**SUBJECT: MATHEMATICS**

**CLASS: JSS 3**

**DATE:**

**TERM: 1st TERM**

**REFERENCE BOOKS**

- New General Mathematics by M. F Macrae et al bk 3
- Essential Maths by AJS OluwasanmiBk 3

WEEK SIX

**TOPIC: FACTORISATION**

**CONTENT**

- Factorisation of simple expression
- Difference of two squares
- Factorisation of quadratic expression

**FACTORISATION OF SIMPLE EXPRESSION**

To factorise an expression completely, take the HCFoutside the bracket and then divide each term with the HCF.

Example:

Factorise the following completely.

- 8xy + 4x2y
- 6ab – 8a2b + 12ab

Solution:

- 8xy + 4x2y

8xy = 2 X 2 X 2 X xX y

4x2y = 2 X 2 X xXxX y

HCF = 4xy

8xy + 4x2y = 4xy(8xy4xy + 4x2y4xy)

= 4xy( 2 + x)

- 9a2bc3 – 12ab2c2

9a2bc3 = 3 X 3 X a X a X c X c X c

12ab2c2 = 2 X 2 X 3 X b X b X c X c

HCF = 3abc2

= 3abc2(3ac – 4b)

**EVALUATION**

Factorise the following expression

- 9x2yz2 – 12x3z3
- 14cd + 35cd2f
- 20m2n – 15mn2

**FACTORISATION BY GROUPING**

To factorise an expression containing four terms, you need to group the terms into pairs.Thenfactorise each pair of terns.

Example:

Factoriseab – 2cb + 2cf – af

Solution:

Group ab and af together and 2cb and 2cf together

i.eab – 2cb + 2cf –af = ab – af – 2cb + 2cf

= a( b – f ) -2c( b – f )

= (a – 2c)( b – f)

**EVALUATION**

Factorise these expressions;

- 16uv – 12vt + 20mu – 15mt
- ap +aq +bq + bp
- mn – pq-pn +mq

**FACTORISATION OF QUADRATIC EXPRESSIONS**

A quadratic expression has two (2) as its highest power; hence this at times is called a polynomial of the second order. The general representation of quadratic expression is ax 2 + bx + c where a ≠ 0. From above expression, a, b, and c stands for a number.

**NOTE**

- if ax 2 +bx + c= 0, this is known as quadratic equation
- a is coefficient of x2, b is coefficient of x and c is a constant term.
- When an expression contains three terms, it is known as trinomial.
- To be able to factorize trinomial, we need to convert it to contain four terms.

Examples: factorization of trinomial of the form x2 +bx + c.

- Factorise x2 +7x +6

Steps:

- Multiply the 1st and the last term (3rd term) of the expression.
- Find two factors of the above multiple such that if added gives the second term (middle) and when multiplied gives the result in step 1.
- Replace the middle term with these two numbers and factorise by grouping.

Solution to example:

X2 x 6 = 6x2

Factors: 6 and 1

X2 + 6x + x + 6

X(x+6) +1(x+6)

(x+6)(x+1)

**EVALUATION**

- z2 – 2z + 1
- x2 +10x – 24

**FACTORISATION OF QUADRATIC EQUATIONS OF THE FORM ax**** 2**** +bx +c**

Example: 5x 2 -9x +4

Solution:

Product: 5x 2 x 4 = 20x 2

Factors: -5 and -4

Sum: -5-4 = -9

Hence, 5x 2 – 9x + 4

5x2 -5x -4x +4

5x(x-1)-4(x-1)

(5x-4)(x-1)

**EVALUATION**

- 2x 2 +13x +6
- 13d 2 – 11d – 2

**FACTORISATION OF TWO SQUARES**

To factorise two squares with difference, we need to remember the law guiding difference of two squares i.e. x 2 – y2 = (x + y) (x- y).

Examples**:**

- P 2 – Q2 = (P+Q) (P-Q)
- 36y 2 - 1= 6 2 y 2 - 1 2

= (6y)2 - 1 2 = ( 6y+1) (6y-1).

**EVALUATION**

- 121- y 2
- x2y2 - 42

**READING ASSIGNMENT**

Essential Mathematics for J.S.S.3 Pg29-36

Exam focus for J.S.S CE Pg101-105-

**WEEKEND ASSIGNMENT**

- The coefficient of x 2 in x 2 + 3x -5 is A. 3 B. 1 C. -5 D. 2
- Simplify e 2 – f 2 A. (e+f)(e-f) B. (e+f)(f+e) C. (e-f)(f-e) D. e+f
- Factorize x 2 +x -6 A. (x+3)(x+2) B. (x-2)(x+3) C. (x+1)(x+5) D. x + 2
- Solve by grouping 5h 2 -20h + h – 4 A. (h-4)(5h+1) B. (h+4)(5h-1) C. (h+2)(h-5) D. h - 4
- 49m 2 – 64n 2 when factorised will be A. (7m+8n)(8m+7n) B. (8m-7n)(8m+7n)

- (7m-8n)(7m+8n) D. 7m – 8n

**THEORY**

Factorise the following

- 4p2 – 12p +9q2
- f 2 – 2f + 1

**FACTORISATION OF QUADRATIC EXPRESSIONS**

Aquadratic expression has two (2) as its highest power; hence this at times is called a polynomial of the second order. The general representation of quadratic expression is ax 2 + bx + c where a ≠ 0. From above expression, a, b, and c stands for a number.

**NOTE**

- if ax 2 +bx + c= 0, this is known as quadratic equation
- a is coefficient of x2, b is coefficient of x and c is a constant term.
- When an expression contains three terms, it is known as trinomial.
- To be able to factorize trinomial, we need to convert it to contain four terms.

Examples: factorization of trinomial of the form x2 +bx + c.

- Factorise x2 +7x +6

Steps:

- Multiply the 1st and the last term (3rd term) of the expression.
- Find two factors of the above multiple such that if added gives the second term (middle) and when multiplied gives the result in step 1.
- Replace the middle term with these two numbers and factorise by grouping.

**Solution to example**:

X2 x 6 = 6x2

Factors: 6 and 1

X2 + 6x + x + 6

X(x+6) +1(x+6)

(x+6)(x+1)

**EVALUATION**

- z2 – 2z + 1
- x2 +10x – 24

**Factorisation of quadratic equations of the form ax**** 2**** +bx +c**

Example: 5x 2 -9x +4

Solution:

Product: 5x 2 x 4 = 20x 2

Factors: -5 and -4

Sum: -5-4 = -9

Hence, 5x 2 – 9x + 4

5x2 -5x -4x +4

5x(x-1)-4(x-1)

(5x-4)(x-1)

**EVALUATION**

- 2x 2 +13x +6
- 13d 2 – 11d – 2

**FACTORISATION OF TWO SQUARES**

**To **factorise two squares with difference, we need to remember the law guiding difference of two squares i.e. x 2 – y2 = (x + y) (x- y).

**Examples:**

- P 2 – Q2 = (P+Q) (P-Q)
- 36y2 - 1= 6 2 y 2 - 1 2= (6y) 2 - 1 2 = ( 6y+1) (6y-1).

**EVALUATION**

- 121- y 2
- x2y2 - 42

**READING ASSIGNMENT**

Essential Mathematics for J.S.S.3 Pg29-36

Exam focus for J.S.S CE Pg101-105-

**WEEKEND ASSIGNMENT**

- The coefficient of x 2 in x 2 + 3x -5 is (a) 3 (b) 1 (c) -5
- Simplify e 2 – f 2 (a) (e+f)(e-f) (b) (e+f)(f+e) (c) (e-f)(f-e)
- Factorize x 2 +x -6 (a) (x+3)(x+2) (b) (x-2)(x+3) (c) (x+1)(x+5)
- Solve by grouping 5h 2 -20h + h – 4 (a) (h-4)(5h+1) (b) (h+4)(5h-1) (c) (h+2)(h-5)
- 49m 2 – 64n 2 when factorised will be (a) (7m+8n)(8m+7n) (b) (8m-7n)(8m+7n)

(c) (7m-8n)(7m+8n)

**THEORY**

Factorise the following

**4p2 – 12p +9q2**

- f 2 – 2f + 1

**FACTORISATION OF QUADRATIC EXPRESSIONS**

A quadratic expression has two (2) as its highest power; hence this at times is called a polynomial of the second order. The general representation of quadratic expression is ax 2 + bx + c where a ≠ 0. From above expression, a, b, and c stands for a number.

NB:

- if ax 2 +bx + c= 0, this is known as quadratic equation
- a is coefficient of x2, b is coefficient of x and c is a constant term.
- When an expression contains three terms, it is known as trinomial.
- To be able to factorize trinomial, we need to convert it to contain four terms.

Examples: **factorization of trinomial of the form x****2**** +bx + c.**

- Factorise x2 +7x +6

Steps:

- Multiply the 1st and the last term (3rd term) of the expression.
- Find two factors of the above multiple such that if added gives the second term (middle) and when multiplied gives the result in step 1.
- Replace the middle term with these two numbers and factorise by grouping.

**Solution to example**:

X2 x 6 = 6x2

Factors: 6 and 1

X2 + 6x + x + 6

X(x+6) +1(x+6)

(x+6)(x+1)

**Evaluation**: 1. z2 – 2z + 1

- x2 +10x – 24

**Factorisation of quadratic equations of the form ax**** 2**** +bx +c**

Example: 5x 2 -9x +4

Solution:

Product: 5x 2 x 4 = 20x 2

Factors: -5 and -4

Sum: -5-4 = -9

Hence, 5x 2 – 9x + 4

5x2 -5x -4x +4

5x(x-1)-4(x-1)

(5x-4)(x-1)

**Evaluation:**

- 2x 2 +13x +6
- 13d 2 – 11d – 2

**FACTORISATION OF TWO SQUARES**

Tofactorise two squares with difference, we need to remember the law guiding difference of two squares i.e. x 2 – y2 = (x + y) (x- y).

**Examples:**

- P 2 – Q2 = (P+Q) (P-Q)
- 36y 2 - 1= 6 2 y 2 - 1 2

= (6y)2 - 1 2 = ( 6y+1) (6y-1).

**Evaluation:**

- 121- y 2
- x2y2 - 42

**READING ASSIGNMENT**

Essential Mathematics for J.S.S.3 Pg29-36

Exam focus for J.S.S CE Pg101-105

**WEEKEND ASSIGNMENT**

- The coefficient of x 2 in x 2 + 3x -5 is (a) 3 (b) 1 (c) -5
- Simplify e 2 – f 2 (a) (e+f)(e-f) (b) (e+f)(f+e) (c) (e-f)(f-e)
- Factorize x 2 +x -6 (a) (x+3)(x+2) (b) (x-2)(x+3) (c) (x+1)(x+5)
- Solve by grouping 5h 2 -20h + h – 4 (a) (h-4)(5h+1) (b) (h+4)(5h-1) (c) (h+2)(h-5)
- 49m 2 – 64n 2 when factorised will be (a) (7m+8n)(8m+7n) (b) (8m-7n)(8m+7n)

(c) (7m-8n)(7m+8n)

**THEORY**

**Factorise the following**

**4p2 – 12p +9q2**

- f 2 – 2f + 1

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