Chapter 1 Factorisation.

Standard Competence: 4. Understand and doing the operation of algebraic ,function, linear equation, linear equation system and using it in problems solving. Base Competence: 4.1. Solving the operation of algebraic terms. 4.2. Determining the factors of algebraic expressions. 4.3. Solving the operation of algebraic fractions. Indicators: a. Explaining the definition of coefficient, variable constant, one term, two terms and three terms in the same variable or not. b. Solving the operation of addition, subtraction, multiplication and the power of one term or two terms. c. Solving the operation of division by the like terms or unlike terms. d. Factorizing the terms of algebraic to three terms. e. Simplifying the algebraic expressions. f. Solving the power of constant and terms. g. Solving the operation of addition, subtraction, multiplication, division and the power of algebraic fractions with the denominator one term or two terms.

A.

SOLVING THE OPERATION OF ALGEBRAIC

1. The definition of coefficient, variable, constant and terms. Coefficient is a multiplying factors. Thus in 3x, 3 is the coefficient of x in -4pq, -4q is the coefficient of q Variable is a quantity whose value changes. Constant is a quantity whose value constant. Terms are parts of the expressions which are connected by plus or minus. Example : In Algebraic expression : 6 bananas + 3 apples – 2 bananas – 5 apples – 5 We can write : 6b – 3a – 2b – 5a - 5 Note: 6 is coefficient of b, 3 is coefficient of a, -2 is coefficient of b a and b is variables 5 is constant 6b and -2b are like terms 6b and - 3a are unlike terms

2. Addition and subtraction of algebraic. The algebraic expression can be simplified if only have the like terms. Example: Simplify : 3x + 7y –11x + 5y = 3x – 11x + 7y + 5y ( Grouping the like terms ) = - 8x + 12y

TASK 1 1.

Simplify :

2 mangoes + 4 apples – 5 mangoes + 3 apples + Rp5,000 = …….. - ………. + ………+ ……… + ………. = …………………

2.

Add 5x + 7 and 3(4x – 4) = ………. + …………… = ……………………….. = ………………………...

3.

Subtract 5x + 7 of 3(6x + 8) = ………….. - .…………. = ………………………… = …………………………

4.

Subtract 5x + 7 with 3(6x + 8) = ………….. - ………….. = ………………………… = …………………………

5.

Add 2p + 5 and 3(2p – 3) = ……….… + …………. = ………………………... = ………………………..

TASK 2 1.

Find the number of terms in the following algebraic expressions a. 3x + 5y – 7 d. 4xy + 7x – 5y + 8 b. 5p2 – 8p + 3 e. 2x2y – x c. 5x –2 f. –2m + 7n – 4

2.

Find the coefficients and variables in the following algebraic expressions : a. 2x + 5y – 11 c. 4m2 + 5mn - 7n b. –5xy + 7x – 10y d. 15x2y – 5x2 – 10

3.

Simplify : a. 3m + 7n – 10m + 5 c. 4(k – 5) – 2(k + 11) d. –5(p2 + 7) + 2(3p2 – 15)

b. 2x + 11x – 5x – 7x e. 12p + 15q – 3q + 4p f. 14m – 3(2m – 3)

4.

5.

Add : a. 4m + 5 and 7 – 9m b. 15k – 7 and 8k + 5m – 3

c. 5x – 11 and 2y – 11x + 7 d. 2(3x + 3) and 4(3x – 5)

Subtract : a. 5x – 8 of 7x + 10 b. 5(2x + 2) of -2x + 1

c. 2x + 12 of 15x – 5 d. 4m + 3 by 2(5x – 1)

3. Multiplication of one term, two terms and three terms

TASK 3 1.

Expand : a. 3(2x + 5y ) b. –3k(5kl + 4)

= …… + …… = …… ……

2. 3.

5 (4a + 3b – 7c) 4 (-3x + 4y – 6y)

= …………………………….. = ……………………………..

4.

(3a + 7)(2a – 8)

= …… - …. + …. - …..

(distributive property)

= ……………………………… 5.

(4m + 7)

2

= …… + ……+ …… + ….… = ……………………………. Special Product : (a + b)2 = a2 + 2ab + b2 2 (a – b) = a2 – 2ab + b2 (a + b)(a – b) = a2 – b2

4. The power of one term, two terms and three terms. The power of one term can use the properties of the power / indices. 2 2 2 (ab)(cab=)a c=b ca b

then, the power of two terms and three terms and the next can use the Pascal`s Triangle for finding its coefficient 1 1

1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1

second power…..(a +b) 2 = a2 + 2ab + b2 3-th power…… (a – b) 3 = a3 - 3a2b + 3ab2 + b3 4-th power

TASK 4 Expand : 1. (2x + 3) 3

2.

3.

4.

5.

(3p – 2) 4

= …….. + ………. + ……… + ……… = …….. + ………. + ……… + ……… = ……... + ……… + ……… + ……… = ……. …….. ……… ……… = ……. …….. ……… ……… = ……. ……. ……… ………

2 m + 3 ) 3 = ……. 3 = ……. = ……. (5p + 7) 2 = ……. = ……. = ……. 2 3 ( k - 2 ) = ……. 5 = ……. = ……. (

…….

……… ………

……. ……. …….. …….. …….

……… ……… ……… ……… ………

…….

……… ………

……. …….

……… ……… ……… ………

……… ……… ……… ……… ………

TASK 5 1. State as addition of terms : a. (x +2)(x – 4) b. (x + 7)(2x – 8) c. (2m – 5)(3m + 7)

d. 2p(4p – 3)(p + 5 ) e. 4k(2k + 7m – 11) f (2n + 4)(4n – 5m + 6)

2. Expand and simplify : a. (2x + 7)2 – (x – 3) 2 b. 2(4m – 8) 2 2 2 ) c. (15n 3n

d. (5x + 2y – 3)2 e. (3x + 7) 3 1 f. (x + )2 x

3. Find the area of rectangle ( in x, y or y), if its length and width are : a. (3x + 7) cm and 2x cm c. 5z cm and (4z – 2) cm b. (2y – 3) cm and (2y + 3) cm d. (2x – y) cm and (2x + 8) cm 4. Find the volume of cube if its edges are : a. (5x – 2)cm c. (-3y + 4) cm b. 2x cm d. (4x – 5y) cm 5. Simplify : a.16a2b : 12ab2 b. –4x2y : -2xy c. 5x2y3 : xy2

d. 32m2n X 3m4n2 e. (2m3n2 X 4mn ) : m2n f. 4mn X 8mn2 : 2mn

B.

FACTORISATIONS. When a number or an expression is the product of two or more numbers on expression, then each of latter is called Its factors Example : 15 x2 – 4

= 5X3 = (x – 2)(x + 2)

then 5 and 3 are factors of 15 then (x – 2) and (x + 2) are factors of x2 – 4

There are 4 forms of factorizations : 1. Form 1

ab + ac = a(b + c) ab – ac = a(b – c)

2. Form 2.

a2 – b2 = (a –b)(a + b)

3. Form 3

ax2 + bx + c , a = 1 x + bx + c = (x +p)(x +q) 2

Note : p . q = c p+ q = b

Example : Factorize : x2 + 7x + 12 Solution : x2 + 7x + 12 = (x + … )(x + … ) = (x + 3)(x + 4) 4. Form 4

ax2 + bx + c , a ≠ 1 (ax + p)(ax + q) ax2 + bx + c = a

12 = 1 x 12 = 2x6 = 3x4

3+4 = 7

Note : p . q = a . c p+ q = b

Example : Factorize : 3x2 + 10x - 8 Solution : (3x + … )( 3x + …) 3 (3x - 2)(3x + 12) = 3 (3x - 2). 3 (x + 4) = 3 = (3x – 2)(x + 4)

3x2 + 10x – 8 =

24 = 1 x -24 = 2 x -12 = -2x12

-2 + 12 =10

TASK 6 1. Factorize : a. 4x2 + 10x b. 12m2n – 15mn 2. Factorize : a. 4x2 – 25 b. 8p2 – 98 3. Factorize : a. x2 + 3x – 10 b. p2 – 2p – 15

= ……………… = ……………… = ……………… = ………………

4. Factorize : a. 8m2 - 14m + 4 b. 2q2 + 11q + 24 5. Factorize : a. d2 + 4d – 21 b. 4k2 –13k + 6

= ………………. = ………………. = ………………. = ……………….

= ……………… = ………………

TASK 7 1. Factorize : a. 15x2 + 20 b. 16m2n – 4m c. 20k2 – 4km + 14m

d. 38xy2 + 98x2y e. 4p(2x – 5) + p(2x – 5) f. 2(3x – 7) - (3x – 7)

2. Factorize : a. x2 – 25y2 b. 4m2 – (m+3) 2 c. 16k2 – 25

d. 9 – 25n2 e. x2 –16 f. 8x2 – 162

3. Factorize : a. x2 + 7x + 10 b. x2 + 2x + 1 c. p2 – 6p – 7 d. q2 – q – 20 e. m2 + 10m – 75 f. m2 – 2.4m + 1.44 g. k2 – 18k + 81

h. x2 + 3x – 4 i. 60 – 7y – y2 j. y2 + 9y + 14 k. a2 – 10a + 25 l. x2 + 12x + 36 m. 21 + 4x – x2 n. a2 – 3a – 28

4. Factorize : a. 4m2 + 27m – 7 b. 2x2 – 13x + 15 c. 6p2 + 13p + 6 d. 9k2 – 6k + 1

e. 8a2 + 10a – 3 f. 2x2 – x - 3 g. 36m2 – 13mn + n2 h. 12- 43x + 10x2

5.

Factorize : a. x4 – 4y4 b. x2 – (x – 2) 2 c. 4m2n + 28mn2 d. x2 – 9x – 36

e. x3 – y3 f. x3 + 27 g. 8k2 – 19k - 25 e. 21 + 11x – 2x2

C.

ALGEBRAIC FRACTION The numerators and denominators of a algebraic fractions can be factorize. Example : 1.

Simplify :

x + 5 2x − 1 + 3 2

Solution :

x + 5 2x − 1 2 ( x + 5) 3 (2 x − 1) = . + + . 3 2 2 3 3 2 =

2 x + 10 6 x − 3 + 6 6

=

8x + 3 6

(make the same denominators )

TASK 8 1. Simplify :

4 x + 10 8

= …………. = ………….

2. Simplify :

2 1 − 2x − 1 x + 3

= …………. - …………… = ………….. - …………… = ………….. - …………… = ………….. - ……………

3. Simplify :

4 x 2 − 20 x + 25 5 − 2x

= …………………. = …………………. = ………………….

3 x 2 + 10 x − 8 4. Simplify : 6 x 2 − 13 x + 6

= …………………. = ………………… = …………………

5. Simplify : 3 + x +

2x − 4 2x − 2

= ………………… = ………………… = …………………

TASK 9 Simplify : 5 x − 15 y 1. 2x − 6 y

2.

2a − 2b b−a

3.

4 w2 − 25 5 − 2w

4.

a 2bc + ab 2c − abc 2 4abc

5.

5 x − 15 x2 − 9

6.

3 x + 12 x + 4 : 10 x 5x

7.

P 4 − 16 ( x + 2)( x − 2)

8.

x +1 x + 3 − x+2 x+4

4 3 − x + 3 x +1

1 1 + x y 10. x y − y x

9.

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