# Lesson Notes By Weeks and Term - Junior Secondary School 3

FACTORISATION

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 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.

1. 8xy + 4x2y
2. 6ab – 8a2b + 12ab

Solution:

1. 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)

1. 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

1. 9x2yz2 – 12x3z3
2. 14cd + 35cd2f
3. 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;

1. 16uv – 12vt + 20mu – 15mt
2. ap +aq +bq + bp
3. mn – pq-pn +mq

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

1. if ax 2 +bx + c= 0, this is known as quadratic equation
2. a is coefficient of x2, b is coefficient of x and c is a constant term.
3. When an expression contains three terms, it is known as trinomial.
4. 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.

1. Factorise x2 +7x +6

Steps:

1. Multiply the 1st and the last term (3rd term) of the expression.
2. 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.
3. 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
2. 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

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

1. 2x 2 +13x +6
2. 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:

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

= (6y)2  - 1 = ( 6y+1) (6y-1).

EVALUATION

1. 121- y 2
2. x2y2 - 42

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

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

WEEKEND ASSIGNMENT

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

THEORY

Factorise the following

1. 4p2 – 12p +9q2
2. f 2 – 2f + 1

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

1. if ax 2 +bx + c= 0, this is known as quadratic equation
2. a is coefficient of x2, b is coefficient of x and c is a constant term.
3. When an expression contains three terms, it is known as trinomial.
4. 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.

1. Factorise x2 +7x +6

Steps:

1. Multiply the 1st and the last term (3rd term) of the expression.
2. 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.
3. 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
2. 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

1. 2x 2 +13x +6
2. 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:

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

EVALUATION

1. 121- y 2
2. x2y2 - 42

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

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

WEEKEND ASSIGNMENT

1. The coefficient of x 2  in x 2  + 3x -5 is  (a) 3 (b) 1 (c) -5
2. Simplify e 2 – f 2  (a) (e+f)(e-f) (b) (e+f)(f+e) (c) (e-f)(f-e)
3. Factorize x 2 +x -6  (a) (x+3)(x+2) (b) (x-2)(x+3) (c) (x+1)(x+5)
4.  Solve by grouping 5h 2 -20h + h – 4 (a) (h-4)(5h+1) (b) (h+4)(5h-1) (c) (h+2)(h-5)
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

1. f 2 – 2f + 1

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:

1. if ax 2 +bx + c= 0, this is known as quadratic equation
2. a is coefficient of x2, b is coefficient of x and c is a constant term.
3. When an expression contains three terms, it is known as trinomial.
4. 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.

1. Factorise x2 +7x +6

Steps:

1. Multiply the 1st and the last term (3rd term) of the expression.
2. 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.
3. 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

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:

1. 2x 2 +13x +6
2. 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:

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

= (6y)2  - 1 = ( 6y+1) (6y-1).

Evaluation:

1. 121- y 2
2. x2y2 - 42

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

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

WEEKEND ASSIGNMENT

1. The coefficient of x 2  in x 2  + 3x -5 is  (a) 3 (b) 1 (c) -5
2. Simplify e 2 – f 2  (a) (e+f)(e-f) (b) (e+f)(f+e) (c) (e-f)(f-e)
3. Factorize x 2 +x -6  (a) (x+3)(x+2) (b) (x-2)(x+3) (c) (x+1)(x+5)
4. Solve by grouping 5h 2 -20h + h – 4 (a) (h-4)(5h+1) (b) (h+4)(5h-1) (c) (h+2)(h-5)
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

1. f 2 – 2f + 1