LESSON NOTES 2021 2022 Subject: Mathematics Chapter: 15 ...

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LESSON NOTES 2021 – 2022 Subject: Mathematics Chapter: 15 Factorisation Grade: VIII NOTE: THE CONCEPT MAP AND ALL THE EXERCISE QUESTIONS TO BE SOLVED IN NOTEBOOK. Concept Map: Factors: A factor is an exact divisor of that number. e.g.: 1, 2, 4 are the factors of 4. 1 is the factor of every number. Every number is a factor of itself. Exercise 15A 1. Find the greatest common factor of the given monomials. i) 3, 6 Greatest common factor: 3 ii) βˆ’ 5 Greatest common factor: iii) 18 7 , βˆ’9 3 Greatest common factor: 9 3 Algebraic Expression Factors Factorization Multiplication

Transcript of LESSON NOTES 2021 2022 Subject: Mathematics Chapter: 15 ...

Page 1: LESSON NOTES 2021 2022 Subject: Mathematics Chapter: 15 ...

LESSON NOTES 2021 – 2022

Subject: Mathematics Chapter: 15 Factorisation

Grade: VIII

NOTE: THE CONCEPT MAP AND ALL THE EXERCISE QUESTIONS TO BE SOLVED IN NOTEBOOK. Concept Map:

Factors: A factor is an exact divisor of that number. e.g.: 1, 2, 4 are the factors of 4.

1 is the factor of every number.

Every number is a factor of itself.

Exercise 15A

1. Find the greatest common factor of the given monomials.

i) 3π‘š, 6𝑛

Greatest common factor: 3

ii) π‘Žπ‘₯ βˆ’ 5π‘Žπ‘

Greatest common factor: π‘Ž

iii) 18π‘₯7, βˆ’9π‘₯3

Greatest common factor: 9π‘₯3

Algebraic Expression Factors

Factorization

Multiplication

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LESSON NOTES 2021 – 2022

Subject: Mathematics Chapter: 15 Factorisation

Grade: VIII

iv) 10π‘₯𝑦, 15π‘₯2𝑦3

Greatest common factor: 5π‘₯𝑦

v) 38π‘Ž3π‘₯5𝑦, 15π‘₯2𝑦3

Greatest common factor: 19π‘Ž3π‘₯3

vi) 9π‘₯𝑦2, βˆ’6π‘₯𝑦, βˆ’3π‘₯

Greatest common factor: 3π‘₯

vii) 14π‘₯3𝑦3, 21π‘₯5𝑦2, 35π‘₯6𝑦6

Greatest common factor: 7π‘₯3𝑦2

viii) 𝑝2π‘ž8π‘Ÿ3, βˆ’π‘6π‘ž7π‘Ÿ3, π‘π‘ž5π‘Ÿ2

Greatest common factor: π‘π‘ž5π‘Ÿ2

ix) 24π‘Ž2𝑏, βˆ’16π‘Žπ‘π‘, 32π‘Ž2𝑏2𝑐3, 36π‘Žπ‘2𝑐2

Greatest common factor: 4π‘Žπ‘

Factorise the following expressions.

2. π‘Ž2 βˆ’ π‘Žπ‘₯

=π‘Ž(π‘Ž βˆ’ π‘₯)

3. 3π‘Ž2 βˆ’ 9

=3(π‘Ž2 βˆ’ 3)

4. 5π‘Žπ‘₯ βˆ’ 5π‘Ž2π‘₯2

=5π‘Žπ‘₯(1 βˆ’ π‘Žπ‘₯)

5. πœ‹π‘…2 βˆ’ πœ‹π‘Ÿ2

=πœ‹(𝑅2 βˆ’ π‘Ÿ2)

6. 10π‘₯𝑦 βˆ’ 15π‘₯2𝑦2

=5π‘₯𝑦(2 βˆ’ 3π‘₯𝑦)

7. 8π‘Ž4𝑏2𝑐3 + 12π‘Ž2𝑏𝑐3

=4π‘Ž2𝑏𝑐3(2π‘Ž2𝑏 + 3)

8. π‘₯2 βˆ’ π‘₯𝑦 + π‘₯𝑦2

=π‘₯(π‘₯ βˆ’ 𝑦 + 𝑦2)

9. 5π‘₯4 βˆ’ 10π‘Ž2π‘₯2 βˆ’ 15π‘Ž2π‘₯3

=5π‘₯2(π‘₯2 βˆ’ 2π‘Ž2π‘₯2 βˆ’ 3π‘Ž2π‘₯)

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Subject: Mathematics Chapter: 15 Factorisation

Grade: VIII

10. 3π‘₯3𝑦2 + 9π‘₯2𝑦 + 12π‘₯𝑦

=3π‘₯𝑦(π‘₯2𝑦 + 3π‘₯ + 4)

11. 60π‘₯2𝑦 + 25𝑦2 + 20𝑦2𝑧

Refer que, no. 10

12. π‘Žπ‘₯3𝑦2 + 𝑏π‘₯2𝑦 + 𝑐π‘₯𝑦𝑧

Refer que, no. 10

13. 8π‘₯4 𝑦 + 2π‘₯𝑦4 βˆ’ 24π‘₯2𝑦3 + 18π‘₯3𝑦2

Refer que, no. 10

14. 3π‘₯(π‘Ž βˆ’ 𝑏) βˆ’ 6𝑦(π‘Ž βˆ’ 𝑏)

=3(π‘Ž βˆ’ 𝑏)(π‘₯ βˆ’ 2𝑦)

15. (π‘₯ βˆ’ 𝑦)2 βˆ’ 2(π‘₯ βˆ’ 𝑦)

=(π‘₯ βˆ’ 𝑦)(π‘₯ βˆ’ 𝑦 βˆ’ 2)

16. βˆ’3(π‘₯ βˆ’ 3𝑦) + 6(π‘₯ βˆ’ 3𝑦)2

=βˆ’3(π‘₯ βˆ’ 3𝑦)(1 βˆ’ 2(π‘₯ βˆ’ 3𝑦))

=βˆ’3(π‘₯ βˆ’ 3𝑦)(1 βˆ’ 2π‘₯ + 6𝑦)

17. (π‘₯ + 3)3 βˆ’ 2(π‘₯ = 3)2 + 3(π‘₯3)

=(π‘₯ + 3)[(π‘₯ + 3)2 βˆ’ 2(π‘₯ + 3) + 3]

=(π‘₯ + 3)[π‘₯2 + 6π‘₯ + 9 βˆ’ 2π‘₯ βˆ’ 6 + 3]

=(π‘₯ + 3)(π‘₯2 + 4π‘₯ + 6)

18. π‘₯(π‘Ž + 𝑏)4 + 𝑦(π‘Ž + 𝑏)3 + 𝑧(π‘Ž + 𝑏)2

=(π‘Ž + 𝑏)2 [ π‘₯(π‘Ž + 𝑏)2 + 𝑦(π‘Ž + 𝑏) + 𝑧]

=(π‘Ž + 𝑏)2 [ π‘₯( π‘Ž2 + 2π‘Žπ‘ + 𝑏2) + π‘¦π‘Ž + 𝑦𝑏 + 𝑧]

=(π‘Ž + 𝑏)2 [ π‘₯π‘Ž2 + 2π‘₯π‘Žπ‘ + π‘₯𝑏2 + π‘¦π‘Ž + 𝑦𝑏 + 𝑧]

19. 25(π‘₯ βˆ’ 𝑦)2 βˆ’ 15(π‘₯ βˆ’ 𝑦) βˆ’ π‘₯ + 𝑦

Refer Que 18

20. π‘₯2 + π‘₯𝑦 + π‘₯𝑧 + 𝑦𝑧

=π‘₯(π‘₯ + 𝑦) + 𝑧(π‘₯ + 𝑦)

=(π‘₯ + 𝑦)(π‘₯ + 𝑧)

21. π‘Ž2 βˆ’ π‘Žπ‘ + π‘Žπ‘ βˆ’ 𝑏𝑐

=Refer Que 20

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Subject: Mathematics Chapter: 15 Factorisation

Grade: VIII

22. π‘Žπ‘₯ βˆ’ 𝑏π‘₯ βˆ’ π‘Žπ‘§ + 𝑏𝑧

=π‘₯(π‘Ž βˆ’ 𝑏) βˆ’ 𝑧(π‘Ž βˆ’ 𝑏)

=(π‘Ž βˆ’ 𝑏)(π‘₯ βˆ’ 𝑧)

23. 4π‘₯𝑦 βˆ’ 𝑦2 + 8π‘₯𝑧 βˆ’ 2𝑦𝑧

=𝑦(4π‘₯ βˆ’ 𝑦) + 2𝑧(4π‘₯ βˆ’ 𝑦)

=(4π‘₯ βˆ’ 𝑦)(𝑦 + 2𝑧)

24. 𝑏π‘₯ + 2π‘₯𝑧 + 2𝑏𝑦 + 4𝑦𝑧

=π‘₯(𝑏 + 2𝑧) + 2𝑦(𝑏 = 2𝑧)

=(𝑏 + 2𝑧)(π‘₯ + 2𝑦)

25. 2π‘₯2 + 3π‘₯𝑦 βˆ’ 2π‘₯𝑧 βˆ’ 3𝑦𝑧

Refer Que 20

26. 𝑝2π‘₯ βˆ’ π‘π‘žπ‘¦ βˆ’ 2𝑝π‘₯ βˆ’ 2π‘žπ‘¦

=Refer Que 20

27. π‘Žπ‘2 βˆ’ (π‘Ž βˆ’ 𝑏)𝑐 βˆ’ 𝑏

=π‘Žπ‘2 βˆ’ π‘Žπ‘ + 𝑏𝑐 βˆ’ 𝑏

=π‘Žπ‘(𝑐 βˆ’ 1) + 𝑏(𝑐 βˆ’ 1)

=(π‘Žπ‘ + 𝑏)(𝑐 βˆ’ 1)

28. π‘₯3 βˆ’ π‘₯2 + π‘₯ βˆ’ 1

=π‘₯2(π‘₯ βˆ’ 1) + 1(π‘₯ βˆ’ 1)

=(π‘₯ βˆ’ 1)(π‘₯2 + 1)

29. π‘₯3 + 2π‘₯ βˆ’ 2π‘₯2 βˆ’ 4

=π‘₯(π‘₯2 + 2) βˆ’ 2(π‘₯2 + 2)

=(π‘₯2 + 2) (x-2)

30. π‘Žπ‘2 βˆ’ 𝑏𝑐2 βˆ’ π‘Žπ‘ + 𝑐2

= 𝑏(π‘Žπ‘ βˆ’ 𝑐2) βˆ’ 1(π‘Žπ‘ βˆ’ 𝑐2)

= (π‘Žπ‘ βˆ’ 𝑐2)(𝑏 βˆ’ 1)

31. 8 βˆ’ 4π‘₯ βˆ’ 2π‘₯3 + π‘₯4

= 4(2 βˆ’ π‘₯) βˆ’ 𝑐π‘₯3(2 βˆ’ π‘₯)

= (4 βˆ’ π‘₯3)(2 βˆ’ π‘₯)

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Subject: Mathematics Chapter: 15 Factorisation

Grade: VIII

32. π‘₯3 βˆ’ 2π‘₯𝑦3 βˆ’ 4𝑦3 + 2π‘₯2𝑦

=π‘₯(π‘₯2 βˆ’ 2𝑦2) + 2𝑦 (βˆ’2𝑦2 + π‘₯2)

=(π‘₯2 βˆ’ 2𝑦2)(π‘₯ + 2𝑦)

33. π‘Žπ‘₯𝑦 + 𝑏𝑐π‘₯𝑦 βˆ’ π‘Žπ‘§ βˆ’ 𝑏𝑐𝑧

Refer Que 22

34. 2π‘Žπ‘₯2 + 3π‘Žπ‘₯𝑦 βˆ’ 2𝑏π‘₯𝑦 βˆ’ 3𝑏𝑦2

Refer Que 25

35. π‘Žπ‘₯ + 𝑏π‘₯ βˆ’ π‘Žπ‘¦ βˆ’ 𝑏𝑦 + π‘Žπ‘§ + 𝑏𝑧

=π‘₯(π‘Ž + 𝑏) βˆ’ 𝑦(π‘Ž + 𝑏) + 𝑧(π‘Ž + 𝑏)

=(π‘Ž + 𝑏)(π‘₯ βˆ’ 𝑦 + 𝑧)

36. π‘Žπ‘₯ βˆ’ 𝑏π‘₯ + 𝑏𝑦 + 𝑐𝑦 βˆ’ 𝑐π‘₯ βˆ’ π‘Žπ‘¦

Refer Que 35

37. (π‘Žπ‘₯ + 𝑏𝑦)2 βˆ’ (𝑏π‘₯ + π‘Žπ‘¦)2

=[π‘Žπ‘₯ + 𝑏𝑦 + 𝑏π‘₯ + π‘Žπ‘¦][(π‘Žπ‘₯ + 𝑏𝑦) βˆ’ (𝑏π‘₯ βˆ’ π‘Žπ‘¦)]

=[π‘₯(π‘Ž + 𝑏) + 𝑦(π‘Ž + 𝑏)][π‘₯(π‘Ž βˆ’ 𝑏) βˆ’ 𝑦(π‘Ž βˆ’ 𝑏)]

=(π‘Ž + 𝑏)(π‘₯ + 𝑦)(π‘Ž βˆ’ 𝑏)(π‘Ž βˆ’ 𝑦)

Factorization Using Identifiers

Factorization when the given expression is a perfect square.

We know that

A] π‘Ž2 + 2π‘Žπ‘ + 𝑏2 = (π‘Ž + 𝑏)2

B] π‘Ž2 βˆ’ 2π‘Žπ‘ + 𝑏2 = (π‘Ž βˆ’ 𝑏)2

Exercise 15B

Factorise the following algebraic expression

1. π‘₯2 βˆ’ 16π‘₯ + 64

= (π‘₯)2 βˆ’ 2 Γ— 8 Γ— π‘₯ + (8)2

= (π‘₯ βˆ’ 8)2

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LESSON NOTES 2021 – 2022

Subject: Mathematics Chapter: 15 Factorisation

Grade: VIII

2. 1 βˆ’ 2π‘₯ + π‘₯2

= (1)2 βˆ’ 2 Γ— 1 Γ— π‘₯ + (π‘₯)2

= (1 βˆ’ π‘₯)2

3. 9 + 6𝑝 + 𝑝2

Refer que 2

4. π‘Ž2 βˆ’ 10π‘Ž + 25

Refer que 1

5. 1 βˆ’ 6π‘₯ + 9π‘₯2

= (1)2 βˆ’ 2 Γ— 3 Γ— π‘₯ + (3π‘₯)2

= (1 βˆ’ 3π‘₯)2

6. 49π‘₯2 βˆ’ 14π‘₯ + 1

= (7π‘₯)2 βˆ’ 2 Γ— 7 Γ— π‘₯ + (1)2

= (7π‘₯ βˆ’ 1)2

7. 49π‘₯2 βˆ’ 84π‘₯𝑦 + 36𝑦2

= (7π‘₯)2 βˆ’ 2 Γ— 7π‘₯ Γ— 6𝑦 + (6𝑦)2

= (7π‘₯ + 6𝑦)2

8. 4𝑦2 βˆ’ 8𝑦 + 4

Refer que 6

9. (π‘Ž + 𝑏)2 βˆ’ 4π‘Žπ‘

= π‘Ž2 βˆ’ 2π‘Žπ‘ + 𝑏2 βˆ’ 4π‘Žπ‘

= π‘Ž2 βˆ’ 2π‘Žπ‘ + 𝑏2

= (π‘Ž βˆ’ 𝑏)2

10. π‘₯4 βˆ’ 2π‘₯2𝑦2 + 𝑦4

= ((π‘₯)2)2 + 2 Γ— π‘₯2 Γ— 𝑦2 + ((𝑦)2)2

= (π‘₯2 + 𝑦2) 2

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LESSON NOTES 2021 – 2022

Subject: Mathematics Chapter: 15 Factorisation

Grade: VIII

11. π‘₯2

4+ 2π‘₯2𝑦2 + 𝑦4

= (π‘₯2)2 + 2 Γ— π‘₯2 Γ— (𝑦2) 2

= (π‘₯

2+ 1) 2

12. π‘₯2𝑦2 βˆ’ 6π‘₯𝑦𝑧 + 𝑦2

= (π‘₯𝑦)2 βˆ’ 2 Γ— π‘₯𝑦 Γ— 3𝑧 + (3𝑧)2

= (π‘₯𝑦 βˆ’ 3𝑧) 2

13. π‘Ž2

4βˆ’ 3π‘Žπ‘3 + 9𝑏6

= (π‘Ž

2) 2 βˆ’ 2 Γ—

π‘Ž

2Γ— 3𝑏3 + (3𝑏3) 2

= (π‘Ž

2βˆ’ 3𝑏3) 2

14. (π‘Ž2

9) + (

𝑏2

64) + (

π‘Žπ‘

12)

=π‘Ž2

9+

π‘Žπ‘

12+

𝑏2

64

= (π‘Ž

3) 2 + 2 Γ—

π‘Ž

3Γ—

𝑏

8+ (

𝑏

8) 2

= (π‘Ž

3βˆ’

𝑏

8) 2

15. π‘Ž2

16βˆ’

π‘Žπ‘

6+

𝑏2

9

Refer que 1

* When the given expression is a difference of two squares:

In this case we use identity

π‘Ž2 βˆ’ 𝑏2 = (π‘Ž + 𝑏)(π‘Ž βˆ’ 𝑏)

Exercise 15C

Factorise the following Algebraic Expressions:-

1. 4π‘₯2 βˆ’ 9𝑦2

= (2π‘₯)2 βˆ’ (3𝑦)2

= (2π‘₯ + 3𝑦)(2π‘₯ βˆ’ 3𝑦)

2. 81𝑝2 βˆ’ 121

= (9𝑝)2 βˆ’ (11)2

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Subject: Mathematics Chapter: 15 Factorisation

Grade: VIII

3. 100 βˆ’ 49π‘Ž2

= (10)2 βˆ’ (7π‘Ž)2

= (10 βˆ’ 7π‘Ž)(10 + 7π‘Ž)

4. π‘Ž5 βˆ’ 16π‘Ž3

= π‘Ž3[π‘Ž2 βˆ’ 16]

= π‘Ž3[π‘Ž2 βˆ’ 42]

= π‘Ž3(π‘Ž βˆ’ 4)(π‘Ž + 4)

5. 12π‘₯2 βˆ’ 75𝑦2

= (√12π‘₯)2

βˆ’ (√75𝑦)2

= (√4 Γ— 3π‘₯)2

βˆ’ (√25 Γ— 3𝑦)2

= (2√3π‘₯) 2 βˆ’ (5√3𝑦) 2

= [(2√3π‘₯) βˆ’ (5√3𝑦) ] [ (2√3π‘₯) + (5√3𝑦)]

6. π‘₯3 βˆ’ 64π‘₯

Refer que 4

7. 15π‘₯5 βˆ’ 135π‘₯3

Refer que 4

8. 81𝑝7 βˆ’ 𝑝3π‘ž4

= 𝑝3(81𝑝4 βˆ’ π‘ž4)

= 𝑝3 [ (9𝑝)2 βˆ’ (π‘ž2) 2 ]

= 𝑝3 (9𝑝 βˆ’ π‘ž2)(9𝑝 + π‘ž2)

9. π‘₯2

36βˆ’

𝑦2

25

= (π‘₯

6)2 βˆ’ (

𝑦

5)2

= (π‘₯

6+

𝑦

5) (

π‘₯

6βˆ’

𝑦

5)

10. 49x2

144βˆ’

16x2

25

Refer que. 9

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Subject: Mathematics Chapter: 15 Factorisation

Grade: VIII

11. π‘₯2

8βˆ’

𝑦2

32

Refer que. 9

12. 1 βˆ’ (π‘₯ βˆ’ 𝑦)2

= (1)2 βˆ’ (π‘₯ βˆ’ 𝑦)2

= (1 + (π‘₯ βˆ’ 𝑦))(1 βˆ’ (π‘₯ βˆ’ 𝑦))

= (1 + π‘₯ βˆ’ 𝑦)(1 βˆ’ π‘₯ + 𝑦)

13. (2π‘₯ + 3𝑦)2 βˆ’ 16𝑧2

Refer que. 12

14. 4(π‘₯ βˆ’ 𝑦)2 βˆ’ (π‘₯ + 𝑦)2

= (2(π‘₯ βˆ’ 𝑦))2 βˆ’ (π‘₯ + 𝑦)2

= (2π‘₯ βˆ’ 2𝑦 + π‘₯ + 𝑦)(2π‘₯ βˆ’ 2𝑦 βˆ’ π‘₯ βˆ’ 𝑦)

= (3π‘₯ βˆ’ 𝑦)(π‘₯ βˆ’ 3𝑦)

15. 25 βˆ’ π‘Ž2 βˆ’ 2π‘Žπ‘ βˆ’ 𝑏2

= 25 βˆ’ (π‘Ž2 + 2π‘Žπ‘ + 𝑏2)

= 52 βˆ’ (π‘Ž + 𝑏)2

= (5 + (π‘Ž + 𝑏))(5 βˆ’ (π‘Ž + 𝑏))

= (5 + π‘Ž + 𝑏)(5 βˆ’ π‘Ž βˆ’ 𝑏)

16. π‘₯4 + 𝑦4 + π‘₯2𝑦2

Add and subtract 2π‘₯2𝑦2

= (π‘₯)22+ 2π‘₯2𝑦2 + (𝑦)22

+ π‘₯2𝑦2 βˆ’ 2π‘₯2𝑦2

= (π‘₯2 + 𝑦2)2 βˆ’ π‘₯2𝑦2

= [(π‘₯2 + 𝑦2) + π‘₯𝑦][(π‘₯2 + 𝑦2) βˆ’ π‘₯𝑦]

17. π‘₯4 + π‘₯2 + 1 [𝐻𝑖𝑛𝑑: π‘Žπ‘‘π‘‘ π‘Žπ‘›π‘‘ π‘ π‘’π‘π‘‘π‘Ÿπ‘Žπ‘π‘‘ π‘₯2]

Refer que 16.

18. π‘₯4 + 4 [𝐻𝑖𝑛𝑑: π‘Žπ‘‘π‘‘ π‘Žπ‘›π‘‘ π‘ π‘’π‘π‘‘π‘Ÿπ‘Žπ‘π‘‘ 4π‘₯2]

Refer que 16.

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Subject: Mathematics Chapter: 15 Factorisation

Grade: VIII

19. Evaluate

(203)2 βˆ’ (197)2

= (203 + 197)(203 βˆ’ 197)

= (400)(6)

= 2400

20. Evaluate

(6.6)2 βˆ’ (3.4)2

Refer que 19.

21. Simplify

0.75 Γ— 0.75 βˆ’ 0.25 Γ— 0.25

0.75 βˆ’ 0.25

=(0.75)2 βˆ’ (0.25)2

0.75 βˆ’ 0.25

=(0.75 + 0.25)(0.75 βˆ’ 0.25)

(0.75 βˆ’ 0.25)

= 0.75 + 0.25

= 1

* Factorization of Quadratic Trinomial

In order to factories quadratic trinomial π’™πŸ + 𝒑𝒙 + 𝒒 , two numbers a&b have to be found such that a+b=p and ab=q

Then, π’™πŸ + 𝒑𝒙 + 𝒒 = π’™πŸ + (𝒂 + 𝒃)𝒙 + 𝒂𝒃[π’”π’–π’ƒπ’”π’•π’Šπ’–π’•π’Šπ’π’ˆ 𝒑 & 𝒒]

= (𝒙 + 𝒂)(𝒙 + 𝒃)

Exercise 15D

Factorise the following algebraic expression

1. π‘₯2 + 7π‘₯ + 12

= π‘₯2 + 4π‘₯ + 3π‘₯ + 12

= π‘₯(π‘₯ + 4) + 3(π‘₯ + 4)

= (π‘₯ + 4)(π‘₯ + 3)

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2. π‘₯2 + 9π‘₯ + 20

= π‘₯2 + 5π‘₯ + 4π‘₯ + 20

= π‘₯(π‘₯ + 5) + 4(π‘₯ + 5)

= (π‘₯ + 5)(π‘₯ + 4)

3. 𝑝2 βˆ’ 2𝑝 βˆ’ 15

= 𝑝2 βˆ’ 5𝑝 + 3𝑝 βˆ’ 15

= 𝑝(𝑝 βˆ’ 5) + 3(𝑝 βˆ’ 5)

= 𝑝(𝑝 βˆ’ 5) + 3(𝑝 βˆ’ 5)

= (𝑝 βˆ’ 5)(𝑝 + 3)

4. π‘₯2 βˆ’ 18π‘₯ + 65

= π‘₯2 βˆ’ 13π‘₯ βˆ’ 5π‘₯ + 65

= π‘₯(π‘₯ βˆ’ 13) βˆ’ 5(π‘₯ βˆ’ 13)

= (π‘₯ βˆ’ 13)(π‘₯ βˆ’ 5)

5. π‘Ž2 βˆ’ 11π‘Ž βˆ’ 80

Refer que 3

6. 𝑝2 βˆ’ 11𝑝 βˆ’ 102

= 𝑝2 βˆ’ 17𝑝 + 6𝑝 βˆ’ 102

= 𝑝(𝑝 βˆ’ 17) + 6(𝑝 βˆ’ 17)

= (𝑝 βˆ’ 17) + 6(𝑝 βˆ’ 17)

= (𝑝 βˆ’ 17)(𝑝 + 6)

7. π‘š2 βˆ’ 26π‘š + 120

Refer que 4

8. 3π‘₯2 + 10π‘₯ + 8

= 3π‘₯2 + 6π‘₯ + 4π‘₯ + 8

= 3π‘₯(π‘₯ + 2) + 4(π‘₯ + 2)

= (π‘₯ + 2)(3π‘₯ + 4)

9. 3π‘₯2 + 10π‘₯ + 8

Refer que 8

10. 35 βˆ’ 33π‘₯ + 4π‘₯2

= 4π‘₯2 βˆ’ 33π‘₯ + 35

= 4π‘₯2 βˆ’ 28π‘₯ βˆ’ 5π‘₯ + 35

= 4π‘₯(π‘₯ βˆ’ 7) βˆ’ 5(π‘₯ βˆ’ 7)

= (4π‘₯ βˆ’ 5)(π‘₯ βˆ’ 7)

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11. 12π‘₯2 βˆ’ 29π‘₯ + 15

Refer que 10

12. 4π‘₯2 βˆ’ 8π‘₯ + 3

Refer que 10

13. π‘₯2 + 10π‘₯𝑦 + 21𝑦2

= π‘₯2 + 7𝑦π‘₯ + 3π‘₯𝑦 + 21𝑦2

= π‘₯(π‘₯ + 7𝑦) + 3𝑦(π‘₯ + 7𝑦)

= (π‘₯ + 7𝑦)(π‘₯ + 3𝑦)

14. π‘Ž2 βˆ’ π‘Žπ‘ βˆ’ 20𝑏2

= π‘Ž2 βˆ’ 5π‘Žπ‘ + 4π‘Žπ‘ βˆ’ 20𝑏2

= π‘Ž(π‘Ž βˆ’ 5𝑏) + 4𝑏(π‘Ž βˆ’ 5𝑏)

= (π‘Ž βˆ’ 5𝑏)(π‘Ž + 4𝑏)

15. 6 βˆ’ 2π‘₯2 βˆ’ π‘₯

= βˆ’2π‘₯2 βˆ’ π‘₯ + 6

= 2π‘₯2 + 4π‘₯ βˆ’ 3π‘₯ βˆ’ 6

= 2π‘₯(π‘₯ βˆ’ 2) βˆ’ 3(π‘₯ βˆ’ 2)

= (π‘₯ βˆ’ 2)(2π‘₯ βˆ’ 3)

16. 14π‘Ž2 + 11π‘Žπ‘ βˆ’ 15𝑏2

Refer que 10.

17. 36π‘₯2 + 12π‘₯𝑦𝑧 βˆ’ 15𝑦2𝑧2

= 3(12π‘₯2 + 4π‘₯𝑦𝑧 βˆ’ 5𝑦2𝑧2)

= 3(12π‘₯2 + 10π‘₯𝑦𝑧 βˆ’ 6π‘₯𝑦𝑧 βˆ’ 5𝑦2𝑧2)

= 3[2π‘₯(6π‘₯ + 5𝑦𝑧) βˆ’ 𝑦𝑧(6π‘₯ + 5𝑦𝑧)]

= 3(6π‘₯ + 5𝑦𝑧)(2π‘₯ βˆ’ 𝑦𝑧) Division of algebraic Expressions

1. Division of monomial by another monomial:

Quotient of two monomials is the product of the quotient of its numerical coefficients and the quotient of their variables.

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Grade: VIII

2. Division of polynomial by a monomial

In order to divide a polynomial by a monomial, each terms of the polynomial is divided by the monomial. 3. Division of a polynomial by a polynomial

In order to divide a polynomial by another polynomial, both the polynomial are factorised and their common factor is cancelled. Division of a polynomial by another polynomial by long division method. The following steps have to be followed to divide a polynomial by another polynomial by long division method. i) Arrange the term of the divisor and the dividend in descending order of their degrees.

ii) In order to obtain the first term of the quotient, divide the first term of the dividend by the first term

of the divisor.

iii) Multiply all the terms of the divisor by the first term of the quotient and subtract the result obtained

from the dividend to get the reminder.

iv) If the reminder is not equal to zero, then consider it a new dividend and proceed as a before.

v) Continue with the same process still a reminder equal to zero or a polynomial of degree less than the

divisor is obtained.

Exercise 15E

1. Divide the following algebraic expressions

i) πŸ‘πŸ“π’™πŸ”π’šπŸ’ 𝑏𝑦 7π’™πŸπ’šπŸ

=35π‘₯6π’šπŸ’

7π’™πŸπ’šπŸ

= (35

7) (

π‘₯6

π‘₯2) (𝑦4

𝑦2)

= 5π‘₯6βˆ’2𝑦4βˆ’2 = 5π‘₯4𝑦2

ii) 12π‘Ž6𝑏6𝑐5 𝑏𝑦 βˆ’ 3π‘Ž4𝑏2𝑐

=12π‘Ž6𝑏6𝑐5

βˆ’3π‘Ž4𝑏2𝑐

= (12

βˆ’3) (

π‘Ž6

π‘Ž4)(

𝑏6

𝑏2)(

𝑐5

𝑐)

= (βˆ’4)(π‘Ž6βˆ’4)(𝑏6βˆ’2)(𝑐5βˆ’1)

= βˆ’4π‘Ž2𝑏4𝑐4

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iii) βˆ’63π‘Ž7𝑏8𝑐3 𝑏𝑦 βˆ’9π‘Ž5𝑏5𝑐3

=βˆ’63π‘Ž7𝑏8𝑐3

βˆ’9π‘Ž5𝑏5𝑐3

= (βˆ’63

βˆ’9)(

π‘Ž7

π‘Ž5)(

𝑏8

𝑏5)(

𝑐3

𝑐3)

= 7π‘Ž2𝑏3

iv) 16𝑏2𝑦π‘₯2𝑏𝑦 βˆ’ 2π‘₯𝑦 Refer que no. iii

v) 4(π‘₯4𝑦4 βˆ’ 2π‘₯3𝑦2 + 3π‘₯𝑦2)𝑏𝑦 βˆ’ 2π‘₯𝑦 Refer que no. iii

vi) βˆ’3π‘₯2 + 9

2π‘₯𝑦 βˆ’ 6π‘₯𝑧 𝑏𝑦

βˆ’3

2π‘₯

=βˆ’3π‘₯2 +

92 π‘₯𝑦 βˆ’ 6π‘₯𝑧

βˆ’32 π‘₯

=βˆ’3π‘₯2

βˆ’32 π‘₯

+

92 π‘₯𝑦

βˆ’32 π‘₯

βˆ’6π‘₯𝑧

βˆ’32 π‘₯

= βˆ’3(2

βˆ’3) (

π‘₯2

π‘₯) +

9

2 (

βˆ’2

3) (

π‘₯𝑦

π‘₯) βˆ’ 6(

2

βˆ’3)(

π‘₯𝑧

π‘₯)

= 2π‘₯ βˆ’ 3𝑦 + 4𝑧

2. Divide the following

i) (7π‘₯ βˆ’ 35)𝑏𝑦7

=7π‘₯ βˆ’ 35

7

=7π‘₯

7βˆ’

35

7

= π‘₯ βˆ’ 5

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ii) (6π‘₯ + 33) Γ· (2x + 11)

=(6π‘₯ + 33)

(2x + 11)

=3(2π‘₯ + 11)

(2x + 11)

= 3

iii) 72π‘Žπ‘(π‘Ž+5) (π‘βˆ’4)Γ· 36π‘Ž(𝑏 βˆ’ 4)

=72ab(a + 5)(b βˆ’ 4)

36π‘Ž(𝑏 βˆ’ 4)

= 2𝑏(π‘Ž + 5)

iv) 28(π‘₯ + 3)(π‘₯2 + 3π‘₯ + 7) Γ· 7(π‘₯ + 3)

=28(π‘₯ + 3)(π‘₯2 + 3π‘₯ + 7)

7(π‘₯ + 3)

= 28(π‘₯2 + 3π‘₯ + 7)

3. Factorise the expression and divide as directed.

i) (π‘₯2 βˆ’ 23π‘₯ + 132) Γ· (π‘₯ βˆ’ 11)

Let’s factorise (π‘₯2 βˆ’ 23π‘₯ + 132) (π‘₯2 βˆ’ 23π‘₯ + 132) = (π‘₯2 βˆ’ 12π‘₯ βˆ’ 11π‘₯ + 132)

= π‘₯(π‘₯ βˆ’ 12) βˆ’ 11(π‘₯ βˆ’ 12) = (π‘₯ βˆ’ 12)(π‘₯ βˆ’ 11)

(π‘₯2 βˆ’ 23π‘₯ + 132) Γ· (π‘₯ βˆ’ 11)

=(π‘₯2 βˆ’ 23π‘₯ + 132)

(π‘₯ βˆ’ 11)

=(π‘₯ βˆ’ 12)(π‘₯ βˆ’ 11)

(π‘₯ βˆ’ 11)

= (π‘₯ βˆ’ 12)

ii) 3𝑦2 + 12𝑦 βˆ’ 63 Γ· (𝑦 + 7)

Refer que no. i

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iii) 14𝑦𝑧(𝑧2 βˆ’ 5𝑧 βˆ’ 36) Γ· 7𝑧(𝑧 βˆ’ 9)

Let’s factorise (𝑧2 βˆ’ 5𝑧 βˆ’ 36) (𝑧2 βˆ’ 5𝑧 βˆ’ 36) = (𝑧2 βˆ’ 9𝑧 + 4𝑧 βˆ’ 36) = 𝑧(𝑧 βˆ’ 9) + 4(𝑧 βˆ’ 9) = (𝑧 βˆ’ 9)(𝑧 + 4)

14𝑦𝑧(𝑧2 βˆ’ 5𝑧 βˆ’ 36) Γ· 7𝑧(𝑧 βˆ’ 9)

=14𝑦𝑧(𝑧2 βˆ’ 5𝑧 βˆ’ 36)

7𝑧(𝑧 βˆ’ 9)

=14𝑦𝑧(𝑧 βˆ’ 9)(𝑧 + 4)

7𝑧(𝑧 βˆ’ 9)

= 2𝑦(𝑧 + 4)

iv) 15π‘Žπ‘(25π‘Ž2 βˆ’ 16𝑏2) Γ· 5π‘Žπ‘(5π‘Ž + 4𝑏)

Refer que. No. iii

4. Divide the following polynomials by binomials.

i) π‘₯2 βˆ’ 11π‘₯ + 30 𝑏𝑦 (π‘₯ βˆ’ 5)

Let’s factorise π‘₯2 βˆ’ 11π‘₯ + 30 = π‘₯2 βˆ’ 6π‘₯ βˆ’ 5π‘₯ + 30 = (π‘₯ βˆ’ 6)(π‘₯ βˆ’ 5)

π‘₯2 βˆ’ 11π‘₯ + 30 𝑏𝑦 (π‘₯ βˆ’ 5)

=π‘₯2βˆ’11π‘₯+30

(π‘₯βˆ’5)

=(π‘₯βˆ’6)(π‘₯βˆ’5)

(π‘₯βˆ’5)

= (π‘₯ βˆ’ 6)

ii) 2𝑦2 + 17𝑦 + 21 𝑏𝑦 2𝑦 + 3

Let’s factorise 2𝑦2 + 17𝑦 + 21 = 2𝑦2 + 14𝑦 + 3𝑦 + 21

= 2𝑦(𝑦 + 7) + 3(𝑦 + 7) = (𝑦 + 7)(2𝑦 + 3)

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2𝑦2 + 17𝑦 + 21 𝑏𝑦 2𝑦 + 3

=2𝑦2 + 17𝑦 + 21

2𝑦 + 3

=(𝑦 + 7)(2𝑦 + 3)

2𝑦 + 3

= 𝑦 + 7

iii) 6π‘Ž2 βˆ’ 7π‘Ž βˆ’ 3 𝑏𝑦 2π‘Ž βˆ’ 3

Let’s factorise 6π‘Ž2 βˆ’ 7π‘Ž βˆ’ 3 = 6π‘Ž2 βˆ’ 9π‘Ž + 2π‘Ž βˆ’ 3 = 3π‘Ž(2π‘Ž βˆ’ 3) + (2π‘Ž βˆ’ 3) = (2π‘Ž βˆ’ 3)(3π‘Ž + 1)

6π‘Ž2 βˆ’ 7π‘Ž βˆ’ 3 𝑏𝑦 2π‘Ž βˆ’ 3

=6π‘Ž2 βˆ’ 7π‘Ž βˆ’ 3

2π‘Ž βˆ’ 3

=(2π‘Ž βˆ’ 3)(3π‘Ž + 1)

2π‘Ž βˆ’ 3

= 3π‘Ž + 1

iv) 15 + 3π‘₯ βˆ’ 7π‘₯2 βˆ’ 4π‘₯3 𝑏𝑦 (5 βˆ’ 4π‘₯)

π‘₯2 + 3π‘₯ + 3 βˆ’4π‘₯ + 5 βˆ’4π‘₯3 βˆ’ 7π‘₯2 + 3π‘₯ + 15 βˆ’4π‘₯3 + 5π‘₯2 + -

-12π‘₯2 + 3π‘₯ + 15 -12π‘₯2 + 15π‘₯ + - βˆ’12π‘₯ + 15 βˆ’12π‘₯ + 15 + - 0 0

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5. Divide the following polynomials by trinomials.

i) 𝑦2 βˆ’ 6𝑦2 + 11𝑦 βˆ’ 6 𝑏𝑦 𝑦2 βˆ’ 4𝑦 + 3

𝑦 βˆ’ 2 𝑦2 βˆ’ 4𝑦 + 3

𝑦3 βˆ’ 6𝑦2 + 11𝑦 βˆ’ 6 𝑦3 βˆ’ 4𝑦2 + 3

_ + _ -2𝑦2 + 11𝑦 βˆ’ 9 -2𝑦2 + 8𝑦 βˆ’ 6 + _ + 3𝑦 βˆ’ 3

ii) π‘₯3 βˆ’ 14π‘₯2 + 37π‘₯ βˆ’ 26 𝑏𝑦 π‘₯2 βˆ’ 12π‘₯ + 13 Refer que no. i

6. The product of two expressions is𝑦2 βˆ’ 𝑧2 + 2π‘₯𝑦 βˆ’ 2π‘₯𝑧. If one of them is 2π‘₯ + 𝑦 + 𝑧 find the other.

𝑦 βˆ’ 𝑧 2π‘₯ + 𝑦 + 𝑧 2π‘₯𝑦 βˆ’ 2π‘₯𝑧 + 𝑦2 βˆ’ 𝑧2 2π‘₯𝑦 + 𝑦2 + 𝑦𝑧

- - -

βˆ’2π‘₯𝑧 βˆ’π‘§2 βˆ’ 𝑦𝑧 βˆ’2π‘₯𝑧 βˆ’π‘§2 βˆ’ 𝑦𝑧 + + +

0 0 0

7. Find the value of a so that 1 βˆ’ 7π‘₯ is a factor βˆ’14π‘₯3 βˆ’ 47π‘₯2 βˆ’ 14π‘₯ + π‘Ž

Refer que 6.