lessson plans - 2012 grade 9. mills... · mini lesson (15 mins.) students will be given a list of...

Sharlene Mills (2012-2013)

LESSSON PLANS - 2012 GRADE 9

Teacher : Sharlene Mills Date : September 17, 2012 Subject : Mathematics Class : 9T Topic : ALGEBRA / LAWS Duration : 100 minutes Aim : What is the effectiveness of number properties to solving problems? Pre-requisite : Students should know:

(i) What are algebraic expressions (ii) Know and use these concepts: variables, terms, factors of a term,

expression, coefficient, constant, and like terms. (iii) Translate verbal expressions/phrases to algebraic symbols, terms,

and/or expressions and vice versa. (iv) Express terms in there simplest forms.

Objectives : After class discussion, students should be able to:

(i) Identify the algebraic laws when working with numbers : commutative, associative and distributive law

(ii) Apply the algebraic laws to solving problems (iii) Use patterns to represent and solve problems

MOTIVATION ( 12 mins.)

2 minutes – Which 3 numbers have the same answer whether they’re added or multiplied together? Answer: 1, 2 and 3. 10 minutes – Students will be given a work sheet to complete. This work sheet is to test the pre-requisite. Please find attached

MINI LESSON (15 mins.)

Teacher will discuss with students the questions on the worksheet. Students will be asked to give reasons for answers. Teacher will guide students through the discovery of the three laws of algebra. To do this, students will be given a worksheet for them to tick whether each statement is true or false. Students will be asked to give reason for answer. During, the discussion the laws will be derived through guided discovery.


E.g. 12 + 4 = 4 + 12 but 3−2 ≠ 2 − 3, also

14× 3 = 3 × 14 but 24÷ 6 ≠ 6 ÷ 24

Note, for multiplication and addition, no matter the order of the numbers, once you are adding or multiplying a set of two or more numbers the result is the same – Commutative Law

GROUP (7 mins.)

Explore and give reason for answers.

1. Is this correct, to find the answer for 120 × 22, this can be done by: (100 × 22) + (20 × 22) + (0 × 22) .

2. (100 × 22) + (20 × 22) + (0 × 22) = 100 × 22 + 20 × 22 + 0 × 22

3. Which law can be applied to this question?

Each group will be given a question relating to a law to explore. WHOLE CLASS SHARING ( mins.)

Students will share with class their answer and the law they discovered from the questions given.

INDEPENDENT WORK (10 mins.)

Halo Supermarket as the following items on special. 1 orange for $55 1 pine for $120

1 apple for $42 The size of each fruit does not matter, they all cost the same amount of money. Mr. Brown went to the supermarket and bought 3 large oranges, 2 small pines, 2 large pines and 4 large apples. Ms. Reid went to the supermarket and bought 3 large apples, 1 small apple, 2 small oranges, 1 small orange, and 4 small pines. They both have only a $1000 bill to pay for their items. Can the $1000 pay for the goods, if so, who will receive the most change and why. A real life problem relating to the three laws will be given.

LESSON EVALUATION (8 mins.)

Define the following terms and give one example of each.

1. Commutative Law 2. Distributive Law 3. Associative Law


EVALUATION

Teacher : Sharlene Mills Date : September 19, 2012 Subject : Mathematics Class : 9T


Topic : ALGEBRA Duration : 100 minutes Aim : What is the effectiveness of number properties to solving problems? Pre-requisite : Students should know:

(v) What are algebraic expressions (vi) Know and use these concepts: variables, terms, factors of a term,

expression, coefficient, constant, and like terms. (vii) Translate verbal expressions/phrases to algebraic symbols, terms,

and/or expressions and vice versa. (viii) Substitute numbers for letters in algebraic expressions to simplify

them (ix) Express terms in there simplest forms.


(i) Identify the algebraic laws when working with numbers : commutative, associative and distributive law

(ii) Apply the algebraic laws to solving problems (iii) Add like and unlike terms ( not integers only natural numbers) (iv) Subtract like and unlike terms ( not integers only natural numbers)

MOTIVATION ( 3 mins..)

How many squares are there in this diagram? Answer: 14

MINI LESSON (15 mins.)

Students will be given a list of problems relating to the laws. They will be asked to identify the pattern in the numbers and give reason for answer. During, the discussion the laws will be derived through guided discovery. E.g. 12 + 4 = 4 + 12 but 3−2 ≠ 2 − 3, also

14× 3 = 3 × 14 but 24÷ 6 ≠ 6 ÷ 24

Note, for multiplication and addition, no matter the order of the numbers, once you are adding or multiplying a set of two or more numbers the result is the same – Commutative Law

GROUP (7 mins.)

Explore and give reason for answers.

1. Is this correct, to find the answer for 120 × 22, this can be done by:


(100 × 22) + (20 × 22) + (0 × 22) .

2. (100 × 22) + (20 × 22) + (0 × 22) = 100 × 22 + 20 × 22 + 0 × 22

3. Which law can be applied to this question?

Each group will be given a question relating to a law to explore. WHOLE CLASS SHARING ( mins.)

Students will share with class their answer and the law they discovered from the questions given.

INDEPENDENT WORK (10 mins.)

Halo Supermarket as the following items on special.

1 orange for $55 1 pine for $120

1 apple for $42 The size of each fruit does not matter, they all cost the same amount of money. Mr. Brown went to the supermarket and bought 3 large oranges, 2 small pines, 2 large pines and 4 large apples. Ms. Reid went to the supermarket and bought 3 large apples, 1 small apple, 2 small oranges, 1 small orange, and 4 small pines. They both have only a $1000 bill to pay for their items. Can the $1000 pay for the goods, if so, who will receive the most change and why. A real life problem relating to the three laws will be given.

GROUP ( mins.) Each group will be given a work sheet. On the work sheet, there will be shapes and symbols used to represent algebraic terms. Students will be asked to use these shapes and symbols to derive algebraic expression. Using these shapes and symbols, students will be asked to group all the like terms. After grouping like terms, they will be asked to create one expression to represent each grouping. Create one expression to represent all groups Eg. r , s m


4m, 4s 5r Algebraic expression: 4푚 + 4푠 + 5푟 Question: Which is correct: 4푚 + 4푠 + 5푟 = 4푚 + 4푠 + 5푟 or

4푚 + 4푠 + 5푟 = 13푚푠푟 Give reason for answer.

WHOLE CLASS

The question given will be used to derive the concept of adding or substracting algebraic expressions. Each group will explain the reason for their answer. Discuss to develop the concepts of: like and unlike terms. When simplifying algebraic expressions u can only add or substract like terms.

LESSON EVALUATION

Students will be given 5 problems relating to adding or subtracting algebraic expression.

EVALUATION

Teacher : Sharlene Mills Date : Subject : Mathematics Class : 9T Topic : ALGEBRA/Simplifying algebraic expressions (Addition & Subtraction) Duration : 50 minutes Aim : What is the effectiveness of number properties to solving problems?


Pre-requisite : Students should know how to add or subtract directive numbers.


(i) Evaluate algebraic expressions. (ii) Identify like terms and unlike terms. (iii) Simplify algebraic expressions by combining like terms

MOTIVATION

Find the hidden word or phrase Expected Answer: You are a cut above the rest.

MINI LESSON

Teacher will review with students the concept of adding or subtracting directive numbers. (i) If you owe some one $10 and you have $5. When you give back

the $5 you have, do you still owe the person or do you have change left.

When you give back the $5 you still owe the person $5 When working with directed numbers, negative (owe), positive (own or

have) So −9 − 3 (you owe some 9 and you borrow 3 more, how much do you owe in all – 12) −9 − 3 = −12

12 − 15 ( you have 12 but you owe 15, when you give back the 12, you still owe 3.

12 − 15 = −13

−10 + 13 (owe 10 but you have 13, give back the 10, you still have 3 left. −10 + 13 = 3

−20 + 6 = −14 (owe 20 but have 6, give back the 6, still owe 14)

Using this concept how can you simplify: −3푎 + 6푏 − 5푎 − 12푏

−3푎 + 6푏 − 5푎 − 12푏 (group like terms) −3푎 − 5푏 + 6푏 − 12푏 (simplify) −3푎 − 5푏 = −8푏 (owe 3bs borrowed 5 more bs, owe 8 in all)

+6푏 − 12푏 = −6푏 (have six bs, but borrowed 12, give back 6, still owe 6)

Teacher will use tangible objects to assist students to understand concept.

U R A C U T THE R E S T


INDEPENDENT WORK

Students will be given a work sheet to complete. Please find attached the worksheet.

LESSON EVALUATION

Students will be asked to write a summary of what they learnt in class.

EVALUATION Teacher : Sharlene Mills Subject : Mathematics Topic : ALGEBRA - Multiplying algebraic expressions. Duration : 100 minutes Aim : Use patterning to understand the concept of dividing or multiplying

algebraic expression.


Pre-requisite : Students should know how to divide or multiply directive numbers.


(i) Simplify expressions by multiplication using the correct steps. (ii) Multiply any two polynomials. (iii) Use patterns to help them solve problems.

MOTIVATION

+ = _ = - = + - = Solve + = _ = - = + - =


MINI LESSON

Teacher will review with students multiplication of integer numbers

Students will be asked to observe and analyze the pattern in the tables below

Table 1 Table 2 4 × 3 = 12 (−3) × 3 = −9 4 × 2 = 8 (−3) × 2 = −6 4 × 1 = 4 (−3) × 1 = −3 4 × 0 = 0 (−3) × 0 = 0 4 × (−1) = −4 (−3) × (−1) = 3 4 × (−2) = −8 (−3) × (−2) = 6 4 × (−3) = −12 (−3) × (−3) = 9 4 × (−4) = (−3) × (−4) = 4 × (−5) = (−3 × (−5) =

From the table above, the pattern observed : − × − = − − × + = − + × − = − + × + = +

Review the commutative law This law states that no matter the order of the term, once you

are adding or multiplying a set of two or more terms the result is the same, so: 푥 × 푦 = 푦 × 푥 = 푥y Therefore the order in which we multiply algebraic terms is not important.

Look at the pattern, what are your observations?

3 × 5푥 = 3 × 5 × 푥 = 15푥

3푥 × 4푦 = 3 × 4 × 푥 × 푦 = 12푥푦

5푥 × 3푥 = 5 × 3 × 푥 × 푥 = 15푥

5푦 × (−3푥) = 5 × (−3) × 푦 × 푥 = −15푥푦 표푟 − 15푦푥

−4푦 × 3푦 = −4 × 3 × 푦 × 푦 = −12푦

−3푦 × (−2푥) = −3 × (−2) × 푦 × 푥 = 6푥푦 표푟 6푦푥

3푝푞 × (−2푝 푞) × 2푝푞 =

3 × (−2) × 2 × 푝 × 푝 × 푝 × 푝 × 푞 × 푞 × 푞 × 푞 −12푝 푞


INDEPENDENT WORK LESSON EVALUATION EVALUATION

Discovery Unlike terms can multiply (you cannot add or subtract unlike term).

No matter the number of terms, answer as only one expression When multiplying, re-group expression (like terms), then multiply

(number times number, letter times letter)

+ term × a + term gives + answer − term × a + term gives − answer + term × a − term gives − answer − term × a − term gives + answer

Some answers have a letter with a number above it. When this occurs, the letter is called the base and the number is called the power (푝 ) The magnitude of the power is dependent on the number of times we multiply the base.

Students will be given questions to relating to concept learnt. For students to get a full understanding of concept, the teacher will give multiple questions for them to practice being that there are many rules they need to apply. Students will be asked to complete the puzzle below. This is to evaluate their understanding of working with integer numbers.

Try to fill in the missing numbers. Use the numbers 1 through 9 to complete the equations. Each number is only used once.


Teacher : Sharlene Mills Subject : Mathematics Topic : ALGEBRA - Division of Algebraic Expression Duration : 50 minutes Objectives : After class discussion, students should be able to:

(i) Simplify expressions by division using the correct steps. (ii) Divide any two polynomials. (iii) Use patterns to help them solve problems.

MOTIVATION

Find the hidden word or phrase Expected Answer: More than visit or multiple visits.

MINI LESSON

Look at the table and try and identify the pattern.

Table 1

12 ÷ 3 = 4 8 ÷ 2 = 4 4 ÷ 1 = 4 4 ÷ (−1) = −4 8 ÷ (−2) = −4 12 ÷ (−3) = −4 16 ÷ (−4) = 20 ÷ (−5) =

From the tables above, the observations are:

− ÷ − = − − ÷ + = − + ÷ − = − + ÷ + = +

Look at the following solution of each problem and try to evaluate them

푥 ÷ 푥 = × ×

= 1

Table 2 −12 ÷ 3 = −4 −8 ÷ 2 = −4 −4 ÷ 1 = −4 −4 ÷ (−1) = 4 −8 ÷ (−2) = 4 −12 ÷ (−3) = 4 −16 ÷ (−4) = −20 ÷ (−5) =

vvvvviiiiisssssiiiiittttt


10푎 ÷ 5푏 = × ×

= 15푝 푞 ÷ 5푝푞 = × × × × ×

× × × = 표푟 3푝푞

−12푝 푞 ÷ 8푝 푞 = × × × × ×

× × × × × × × = −

−24푝 푞 ÷ (−18푝 푞 = × × × × × × × × ×

× × × × × × × × =

From the solution of the problems above, to divide algebraic expressions

Rewrite the expression in a common fraction form Each algebraic expression is expanded. Reduce/simplify terms by dividing both numerator and denominator by common factors Answer is written in fraction, algebraic term or a number in its simplest form.

PARTNER/ SMALL GROUP

Working in groups students will be asked to solve the following problems (i) The sum of the length of the sides of a square is 28푥. (ii) If the total distance around a regular pentagon is 40푣. What is the length of each side

of the pentagon.

INDEPENDENT WORK

Students will be given activity from their workbook pertaining to concept learnt

LESSON EVALUATION

Students will be given a work sheet to complete

EVALUATION Teacher : Sharlene Mills Subject : Mathematics


Topic : COMPUTATION – Application of the distributive law Duration : 100 minutes Objectives : After class discussion, students should be able to:

(i) Simplify algebraic expressions using the distributive property. (ii) Expand and simplify expressions (iii) Apply all the concepts needed to simplify expressions

MOTIVATION

Change one letter on each line to arrive at the final word. RIDE _______ _______ _______ WASH

MINI LESSON

Question – A rectangle as lengths (푎 + 푏cm) and width (푎cm). What is the area of this rectangle? Using a diagram to illustrate the question given. Label sides of the rectangle

Area of rectangle is length × width (푎 + 푏) × 푎 Separate rectangle into a square and a rectangle

푎 Area of square Area of rectangle

푎푎푎

푎 푏

푏

푎 Therefore the area of the rectangle is 푎 + 푎푏

Note we cannot add unlike terms so our answer as two terms.

Suppose the length of the rectangle was (푎 − 푏)cm and the width was 푎cm.

Area of rectangle is length × width (푎 − 푏) × 푎 Separate rectangle into a square and a rectangle

푎 Area of square Area of rectangle

푎푎푎

푎 −푏

푎

푎푏

푎

푎푏

푎

푎푏

푎

−푎푏


−푏

푎 Therefore the area of the rectangle is 푎 − 푎푏

From the problems given, it clear that the distributive law was used to arrive at the answer for the area of the rectangle (length × width) The distributive law states that each term immediately outside the bracket is multiplied by each and every term inside the bracket. ∴ (푎 + 푏) × 푎 is the same as 푎 × (푎 + 푏) 푎 × 푎 + 푎 × 푏 (푎 is the term immediately outside the bracket) 푎 + 푎푏 (푎 − 푏) × 푎 is the same as 푎 × (푎 − 푏) 푎 × 푎 + 푎 × −푏 (recall + × − = −) 푎 − 푎푏

Let us simplify : (i) 3(푥 + 2푦) = 3 × 푥 + 3 × 2푦 = 3푥 + 6푦

(ii) −4푎(3푎 + 5) = −4푎 × 3푎(−4푎 × 5) = −12푎 − 20푎

(iii) −2(5푥 − 7푦) = −2 × 5푥(−2 × (−7푦)) = −10푥 + 14푦

(iv) −3푥(5푥 − 7푦 + 2) = −3푥 × 5푥/−3푥 × (−7푦)/ −3푥 × 2 = −15푥 + 21푥푦 −6푥

Discovery If the term outside the bracket is positive, then the signs of the terms inside the bracket

remain unchanged once the law is applied.

If the term outside the bracket is negative, then the signs of the terms inside the bracket are all changed once the law is applied

INDEPENDENT WORK

Students will be given activity from there textbook. Exercise 3D (pages 165-166 - A Complete Mathematics Course Book 3)

LESSON EVALUATION

Further practice of concept learnt

EVALUATION


Teacher : Sharlene Mills Subject : Mathematics Topic : ALGEBRA - Simplifying Fractional Expressions Duration : 50 minutes Pre-requisite : Students should be able to add or subtract fractional numbers.



(i) Use real life situations to model and simplify fractional expressions.

MOTIVATION

How can you get ten horses into nine stables, one per stable?

Answer: Place one letter from TEN HORSES into each of the nine stables.

MINI LESSON

Working in groups, students will be asked to find the area of the following shapes.

푥cm 푦 cm 6cm

7cm cm

For students to calculate the area of each shape, they must be knowledge of the concepts used to calculate area of plain shaped. Students should also know the concept of multiplying fractions Area of triangle - × 푏 × ℎ Area of square - 푙푒푛푔푡ℎ × 푤푖푑푡ℎ but the sides of a square are equal so Area = 푙

Area of rectangle - 푙푒푛푔푡ℎ × 푤푖푑푡ℎ

∴ to find the area of each shape

Triangle = 푥 × × 7 = 2푥cm2 ( simplify fraction by using common factors)

Square = 푦 × 푦 = 푦 cm2

Rectangle = × 6 = 표푟 3 cm2

Therefore, to simplify the following

(i) 푞 × 8 = 2푞

(ii) 푥 × (−36푦) = −24푥푦


(iii) − 푚 × 푛 = − 푚푛

(iv) − 푥 × (−12푥) = 푥 표푟 1 푥

Note: (i) the rules for multiplying integers is applied

− term × a + term gives − answer + term × a − term gives − answer − term × a − term gives + answer

(ii) common factors are identified, then the fraction is reduced to its simplest form (iii) The answer gives one term, which implies that unlike terms can be multiplied.

INDEPENDENT WORK

Students will be given questions from their workbook to complete.

EVALUATION Teacher : Sharlene Mills Day 1 Subject : Mathematics Class : 9T Topic : ALGEBRA – Factorizing Algebraic Expressions (Highest Common

Factor – H.C.F) Duration : 100 minutes Aim : Identifying common elements in an algebraic expression. Prerequisites : Students should know:

(i) What is an algebraic expression


(ii) Different elements in an algebraic expression (variable, coefficient, constant, term)

(iii) Simplify an algebraic expression using the Distributive Law

Objectives : After class discussion, students should be able to: Knowledge :(i) Recognize Factorization as the reverse of expansion

(ii) Recognize the difference between a common factor and the greatest common factor

Skill : Identify the H.C.F of any given expression Factorize expression using H.C.F

Attitude : Manipulate any expression in order to arrive at the H.C.F

MOTIVATION

Change one letter on each line to arrive at the final word. SIDE _______ _______ _______ _______ WASH

MINI LESSON

Students will be asked to simplify the following expressions.

(i) 4(2푥 + 푦) = 4 × 2푥 + 4 × 푦 = 8푥 + 4푦

(ii) −4푎(3푎 + 5) = −4푎 × 3푎(−4푎 × 5) = −12푎 − 20푎

Note the procedure – term outside the bracket is multiplied by each and every term inside the bracket

Scenario Ms. Clark has 4 children attending Clan Carthy High School (Monika, Jannen, Andrea and Tina). Mrs. Gayle also has 3 children attending the same school (Teana, Shaday and Amanda). One day Shaday discovered that Tina is her sister. What is common between the women’s children? Key Term – common Factorization – Identifying common elements in an algebraic expression.

Factorize the following expressions (i) 9푥 + 6푦 (ii) −12푎 − 20푎 (iii) −16푥 + 24푥푦


9푥 + 6푦 (identify common element/s or factor (3)

(first term determines the sign of the factor; first term positive, factor is positive)

+ (divide both term by common factor) 3(3푥 + 2푦) (Factorized) −12푎 − 20푎 (common factor −4푎, first term negative, factor negative)

− = −4푎(3푎 + 5) (Factorised)

−16푥 + 24푥푦 (common factor −8푥푦, first term negative, factor is negative)

−16푥−8푥푦

+24푥푦−8푥푦

= −8푥푦(2푦 − 3)

Steps to factorizing expressions using H.C.F

1. Identify the Highest Common factor among the terms)

Note in question 2 both terms are divisible by 2, but2 is not the largest number that both terms can go into. Also in question 3, both terms are divisible 4 and 2, but 8 is larger than these numbers.

2. First term determines the sign of the factor 3. Divided H.C.F by each term 4. The factor is written outside the bracket, remainder inside bracket

INDEPENDENT WORK

Students will be given questions to complete based on concept learnt.

LESSON EVALUATION

What is the relationship between factorization and expansion.

EVALUATION


Teacher : Sharlene Mills Day 2 Subject : Mathematics Class : 9A Topic : ALGEBRA – Factorizing Algebraic Expressions (Highest Common

Factor – H.C.F) Duration : 50 minutes Aim : Identifying common elements in an algebraic expression. Prerequisites : Students should know:

(i) What is an algebraic expression


(ii) Different elements in an algebraic expression (variable, coefficient, constant, term)

(iii) Simplify an algebraic expression using the Distributive Law (iv) Divide and multiply directed numbers (v) What are index numbers


Knowledge : Determine the Highest Common Factor of more than two terms in an expression

Skill : List all factor of an expression with two or more terms.

Attitude : Apply factoring concepts and procedures to complete exercises

MOTIVATION

How many squares can you identify in the diagram below.

Answer : 14 squares

MINI LESSON

Teacher will review with students the steps they discovered in factorizing an expression.

1. Identify the Highest Common factor among the terms) 2. First term determines the sign of the factor 3. Divided H.C.F by each term 4. The factor is written outside the bracket, remainder inside bracket

Question Suppose you were asked to factories the following expressions.

(i) 9푎푏 + 12푎 푏 − 15푎푏 (common factor 3ab) 3푎푏(3 + 4푎 − 5푏) (to get the expression factorized, each term is divided by the factor

(ii) −18푥 푦 푧 + 12푥 푦 푧 − 36푥 푦 푧 (푐표푚푚표푛 푓푎푐푡표푟 − 6푥 푦 푧 ) −6푥 푦 푧 (3푥 푧 − 2푦 + 6푥푦 푧)


Discovery 1. Not only can we factorize expressions with two terms, we can also factorize expressions

with more than two term 2. First term still determines the sign of the factor 3. The variable in each term can have a power of more than two. When this is seen, we

identify the variable with the least power (this is our factor) INDEPENDENT WORK

Students will be given questions to complete based on concept learnt.

LESSON EVALUATION

Why is it important to identify the variable with the smallest powerto determine what is our factor.

EVALUATION Teacher : Sharlene Mills Day : 3 Subject : Mathematics Class : 9T Topic : ALGEBRA – Factorizing Algebraic Expressions (Grouping – Binomial

Expressions) Duration : 100 minutes Aim : Identifying common elements in an algebraic expression. Prerequisites : Students should know:

(i) What is an algebraic expression (ii) Different elements in an algebraic expression (variable, coefficient,

constant, term) (iii) Simplify an algebraic expression using the Distributive Law (iv) Divide and multiply directed numbers (v) What are index numbers


Knowledge : Determine the Highest Common Factor of more than two terms in an expression


Skill : List all factor of an expression with two or more terms.

Attitude : Apply factoring concepts and procedures to complete exercises

MOTIVATION

Figure out the popular phrase HARM good

Answer - More harm than good

MINI LESSON Look at the two algebraic expressions and tell what the difference between the two is. (i) 8푝푞 + 16푝푟 + 20푝푠 (ii) 3푎푥 − 6푎푦 + 푏푥 − 2푏푦

Expected Answer – the first one has three terms while the second one has four terms. To factorize the first expression, we simply identify the highest common factor

then divide. 8푝푞 + 16푝푟 − 20푝푠 = 4푝(2푞 + 4푟 − 5푠)

but to factorize the second expression we cannot jump to indentifying HCF. First we have to group in pairs (3푎푥 − 6푎푦)(+푏푥 − 2푏푦) Factorize each group independent of the other. (3푎푥 − 6푎푦)(+푏푥 − 2푏푦)

(3푎푥 − 6푎푦)– H.C.F = 3푎(+푏푥 − 2푏푦) – H.C.F = +푏 3푎(푥 − 2푦) + 푏(푥 − 2푦)

Note – the expression derived in the brackets are the same (common factors) To factorize (3푎푥 − 6푎푦)(+푏푥 − 2푏푦) = (푥 − 2푦)(3푎 + 푏)

What if you were asked to factorize 4푝푥 − 4푝푦 − 3푞푥 + 3푞푦 (4푝푥 − 4푝푦)(−3푞푥 + 3푞푦) group in pairs 4푝(푥 − 푦) − 3푞(푥 − 푦) for the second pair, the first term is negative so factor I is negative (푥 − 푦)(4푝 − 3푞) Factorized To ensure students understand concept, the teacher will illustrate step by step

solution to other expressions which varies due to the magnitude of each term.


Teacher and students will develop a step by step summary to factorizing algebraic expressions by grouping. Students will be required to use these steps to assist them when factorizing by grouping.

INDEPENDENT WORK

Students will be given questions pertaining to concept learnt so that they can develop an understanding

LESSON EVALUATION

Teacher : Sharlene Mills Day : 1 Subject : Mathematics Class : 9T Topic : ALGEBRA – Factorizing Quadratic Expressions (Perfect Square) Duration : 100 minutes Aim : Identifying common elements in an algebraic expression. Prerequisites : Students should know:

(i) What is an algebraic expression (ii) Different elements in an algebraic expression (variable, coefficient,

constant, term) (iii) Simplify an algebraic expression using the Distributive Law (iv) Divide and multiply directed numbers (v) What are index numbers (vi) What are square numbers


Knowledge (i) Identify a Quadratic Expression (ii) Recognise a perfect square

Skill (i) Determine if a quadratic expression is a perfect square (ii) Factor any Perfect Square Expression


푎 + 푏

푎 + 푏

푎 푎

푏 푏

Attitude (i) Use factoring concepts and procedures to complete given

exercises. (ii) Apply the knowledge and skills learnt to generalize a concept for

factoring quadratic expressions.

MOTIVATION

What is the hidden phrase

MINI LESSON A square as lengths of (푎 + 푏). What is the area of this square? Using a diagram to represent the worded question.

Since 푎 and 푏 are independent variables, let us divided the square

accordingly. Then find the area of each portion. 푎푏 a 푎 푏

To find the area of the square, it is the sum of the area of each portion. ∴ Area of square = 푎 + 푎푏 + 푎푏 + 푏 = 푎 + 2푎푏 + 푏 The expression derived is called a Quadratic Expression. For an expression to be quadratic, it must have a square term. This expression is called a Perfect Square

For an expression to be a perfect square: - The first and last terms ( 풂ퟐand 풃ퟐ) must be perfect squares. - There must be no minus signs before ( 풂ퟐ and 풃ퟐ) - The middle term is 2abor − 2ab

To factorize an expression which is a perfect square: 푥 + 6푥 + 9 First : test to see if the two end terms are square terms, if so rewrite as a square term. 푥 is already written as a square term , so 9 written as a square term is 3

풂ퟐ

풂풃

풂풃

풃ퟐ

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Rewriting expression 푥 + 6푥 + 3

Test the middle term, in doing so, when 2 is multiplied by both end terms we

must get the middle term. If this is proven to be true, then our expression is a perfect square. 푥 + 6푥 + 3 = 푥 + (2)(푥)(3) + 3 Notice that when 2 is multiplied by 푥 and 3 we got the middle term 6푥. This proves that the expression is a perfect square.

To factorize expression 푥 + (2)(푥)(3) + 3 = (푥 + 3)(푥 + 3)표푟 (푥 + 3) Review the area of the square.

Example 2. Factorize 푥 − 8푥 + 16 푥 − 8푥 + 4 (Rewrite end terms as square terms) 푥 − (2)(푥)(4) + 16 (test middle term; proven to be true) Factors (푥 − 4)(푥 − 4)표푟 (푥 − 4) Expression is a perfect square

What if we were to factorize 9푥 + 30 + 25. This expression may look

complicated but the same procedure is used. 9푥 + 30 + 25 3 푥 + 30 + 5 ( rewrite end terms as square numbers.) (3푥) + 2(3푥)(5) + 5 ( test middle term, proven to be true)

30푥 Factors (3푥 + 5)(3푥 + 5) or (3푥 + 5)

GROUP WORK Working in groups, students will be given 6 quadratic expressions to determine

whether or not each expression is a perfect square. If expression is a perfect square, students will be asked to factorize. While factorizing the expression, students need to apply the generalized concept to help solve problems. Example of question:

1. 9푥 + 12푥푦 + 4푦 ( perfect square after testing all criteria) 2. 푦 + 9푦 + 9 (not a perfect square after testing all criteria. Does not prove

middle term concept) INDEPENDENT WORK

Students will be given questions relating to concepts learnt from their textbooks.

LESSON EVALUATION

Write a summary of no more than 6 line on :

(i) What is a quadratic expression (ii) How can you identify a perfect square expression


EVALUATION Teacher : Sharlene Mills Day : 2 Subject : Mathematics Class : 9T Topic : ALGEBRA – Factorizing Quadratic Expressions (Difference of

Two Squares) Duration : 100 minutes Aim : Factorize any given Difference of Two Squares expressions Prerequisites : Students should know:

(i) What is an algebraic expression (ii) Identify a Quadratic Expression (iii) Different elements in an algebraic expression (variable, coefficient,

constant, term) (iv) Simplify an algebraic expression using the Distributive Law (v) Divide and multiply directed numbers (vi) What are index numbers (vii) Rewrite a given number in square term


Knowledge (i) Recognize a Difference of two squares expression (ii) Explain the difference between a perfect square and the difference

of two squares Skill


푎 + 푏

(i) Recall the two main conditions which satisfies a difference of two square

(ii) Determine if a quadratic expression is a difference of two squares (iii) Factor any Difference of Two Squares Expression

Attitude (i) Use factoring concepts and procedures to complete given

exercises. (ii) Apply the knowledge and skills learnt to generalize a concept for

factoring difference of two squares expressions.

MOTIVATION

What is half of 8? (Hint: It's not 4) Answer : 3 (if you slice vertically) or o (if you slice horizontally)

MINI LESSON

A rectangleof length of (푎 + 푏) and width (푎 − 푏), then the area of this rectangle using a diagram is:

푎푏 푎 − 푏 Area of rectangle is sum of each portion.

푎 − 푎푏 + 푎푏 − 푏2 = 푎2 − 푏2

This expression is called Difference of Two Squares

For an expression to be a Difference of Two Squares : - The first and last terms ( 풂ퟐand 풃ퟐ) must be perfect squares. - There must be no middle term ( two terms only and they both are square

terms). Unlike the perfect square which as middle term - The square terms must be separated by a minus sign (difference).

To factorize 푥 − 4 푥 − 2 (End terms can be rewritten as square terms, difference of two squares) Factors (푥 − 2)(푥 + 2)

푎 − 푏

푎(푎 − 푏) 푎 − 푎푏

푏(푎 − 푏) 푎푏 − 푏


Recall area of rectangle. Note that for difference of two squares, one factor is positive and the other is negative.

Factor 4푥 − 64 2 푥 − 8 (2푥) − 8 (Difference of two squares) Factors : (2푥 − 8)(2푥 + 8)

GROUP WORK Students will be given 6 quadratic expressions to determine whether or not each term is a difference of two squares. If term proven to difference of two squares, then factorize.

INDEPENDENT WORK

Students will be given questions from their workbook to relating to concept learnt to factorize.

LESSON EVALUATION

Write the three main conditions which make a quadratic expression a difference of two squares.

Teacher : Sharlene Mills Day : 3 Subject : Mathematics Class : 9T Topic : ALGEBRA – Factorizing Algebraic Expressions Duration : 50 minutes Aim : Factorize any given Algebraic Expressions Prerequisites : Students should know:

(i) What is an algebraic expression (ii) Identify a binomial, perfect, or difference of two squares

expression. (iii) Different elements in an algebraic expression (variable, coefficient,

constant, term) (iv) Simplify an algebraic expression using the Distributive Law (v) Divide and multiply directed numbers (vi) What are index numbers (vii) Rewrite a given number in square term (viii) Factorize any given expression by using the concept of factoring:

H.C.F, binomial, perfect, or difference of two squares expression.

Objective Students know how to factorize any given expression in isolation. For example, given a group of expressions to use H.C.F to factorize, students know how to do this. Teacher wants to test students understanding and application of concepts learnt in previous classes.


Group Work Students will be given 8 algebraic expressions to factorize. Students will be required to:

- identify the type of expression - use the correct method of factorizing, to factor each expression.

Evaluation Teacher : Sharlene Mills Day : 1 Subject : Mathematics Class : 9T Topic : ALGEBRA – Factorizing Algebraic Expressions Duration : 100 minutes Aim : Factorize any given Algebraic Expressions Prerequisites : Students should know:



constant, term) (iv) Simplify an algebraic expression using the Distributive Law (v) Divide and multiply directed numbers (vi) What are index numbers (vii) Rewrite a given number in square term (viii) Factorize any given expression by using the concept of factoring:


Objective In the previous class students were given different kinds of algebraic expressions to factorize. Approximately 80% of the class misuses the concepts for factoring algebraic expression. Teacher will use this class to guide students through factoring any given group of expression.


Motivation Change one letter on each line to arrive at a new word BOLD _______ _______ _______ TILT Mini Lesson

The teacher will write 7 algebraic expressions on the board. (i) 3푥 + 12푦 (ii) −16푥 + 24푥 푦

(iii) 3푎푥 − 6푎푦 + 푏푥 − 2푏푦 (iv) 푥 + 14푥 + 49

(v) 푥 − 22푥 + 121 (vi) 푥 − 81 (vii) 푚 − 8푚

Taking each term, identify the kind of expression

(i) 3푥 + 12푦 (Linear expression)

(ii) −16푥 + 24푥 푦 (Quadratic Expression)

(iii) 3푎푥 − 6푎푦 + 푏푥 − 2푏푦 (Linear Expression) (iv) 푥 + 14푥 + 49 (Quadratic Expression)

(v) 푥 − 22푥 + 121 (Quadratic Expression) (vi) 푥 − 81 (Quadratic Expression) (vii) 푚 − 8푚 (Quadratic Expression)


Now that each expression is identified, test conditions to find out what concept should be used to factorize each. (i) 3푥 + 12푦 (Linear expression)

- since this expression is a linear expression, it can be factorized by using H.C.F or grouping.

- Question: Can we use grouping to factorize? Expected Answer : No, because for grouping to take place we need 4 terms. Therefore to factorize we use H.C.F.

- Students will be asked to factor expression 3푥 + 12푦 = 3(푥 + 4푦) [ common factor is 3]

(ii) −16푥 + 24푥 푦 (Quadratic Expression) - Since this expression is quadratic, we need to test if it’s a perfect square or

a difference of two square. - Question : Are the two end terms square number?

Expected Answer: No. Therefore to factorize we use H.C.F. −16푥 + 24푥 푦 = −8푥 푦(2푦 − 3푥) [common factors −8푥 푦 ] Recall: First term tells the sign of the factor

(iii) 3푎푥 − 6푎푦 + 푏푥 − 2푏푦 (Linear Expression) - Note that this linear expression as four terms, so we use to factorise.

(3푎푥 − 6푎푦) + (푏푥 − 2푏푦) group in pairs 3푎(푥 − 2푦) + 푏(푥 − 2푦) identify common factor of each pair. Similar terms in each bracket. (푥 − 2푦)(3푎 + 푏) Factors

(iv) 푥 + 14푥 + 49 (Quadratic Expression)

- Since expression is quadratic test to see if it is a perfect square. - Why use perfect square? Expression has a middle term.

푥 + 14푥 + 49 푥 + 2(푥)(7) + 7 Conditions satisfy a perfect square. Factors (푥 + 7)(푥 + 7)

(v) 푥 − 22푥 + 121 (Quadratic Expression)

- Expression is quadratic, test to see if it is a perfect square.

푥 − 22푥 + 121


푥 − 2(푥)(11) + 11 Conditions satisfy perfect square Factors (푥 − 11)(푥 − 11)

(vi) 푥 − 81 (Quadratic Expression)

- Note: this expression as no middle term and it is also quadratic, test to see if end terms are square terms.

푥 − 81

푥 − 9 Condition satisfies a difference of two squares Factors (푥 − 9)(푥 + 9) one position, one negative

(vii) 푚 − 8푚 (Quadratic Expression)

- Expression is quadratic, there is no middle term. Test to see if expression is a difference of two squares.

푚 − 8푚 (푚 is a square term, but 8푚 cannot be rewritten as a square term. This means that expression is not a difference of two squares, so to factorize we use H.C.F. 푚 − 8푚 = 푚(푚− 8) [common factor is 푚]

Conclusion When factoring any quadratic expression:

(i) An expression can be quadratic but it neither a perfect nor difference of two squares. Therefore before factoring, test condition.

(ii) Difference of two squares has no middle term but the two terms are square terms separated by a minus sign.

(iii) Perfect square expressions have two square end terms, plus a middle term, but when 2 is multiplied by both end terms, the result gives the middle term

(iv) Factoring by grouping, there must be four term. (v) Factoring by H.C.F, all terms must have a common factor

Independent Work Students will be given similar problems to one listed above to factorize. Evaluation


Teacher : Sharlene Mills Day : 2 Subject : Mathematics Class : 9T Topic : ALGEBRA – Factorizing Quadratic Expressions Duration : 100 minutes Aim : Understanding the beauty of Factoring Quadratic Expressions which are

not Perfect or Difference of Two Squares. Prerequisites : Students should know:



constant, term) (iv) What are index numbers

Objective : After class discussion, students should: Knowledge

(i) Recognize Quadratic Expressions which are notPerfect or Difference of two square.

(ii) Explain the difference between a Quadratic Expression which is a perfect square and one which is not.

(iii) Differentiate between a Quadratic Expression which is a Difference of Two Squares and one which is not


Skill (iii) Recall the main conditions which makes a Quadratic Expression

not a Perfect or a Difference of Two Square (iv) Factor any Quadratic Expression which is none Perfect or

Difference of Two Squares. Attitude (v) Use factoring concepts and procedures to complete given

exercises. (vi) Apply the knowledge and skills learnt to generalize a concept for

factoring none Perfect or Difference of Two Squares expressions.

Motivation Find the hidden word in the puzzle below. Answer : Creative Mini Lesson

Working in groups, each group will be given a sheet of paper with three equations. Students will be asked to compare the equations and write down the differences and similarity.

(i) 푥 + 14푥 + 49

(ii) 푥 + 7푥 + 12 (iii) 푥 − 81

Students will be asked to factorize the expressions that can be factorized and give reason.

(i) 푥 + 14푥 + 49 (perfect square)

푥 + 2(푥)(7) + 7 Conditions satisfy a perfect square.

Factors (푥 + 7)(푥 + 7)

(ii) 푥 − 81 (Quadratic Expression) - Note: this expression as no middle term and it is also quadratic, test to see if end terms are

square terms.

푥 − 81 푥 − 9 Condition satisfies a difference of two squares Factors (푥 − 9)(푥 + 9) one position, one negative

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For the expression number two, there is only one square term which is 풙ퟐ . Therefore, to factorize this expression, we cannot use perfect or difference of two squares.

- Taking the expression 풙ퟐ + ퟕ풙 + ퟏퟐ, identify the coefficients of 풙ퟐand ퟕ풙and the constant

in the expression. 풙ퟐ + ퟕ풙 + ퟏퟐ constant 1(coefficient) 7 (coefficient) middle term - To factorize this expression, our aim is to rewrite this expression as a binomial. That is,

splitting the middle into two parts so that we can use grouping to factorize.

To factor 풙ퟐ + ퟕ풙 + ퟏퟐ: - First : multiply the coefficient of 풙ퟐby the constant +12 (ퟏ × +ퟏퟐ = ퟏퟐ)

- Then to split the middle term, we must find two numbers which are factors of 12, but those

two numbers when u add or subtract them, they must give ten.

Check : 1 × 12 = 12 but 1 ± 12 ≠ 7 2 × 6 = 12 but 2 ± 6 ≠ 7 3 × 4 = 12 and 3 + 4 = 7

Note :3 and 4are the secret numbers , we will replace 7푥 with these 2 numbers so that we can factorize the expression by grouping. New Expression: 푥 + 3푥 + 4푥 + 12 (푥 + 3푥)(+4푥 + 12) grouping

푥(푥 + 3) + 4(푥 + 3) (푥 + 4)(푥 + 3) Factors Now Lets try to factorize this one. 푥 + 3푥 − 10 (1 × −10 = −10 Testing factors of −10 to split middle term 1 × −10 = −10 but 1 ± −10 ≠ 3 2 × −5 = −10 but 2 ± −5 ≠ 3 −2 × 5 = −10 and −2 + 5 = 3 -2 and 5 are the secret numbers, we replace 3푥 with these two numbers 푥 − 2푥 + 5푥 − 10 (푥 − 2푥)(+5푥 − 10)


푥(푥 − 2) + 5(푥 − 2) (푥 + 5)(푥 − 2) Factors Group Work Working in groups, students will be given two expressions to factorize using the concept above.

(i) 푥 − 4푥 − 21 (ii) 푥 − 7푥 + 6

Students will share answers with class. Teacher and students will discuss answer to clarify mis-conceptions. Independent Work Students will be given 4 questions relating to concept learnt Lesson Evaluation Factorize 6− 푥 − 푥 Teacher : Sharlene Mills Subject : Mathematics Topic : ALGEBRA – Factorizing Algebraic Expressions Duration : 100 minutes Aim : Factorize any given Algebraic Expressions Prerequisites : Students should know:

(v) What is an algebraic expression (vi) Identify a binomial, perfect, or difference of two squares

expression. (vii) Different elements in an algebraic expression (variable, coefficient,

constant, term) (viii) Simplify an algebraic expression using the Distributive Law (ix) Divide and multiply directed numbers (x) What are index numbers (xi) Rewrite a given number in square term (xii) Factorize any given expression by using the concept of factoring:


Objective In the previous class students were given different kinds of algebraic expressions to factorize. Approximately 80% of the class misuses the concepts for factoring algebraic expression. Teacher will use this class to guide students through factoring any given group of expression.


Motivation Change one letter on each line to arrive at a new word BOLD _______ _______ _______ TILT Mini Lesson

The teacher will write 7 algebraic expressions on the board. (i) 3푥 + 12푦 (ii) −16푥 + 24푥 푦

(iii) 3푎푥 − 6푎푦 + 푏푥 − 2푏푦 (iv) 푥 + 14푥 + 49

(v) 푥 − 22푥 + 121 (vi) 푥 − 81 (vii) 푚 − 8푚

Taking each term, identify the kind of expression

(viii) 3푥 + 12푦 (Linear expression)

(ix) −16푥 + 24푥 푦 (Quadratic Expression)

(x) 3푎푥 − 6푎푦 + 푏푥 − 2푏푦 (Linear Expression) (xi) 푥 + 14푥 + 49 (Quadratic Expression)

(xii) 푥 − 22푥 + 121 (Quadratic Expression) (xiii) 푥 − 81 (Quadratic Expression) (xiv) 푚 − 8푚 (Quadratic Expression)


Now that each expression is identified, test conditions to find out what concept should be used to factorize each. (viii) 3푥 + 12푦 (Linear expression)

- since this expression is a linear expression, it can be factorized by using H.C.F or grouping.

- Question: Can we use grouping to factorize? Expected Answer : No, because for grouping to take place we need 4 terms. Therefore to factorize we use H.C.F.

- Students will be asked to factor expression 3푥 + 12푦 = 3(푥 + 4푦) [ common factor is 3]

(ix) −16푥 + 24푥 푦 (Quadratic Expression) - Since this expression is quadratic, we need to test if it’s a perfect square or

a difference of two square. - Question : Are the two end terms square number?

Expected Answer: No. Therefore to factorize we use H.C.F. −16푥 + 24푥 푦 = −8푥 푦(2푦 − 3푥) [common factors −8푥 푦 ] Recall: First term tells the sign of the factor

(x) 3푎푥 − 6푎푦 + 푏푥 − 2푏푦 (Linear Expression) - Note that this linear expression as four terms, so we use to factorise.

(3푎푥 − 6푎푦) + (푏푥 − 2푏푦) group in pairs 3푎(푥 − 2푦) + 푏(푥 − 2푦) identify common factor of each pair. Similar terms in each bracket. (푥 − 2푦)(3푎 + 푏) Factors

(xi) 푥 + 14푥 + 49 (Quadratic Expression)

- Since expression is quadratic test to see if it is a perfect square. - Why use perfect square? Expression has a middle term.

푥 + 14푥 + 49 푥 + 2(푥)(7) + 7 Conditions satisfy a perfect square. Factors (푥 + 7)(푥 + 7)

(xii) 푥 − 22푥 + 121 (Quadratic Expression)

- Expression is quadratic, test to see if it is a perfect square.

푥 − 22푥 + 121


푥 − 2(푥)(11) + 11 Conditions satisfy perfect square Factors (푥 − 11)(푥 − 11)

(xiii) 푥 − 81 (Quadratic Expression)

- Note: this expression as no middle term and it is also quadratic, test to see if end terms are square terms.

푥 − 81

푥 − 9 Condition satisfies a difference of two squares Factors (푥 − 9)(푥 + 9) one position, one negative

(xiv) 푚 − 8푚 (Quadratic Expression)

- Expression is quadratic, there is no middle term. Test to see if expression is a difference of two squares.

푚 − 8푚 (푚 is a square term, but 8푚 cannot be rewritten as a square term. This means that expression is not a difference of two squares, so to factorize we use H.C.F. 푚 − 8푚 = 푚(푚− 8) [common factor is 푚]

Conclusion When factoring any quadratic expression:

(xiii) An expression can be quadratic but it neither a perfect nor difference of two squares. Therefore before factoring, test condition.

(xiv) Difference of two squares has no middle term but the two terms are square terms separated by a minus sign.

(xv) Perfect square expressions have two square end terms, plus a middle term, but when 2 is multiplied by both end terms, the result gives the middle term

(xvi) Factoring by grouping, there must be four term. (xvii) Factoring by H.C.F, all terms must have a common factor

Independent Work Students will be given similar problems to one listed above to factorize. Evaluation

lessson plans - 2012 grade 9. mills... · mini lesson (15 mins.) students will be given a list of...

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