Module 2Expanding and Factoring

Yr 10A

Lesson1: Revision of expansions laws and further expansion

Homework Lesson 1: 1 A # 8m-r, 5A # 1, 4a-h,t,w , 5A # 5a,b, d, e; 10c, d, h

Lesson 2: The binomial expansion

Homework lesson 2: 5A # 6a-j , 7 a-f , m 3C#7, 10A#5

Lesson 3: Revision of factorization

Homework Lesson 3: 5B # 1a-d, 3a-e

Lesson 4: Factoring difference between squares

Homework Lesson 4: 5B # 4i-l, 5a-d,I,j , 6 i – l ,

Lesson 5: Factorising expressions with four terms

Homework lesson 5: 5B# 7a – f

Lesson 6: Factorising quadratic trinomials

Homework Lesson 6: 5C # 4m-x, 5a-f, l

Lesson 7: Factorising quadratic trinomials

Homework Lesson 7: 5C # 6a – f

Lesson 8: 10 A - Factorisation of ax2 + bx + c

Homework Lesson 8 (10 A only) : 5D # 2 a-l, 3 m-x, 4 a-f

Lesson 9: Completing the square

Homework Lesson 9: 5E # 1a-f, 4a-l, 6 a-l

Materials

Essential mathematics1.1, 5.1-5.3,5.510A + 5.4

Time Required9 periods

Test date:________________l

Part 1: Expanding

Lesson 1: Basic expansion

Things to Know:

1. To expand an expression, multiply each term inside the brackets by the term outside the brackets.

How it’s done:

To expand 3(2v – 7), multiply both terms inside the brackets by

3.

Be careful when the factor outside the

brackets is negative. Signs will change!

When simplifying an expression after

expanding, remember to add or subtract like terms only.

You try:

1A #8

Expand and simplify 2h(h – 5) + 4(h – 5).

2h(h – 5) + 4(h – 5) = 2h2 – 10h + 4h – 20

= 2h2 – 6h – 20

To check, let h = 6 and substitute:

LHS = 2 × 6(6 – 5) + 4(6 – 5)

= 12 × 1 + 4 × 1

= 16

RHS = 2 × 62 – 6 × 6 – 20

= 2 × 36 – 36 – 20

= 16

Expand by multiplying and removing all brackets.

Then simplify by collecting like terms. Like terms have the same bases raised to the same powers.

To check, substitute numbers for the pronumeral into the expanded form and the original factored form.

You Try:

1A # 9

66826635 3 5

0 3

3

5 3

Expand and simplify 4y(3y + 2) – (y – 5).

4y(3y + 2) – (y – 5) = 12y2 + 8y – y + 5

= 12y2 + 7y + 5

To check, let y = 2 and substitute:

LHS = (4 × 2) × (3 × 2 + 2) – (2 – 5)

= 8 × (6 + 2) – (–3)

= 64 + 3

= 67

RHS = 12 × 22 + 7 × 2 + 5

= 48 + 14 + 5

= 67

When there is more than one pair of brackets, deal with them separately.A minus sign outside a set of brackets changes the sign of every term inside the brackets.When there is no numeral outside brackets, it is understood to be 1.

Simplify where possible by collecting like terms.Check your answer by substituting numbers for the pronumeral to see if you get the same value for the factored form and its expanded form.

You try:

1A #9Expand and simplify:

66826636 4 5

0 4

4

5 4

Expand 3p2(5p4 + 2pq).Expand by multiplying each term in the brackets by the term outside the brackets.Multiply coefficients (numbers).When multiplying powers of the same base, add the indices.

You try:

1A # 9Expand and simplify:

66826637 5 5

0 2

5

5 2

Homework Lesson 1: 1 A # 8 m-r, 5A # 1, 4a-h,t,w , 5a,b

Lesson 2: Expanding 2 brackets

Things to Know:

2. To expand binomial products, multiply each term in the second set of brackets by each term in the first set.

How it’s done:

The rectangle below has length (y + 6) units and width (y + 2)

units.

Total area is the product of (y + 6) and (y + 2).

(y + 6)(y +

2)= y2 + 2y + 6y + 12

= y2 + 8y + 12

Method 1

Split in four separate areas and calculate the area of each part then add them all together.

Method 2You can draw in the arrows shown below or

just picture them mentally. The first line of

working is usually not needed after some

practice.

You try:

5A # 6

5A

5A # 7g

5A # 10

Homework lesson 2: 5A # 6a-j , 7 a-f , m; 10 b, c

Part 2: Factoring

Lesson 3: Extracting a common factor

Things to Know: Describe in words:

1. Factorise expressions with common factors by ‘taking out’ the common factors.a. For example: -5x −20 =-5(x +4) b. 4x2−8x = 4x(x −2)

How it works:

Factorise by taking out common factors.

a -3x −12

b 20a2+30a

c 2(x +1) − a(x +1)

You Try:

5B # 2

How it works:

The highest common factor k2 and 2k is k.

k2 + 2k = k(k + 2)

To check, let k = 1 and substitute:

LHS = k2 + 2k

= 12 + 2 × 1

= 3

RHS = 1(1 + 2)

= 1 × 3

= 3

LHS = RHS

An algebraic expression can have numbers and pronumerals, and their products, as factors of its terms.

First find the highest common factor of all the terms.

Then write each term as a product of this HCF and its remaining factor.

Finally write the HCF in front of the brackets and the remaining factor for each term inside the brackets.

Check your factorisation by substituting numbers for the pronumeral

The highest common factor is q3.

q4 + q3 = q3 × q + q3 × 1

= q3(q + 1)

To check, let q = 1 and then substitute:

LHS = q4 + q3 RH

S = q3(q + 1)

= 14 + 13 = 13(1 + 1)

= 1 + 1 = 1 × 2

= 2 = 2

LHS = RHS

If a pronumeral is a common factor, find the highest power of that pronumeral that will divide into each and every term.

If one of the terms is a factor of all the other terms then it is the highest common factor.

Its co-factor (remaining factor to be written inside the brackets) is 1.

Check your factorisation by substituting numbers for the pronumeral.

8m2 – 4mn + 12m = 4m(2m – n + 3) The highest common factor of the terms in an expression may be a single number or pronumeral or a product of a number and powers of one or more pronumerals.The steps in factorising an expression are:

Consider each term as a product of its factors.

Decide upon the HCF to be placed in front of the brackets.

Write the sum (or difference) for the remaining factors of each term inside the brackets.

You try:

5A # 3Factor:

Homework Lesson 3: 5B # 1a-d, 3a-e

Lesson 4: Factorise a difference of perfect squares

Things to Know: Describe in words:

2. Factorise a difference of perfect squares (DOPS) using

a2− b2= (a + b)(a − b)

3. We use surds when a2 or b2is not a perfect square, such as 1, 4, 9 . . .

For example: 3x2−5 =

How it works:

Factorise 25k2 – 9.

25k2 – 9 = (5k + 3)(5k – 3)

This type of expression is called the difference of two squares.The general rule for factorising such expressions is:a2 – b2 = (a + b)(a – b)

You try:

5B # 4

5B # 6

5B # 6

66828116 1 4 0

3

1 4

3

Homework Lesson 4: 5B # 4i-l, 5a-d,I,j , 6 i – l ,

Lesson 5: Factoring expressions with 4 terms

Things to know:

4. A common factor can consist out of a bracket – do grouping first

Grouping in pairsWhen there are four terms in an expression, you may find pairs of terms containing a common binomial factor.

See the process explained on the right using the expression xy + 30 + 5y + 6x.

Step 1: Look for common factorsxy + 30 + 5y + 6x has no factor common to all four

terms.

Step 2: Rearrange terms into pairs that have a common factorxy + 30 + 5y + 6x = xy + 5y + 6x + 30

Step 3: Factorise the pairs separately

xy + 30 + 5y + 6x

= xy + 5y + 6x + 30

=y(x + 5) + 6(x +

5)

Step 4: Factor out any common term

xy + 30 + 5y + 6x

= xy+ 5y + 6x + 30

=y(x + 5) + 6(x + 5)

= (x + 5)(y + 6)

You may need to group your terms in several different ways before you find a common factor.

You try:

5B #7

Homework lesson 5: 5B# 7a-f

Lesson 6: Factoring monic trinomials

Things to Know: Describe in words:

5. To factorise a monic quadratic, find two numbers that multiply to give the constant term and also add to give the coefficient of the linear term.

How it works:

Factorise y2 + 7y + 6.

y2 + 7y + 6 = (y + 1)(y + 6)

Look for numbers that multiply to give the constant term and add to give the coefficient of the linear term.6 × 1 = 66 + 1 = 7

Factorise n2 – 13n + 30.

n2 – 13n + 30 = (n – 3)(n – 10)

The signs in the expression tell you the signs in the factors.Look for numbers that multiply to give the constant term and add to give the coefficient of the linear term.–10 × –3 = +30–10 + –3 = –13

You try:

5C # 4

5C # 5

66828040 1 4 0

3

1 4

3

66828041 2 4 0

3

2 4

3

Homework Lesson 6: 5C # 4m-x, 5a – f,l

Lesson 7: Factoring perfect squares

Factorise the perfect square 9x2 – 24x + 16.

9x2– 24x + 16 = (3x – 4)2

Check:

2 × 3x × 4 = 24x

Use the middle term (the linear term) to check.

You try:5C # 6

66828118 4 4 0

4

4 4

4

Homework Lesson 7: 5C, 6a – f

Lesson 8: Factoring non-monic trinomials

Steps to follow:1. Find product of ac2. Decompose product to equal middle term3. factor out common factors.4. factor out brackets

How it’s done:Factoring trinomial through decomposition

Factorise 10k2 – 13k – 3.

10k2 – 13k – 3

–30

–30 = 2 × –152 + (–15) = –13

10k2 – 13k – 3 = 10k2 + 2k – 15k – 3

= 2k(5k + 1) – 3(5k + 1)

= (5k + 1)(2k – 3)

Find the product of a and c (the quadratic coefficient and the constant term).Use the factors of this product to split the linear (middle) term.Factorise the terms in pairs.Take the common factor outside.

You try:

66828080 3 3 0

4

3 3

4

5D # 3

Write 7y2 – 35y – 42 in its fully-factored form.

7y2 – 35y– 42 = 7(y2 – 5y – 6)

= 7(y + 1)( y – 6)

The first step when factorising should always be to look for any common factors.

You try:

5D # 5

66828152 1 3 0

2

1 3

2

Homework Lesson 8 : 5D # 2 a-l, 3 m-x, 4 a-f

Lesson 9: Completing the square

Things to know:

6. Some trinomial quadratic expressions can be factorised by forming perfect squares. This method is particularly useful when no rational numbers can be found for use in the factors.

a. The general pattern for monic perfect squares:

b. Steps:i. Complete the square to write expression as the difference between 2 perfect squares

ii. Factor difference between squares

Factorise this expression by completing the square.

x2 – 10x – 6

Step 1

x2 – 10x – 6

= x2 – 10x + – –

Step 1: complete the square

Step 2:Factor as if difference between squares

66832876 2 2 0

4

2 2

4

6

= (x – 5)2 – 31

Step 2:

= (x – 5 + )(x – 5 – )

You try:

5E # 5

Homework Lesson 9: 5E # 1a-f, 4a-l, 6a – l( factor by completing the square)