Factorisation JSS3 Mathematics Lesson Note

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 HCF outside the bracket and then divide each term with the HCF. 

Example:

      X(x+6) +1(x+6)

     (x+6)(x+1)

EVALUATION

1.         z2 – 2z + 1

P       TION OF QUADRATIC EQUATIONS OF THE FORM ax 2 +bx +c

Example:  ; 5x 2 -9x +4

 Solution;   Sum: -5-4 = -9

    He; nce, 5x 2 – 9x + 4

       ;           5×2  -5x -4x +4

     ;             5x(x-1)-4(x-1)

   ;               (5x-4)(x-1)

 ; 

EVALUATION; 

1.         2x;  2 +13x +6
2.         1; 3d 2 – 11d – 2

FACTORISATION OF TWO SQUARES

To factorize two squares with a difference, we need to remember the law guiding; g the 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 2   = ( 6y+1) (6y-1). 

EVALUATION

1.       121- y 2
2.         x2y2 – 42

READING ASSIGNMENT

Essential Mathematics for J.S.S.3 Pg 29-36

Exam focus for J.S.S CE Pg 101-105-

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. (8; m-7n)(8m+7n)

1. ;          (7m-8n)(7m+8n) D. 7m – 8n

 ; 

THEORY

Factorise the following   

1.       4p2 – 12p +9q2
2. f 2 – 2f + 1

 FACTORISATION OF QUADRATIC EXPRESSIONS

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

NOTE

1.         if ax 2 +bx + c= 0, this is known as a quadratic equation
2.         a is the coefficient of x2, b is the coefficient of x and c is a constant term.

3;       When an expression contains three terms, it is known as a trinomial.

4.     To be able to factorize a 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 t; he 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 = 6×2

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

Factorization 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

5×2  -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 factorize two squares with a difference, we need to remember the law guiding the 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 2   = ( 6y+1) (6y-1).

EVALUATION

1.         121- y 2
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

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

FACTORISATION OF QUADRATIC EXPRESSIONS

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

NB:

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

Examples: factorization of trinomials 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 = 6×2

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

Factorization 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

5×2  -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 a 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 2   = ( 6y+1) (6y-1).

 Evaluation:

1.         121- y 2
2.         x2y2 – 42

 READING ASSIGNMENT

Essential Mathematics for J.S.S.3 Pg 29-36

Exam focus for J.S.S CE Pg 101-105

 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 factored 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