can factorise this out.

5 (a + b) + v (a + b) = (a + b) (5 + v)

### Example 3

Factorise: 2x + xz 40y 20yz

Create a strategy

Factorise

Group the expression in pairs and factorise using the law, A (C + D) + B (C + D) = ( A + B) (C + D).

Apply the idea

To factorise an expression with 4 terms, we factorise two pairs of terms, then take out the common factor binomial.

2x + xz 40y 20yz = (2x + xz) + ( 40y 20yz)

Group terms with common factors = x (2 + z) 20y (2 + z)

Factorise each pair = (2 + z) (x 20y)

Factorise

### Idea summary

When we factorise expressions with 4 terms, we may need to break the term into 2 groups and factorise each group individually. If there is a common factor between these two groups, we may need to factorise further.

## Monic quadratic trinomials

### Interactive exploration

Explore online to answer the questions

mathspace.co

Use the interactive exploration in 3.05 to explore the connection between the area model of multiplication and the expansion and factorisation of algebraic terms.

A monic quadratic trinomial is in the form x 2 + bx + c. It has three features:

1. The highest power has a coeﬃcient of 1

2. The highest power is 2

3. The trinomial has 3 terms

We factorise monic quadratic trinomials by ﬁnding the factors of the constant, c, that also add up to the coeﬃcient of x, b. Let’s look at an example.

### Example 4

Factorise: x 2 + 6x + 8

Create a strategy

The constant is 8 and the coeﬃcient of x is 6. We need to ﬁnd two factors of 8 that add to 6.

Apply the idea

The factors of 8 that add to 6 are 2 and 4. x 2 + 6x + 8 = x 2 + 4x + 2x + 8 Rewrite the middle term as 6x = 2x + 4x

= (x 2 + 4x) + (2x + 8) Separate into two groups

= x (x + 4) + 2 (x + 4) Factorise each group

= (x + 2) (x + 4) Factorise using the common factor of (x + 4)

### Idea summary

When factorising a monic quadratic trinomial in the form ax 2 + bx + c, we:

1. Find the factors of c ( p and q) that add up to b

2. Rewrite in terms of the factors: ax 2 + px + qx + c, where px + qx = bx

3. Factorise by grouping.

## Practice questions

### Choose your question path

Explorer

1–3, 4acegi, 5bceg, 6abdfhi, 7ace, 9, 10abdf, 11, 12d, 13

### Understanding

Adventurer

1–3, 4acegi, 5aceg, 6acegik, 7bdf, 9, 10bdfh, 11b, 12bd, 13

Trailblazer

1–3, 4adfhj, 5dfjn, 6acegikl, 7bdf, 8, 9, 10cgh, 11b, 12ac, 13, 14

1 For the expression 3(x + 7) + x(x + 7), what is the common factor?

2 Complete this statement: To factorise x 2 + 9x + 18, we need to ﬁnd two numbers whose product is ⬚ and whose sum is ⬚.

3 Given that p < q, ﬁnd the values of p and q in each pair of equations:

a p + q = 19 b p + q = 2 c p + q = 10 pq = 90 pq = 63 pq = 24

### Fluency

4 Factorise:

a y( y + 5) + 7 ( y + 5)

c p( p 3) + 6 ( p 3)

e 5 (r 3) r (r 3)

g 7t (t + u) + 2u (t + u)

i 5y (4w + 3x) z (4w + 3x)

5 Factorise:

a 8x + xz 16y 2yz

c 7xy + wx + 7yz + wz

e x 2 3x + 8x 24

g 50 + 5y + 10x + xy

i x 2 + 5x + 8x + 40

k 2mp + 6 + 3p + 4m

m 2x + 18yz + 12xy + 3z

b a (a 4) 3 (a 4)

d 5 (q + 4) q (q + 4)

f 6r (2r s) rs (2r s)

h x ( y z) w ( y z)

j 8y ( y 4) + 3 (4 y)

b 24 + 3y + 8x + xy

d x 2 + 2x + 5x + 10

f a 3 + 8a 2 + a + 8

h 12xy + wx + 12yz + wz

j x 2 5x + 10x 50

l 5mp + 6 + 2p + 15m

n 4x + 24yz + 32xy + 3z

6 Factorise:

7 Factorise:

### Reasoning

8 What is the largest possible integer value of k that will allow m 2 + km + 24 to be factorised using whole numbers? Justify your answer.

9 Identify and explain the error:

Problem-solving

10 Factorise each expression by ﬁrst factoring out a numeric factor: a 2x 2 + 14x + 20 b 4x 2 + 24x + 32

11 Factorise:

12 Factorise:

3f ( g + h) + (g + h)2

2y(2x 2 + 3z) (2x 2 + 3z) c (2c d) (c + 5d) 3 (d 2c)

3x 2 (x + 4y) 5y(x + 4y)

13 A square has area t2 + 14t + 49.

a Find the side length of the square. b Find the perimeter of the square.

14 Find an expression for the shaded area in the ﬁgure in terms of x. Write your answer in factorised form.

1. Bubaker chooses 2 June, 9 June, and 16 June without showing Tanesha. The expression they provide to Tanesha is 2x 2 + 50x + 50. The instructions they give are to ﬁnd three days in June that fall in a vertical line.

2. Tanesha factorises the expression and gets 2 (x + 9) (x + 16). The factorised expression gives the three dates shown vertically on the calendar.

3. Tanesha chooses 10 July and 18 July without showing Bubaker. The expression they provide to Bubaker is 2 (8x + 4 3x + 5). The instructions they give are to ﬁnd two days in July that fall on a diagonal.

4. Bubaker expands the expression and gets 10x + 18. They are able to identify the dates on the calendar.

5. Tanesha and Bubaker continue taking turns choosing dates, creating patterns and guessing the days.

Find the four days in June 2023 that form a square with four days on the calendar, given the expression a 2 b3 (1 a 7 b7).

### Investigate

1. How did you create some of your expressions to challenge your partner?

2. As the game went on, did you choose to become more creative when creating expressions? Why or why not?

3. Were there any moments in the game where you or your partner misunderstood or made a mistake with an algebraic expression? What was the mistake, and how did you correct it?

### Discussion

1. Consider other ways we could have designed the game or other rules we could have implemented. How might different rules have changed your strategies or the algebraic expressions you used?

2. How could we incorporate other mathematical concepts or operations into this game? For example, could we introduce roots, powers, or other algebraic functions to make the game more challenging?

3. Create a calendar game that involves different mathematical concepts.

## Chapter 3 review

1 Simplify:

2 Simplify:

3 Simplify:

4 Simplify:

5 A construction company charges $2 000 for each job. They also charge labour costs of $50 per hour and $75 per hour for overtime. If x is the amount of regular hours worked and y is the amount of overtime hours worked, write an expression to represent the total cost of the job.

6 Expand:

7 Expand:

8 Expand and simplify:

9 Complete the equation so that x = 11 is the only solution:

10 Expand and simplify:

( f + 9) ( f 9)

11 Expand and simplify:

(h + 3)2

12 Expand:

a (r 8) (s + 7)

b ( p + 13w) ( p 2 + w)

13 Find the area of a triangle with a base length of (2x 5) mm and a height of (x 1) mm.

14 Find the highest common algebraic factor of each term:

a 13t3 and 7t b 24xy 2 and 10y 2 z

15 Factorise:

16 Factorise:

17 The area of a rectangular playground is 18x 2 + 3x. When factorised, this represents the length and width of the playground.

a Factorise the expression to determine the dimensions of the playground.

b Given that the length is longer than the width for all x > 1, which term is the length and which is the width?

c If one of the longer side of the playground is along the community hall, how much fencing is needed to fence the rest of the playground?

18 Factorise:

a x (x + 5) + 7 (x + 5)

19 Factorise:

b 7x (x 3) + 2 (3 x)

a g 2 + 2g + 6g + 12 b 3a + ab 6c 2bc

c 3xy + 10 + 15x + 2y

20 Factorise:

21 Factorise: a 3x 2 + 21x + 30

7a (a 2 3b) 13b (a 2 3b)