lesson notes 2021 2022 subject: mathematics chapter: 15
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LESSON NOTES 2021 β 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, 6π
Greatest common factor: 3
ii) ππ₯ β 5ππ
Greatest common factor: π
iii) 18π₯7, β9π₯3
Greatest common factor: 9π₯3
Algebraic Expression Factors
Factorization
Multiplication
LESSON NOTES 2021 β 2022
Subject: Mathematics Chapter: 15 Factorisation
Grade: VIII
iv) 10π₯π¦, 15π₯2π¦3
Greatest common factor: 5π₯π¦
v) 38π3π₯5π¦, 15π₯2π¦3
Greatest common factor: 19π3π₯3
vi) 9π₯π¦2, β6π₯π¦, β3π₯
Greatest common factor: 3π₯
vii) 14π₯3π¦3, 21π₯5π¦2, 35π₯6π¦6
Greatest common factor: 7π₯3π¦2
viii) π2π8π3, βπ6π7π3, ππ5π2
Greatest common factor: ππ5π2
ix) 24π2π, β16πππ, 32π2π2π3, 36ππ2π2
Greatest common factor: 4ππ
Factorise the following expressions.
2. π2 β ππ₯
=π(π β π₯)
3. 3π2 β 9
=3(π2 β 3)
4. 5ππ₯ β 5π2π₯2
=5ππ₯(1 β ππ₯)
5. ππ 2 β ππ2
=π(π 2 β π2)
6. 10π₯π¦ β 15π₯2π¦2
=5π₯π¦(2 β 3π₯π¦)
7. 8π4π2π3 + 12π2ππ3
=4π2ππ3(2π2π + 3)
8. π₯2 β π₯π¦ + π₯π¦2
=π₯(π₯ β π¦ + π¦2)
9. 5π₯4 β 10π2π₯2 β 15π2π₯3
=5π₯2(π₯2 β 2π2π₯2 β 3π2π₯)
LESSON NOTES 2021 β 2022
Subject: Mathematics Chapter: 15 Factorisation
Grade: VIII
10. 3π₯3π¦2 + 9π₯2π¦ + 12π₯π¦
=3π₯π¦(π₯2π¦ + 3π₯ + 4)
11. 60π₯2π¦ + 25π¦2 + 20π¦2π§
Refer que, no. 10
12. ππ₯3π¦2 + ππ₯2π¦ + ππ₯π¦π§
Refer que, no. 10
13. 8π₯4 π¦ + 2π₯π¦4 β 24π₯2π¦3 + 18π₯3π¦2
Refer que, no. 10
14. 3π₯(π β π) β 6π¦(π β π)
=3(π β π)(π₯ β 2π¦)
15. (π₯ β π¦)2 β 2(π₯ β π¦)
=(π₯ β π¦)(π₯ β π¦ β 2)
16. β3(π₯ β 3π¦) + 6(π₯ β 3π¦)2
=β3(π₯ β 3π¦)(1 β 2(π₯ β 3π¦))
=β3(π₯ β 3π¦)(1 β 2π₯ + 6π¦)
17. (π₯ + 3)3 β 2(π₯ = 3)2 + 3(π₯3)
=(π₯ + 3)[(π₯ + 3)2 β 2(π₯ + 3) + 3]
=(π₯ + 3)[π₯2 + 6π₯ + 9 β 2π₯ β 6 + 3]
=(π₯ + 3)(π₯2 + 4π₯ + 6)
18. π₯(π + π)4 + π¦(π + π)3 + π§(π + π)2
=(π + π)2 [ π₯(π + π)2 + π¦(π + π) + π§]
=(π + π)2 [ π₯( π2 + 2ππ + π2) + π¦π + π¦π + π§]
=(π + π)2 [ π₯π2 + 2π₯ππ + π₯π2 + π¦π + π¦π + π§]
19. 25(π₯ β π¦)2 β 15(π₯ β π¦) β π₯ + π¦
Refer Que 18
20. π₯2 + π₯π¦ + π₯π§ + π¦π§
=π₯(π₯ + π¦) + π§(π₯ + π¦)
=(π₯ + π¦)(π₯ + π§)
21. π2 β ππ + ππ β ππ
=Refer Que 20
LESSON NOTES 2021 β 2022
Subject: Mathematics Chapter: 15 Factorisation
Grade: VIII
22. ππ₯ β ππ₯ β ππ§ + ππ§
=π₯(π β π) β π§(π β π)
=(π β π)(π₯ β π§)
23. 4π₯π¦ β π¦2 + 8π₯π§ β 2π¦π§
=π¦(4π₯ β π¦) + 2π§(4π₯ β π¦)
=(4π₯ β π¦)(π¦ + 2π§)
24. ππ₯ + 2π₯π§ + 2ππ¦ + 4π¦π§
=π₯(π + 2π§) + 2π¦(π = 2π§)
=(π + 2π§)(π₯ + 2π¦)
25. 2π₯2 + 3π₯π¦ β 2π₯π§ β 3π¦π§
Refer Que 20
26. π2π₯ β πππ¦ β 2ππ₯ β 2ππ¦
=Refer Que 20
27. ππ2 β (π β π)π β π
=ππ2 β ππ + ππ β π
=ππ(π β 1) + π(π β 1)
=(ππ + π)(π β 1)
28. π₯3 β π₯2 + π₯ β 1
=π₯2(π₯ β 1) + 1(π₯ β 1)
=(π₯ β 1)(π₯2 + 1)
29. π₯3 + 2π₯ β 2π₯2 β 4
=π₯(π₯2 + 2) β 2(π₯2 + 2)
=(π₯2 + 2) (x-2)
30. ππ2 β ππ2 β ππ + π2
= π(ππ β π2) β 1(ππ β π2)
= (ππ β π2)(π β 1)
31. 8 β 4π₯ β 2π₯3 + π₯4
= 4(2 β π₯) β ππ₯3(2 β π₯)
= (4 β π₯3)(2 β π₯)
LESSON NOTES 2021 β 2022
Subject: Mathematics Chapter: 15 Factorisation
Grade: VIII
32. π₯3 β 2π₯π¦3 β 4π¦3 + 2π₯2π¦
=π₯(π₯2 β 2π¦2) + 2π¦ (β2π¦2 + π₯2)
=(π₯2 β 2π¦2)(π₯ + 2π¦)
33. ππ₯π¦ + πππ₯π¦ β ππ§ β πππ§
Refer Que 22
34. 2ππ₯2 + 3ππ₯π¦ β 2ππ₯π¦ β 3ππ¦2
Refer Que 25
35. ππ₯ + ππ₯ β ππ¦ β ππ¦ + ππ§ + ππ§
=π₯(π + π) β π¦(π + π) + π§(π + π)
=(π + π)(π₯ β π¦ + π§)
36. ππ₯ β ππ₯ + ππ¦ + ππ¦ β ππ₯ β ππ¦
Refer Que 35
37. (ππ₯ + ππ¦)2 β (ππ₯ + ππ¦)2
=[ππ₯ + ππ¦ + ππ₯ + ππ¦][(ππ₯ + ππ¦) β (ππ₯ β ππ¦)]
=[π₯(π + π) + π¦(π + π)][π₯(π β π) β π¦(π β π)]
=(π + π)(π₯ + π¦)(π β π)(π β π¦)
Factorization Using Identifiers
Factorization when the given expression is a perfect square.
We know that
A] π2 + 2ππ + π2 = (π + π)2
B] π2 β 2ππ + π2 = (π β π)2
Exercise 15B
Factorise the following algebraic expression
1. π₯2 β 16π₯ + 64
= (π₯)2 β 2 Γ 8 Γ π₯ + (8)2
= (π₯ β 8)2
LESSON NOTES 2021 β 2022
Subject: Mathematics Chapter: 15 Factorisation
Grade: VIII
2. 1 β 2π₯ + π₯2
= (1)2 β 2 Γ 1 Γ π₯ + (π₯)2
= (1 β π₯)2
3. 9 + 6π + π2
Refer que 2
4. π2 β 10π + 25
Refer que 1
5. 1 β 6π₯ + 9π₯2
= (1)2 β 2 Γ 3 Γ π₯ + (3π₯)2
= (1 β 3π₯)2
6. 49π₯2 β 14π₯ + 1
= (7π₯)2 β 2 Γ 7 Γ π₯ + (1)2
= (7π₯ β 1)2
7. 49π₯2 β 84π₯π¦ + 36π¦2
= (7π₯)2 β 2 Γ 7π₯ Γ 6π¦ + (6π¦)2
= (7π₯ + 6π¦)2
8. 4π¦2 β 8π¦ + 4
Refer que 6
9. (π + π)2 β 4ππ
= π2 β 2ππ + π2 β 4ππ
= π2 β 2ππ + π2
= (π β π)2
10. π₯4 β 2π₯2π¦2 + π¦4
= ((π₯)2)2 + 2 Γ π₯2 Γ π¦2 + ((π¦)2)2
= (π₯2 + π¦2) 2
LESSON NOTES 2021 β 2022
Subject: Mathematics Chapter: 15 Factorisation
Grade: VIII
11. π₯2
4+ 2π₯2π¦2 + π¦4
= (π₯2)2 + 2 Γ π₯2 Γ (π¦2) 2
= (π₯
2+ 1) 2
12. π₯2π¦2 β 6π₯π¦π§ + π¦2
= (π₯π¦)2 β 2 Γ π₯π¦ Γ 3π§ + (3π§)2
= (π₯π¦ β 3π§) 2
13. π2
4β 3ππ3 + 9π6
= (π
2) 2 β 2 Γ
π
2Γ 3π3 + (3π3) 2
= (π
2β 3π3) 2
14. (π2
9) + (
π2
64) + (
ππ
12)
=π2
9+
ππ
12+
π2
64
= (π
3) 2 + 2 Γ
π
3Γ
π
8+ (
π
8) 2
= (π
3β
π
8) 2
15. π2
16β
ππ
6+
π2
9
Refer que 1
* When the given expression is a difference of two squares:
In this case we use identity
π2 β π2 = (π + π)(π β π)
Exercise 15C
Factorise the following Algebraic Expressions:-
1. 4π₯2 β 9π¦2
= (2π₯)2 β (3π¦)2
= (2π₯ + 3π¦)(2π₯ β 3π¦)
2. 81π2 β 121
= (9π)2 β (11)2
LESSON NOTES 2021 β 2022
Subject: Mathematics Chapter: 15 Factorisation
Grade: VIII
3. 100 β 49π2
= (10)2 β (7π)2
= (10 β 7π)(10 + 7π)
4. π5 β 16π3
= π3[π2 β 16]
= π3[π2 β 42]
= π3(π β 4)(π + 4)
5. 12π₯2 β 75π¦2
= (β12π₯)2
β (β75π¦)2
= (β4 Γ 3π₯)2
β (β25 Γ 3π¦)2
= (2β3π₯) 2 β (5β3π¦) 2
= [(2β3π₯) β (5β3π¦) ] [ (2β3π₯) + (5β3π¦)]
6. π₯3 β 64π₯
Refer que 4
7. 15π₯5 β 135π₯3
Refer que 4
8. 81π7 β π3π4
= π3(81π4 β π4)
= π3 [ (9π)2 β (π2) 2 ]
= π3 (9π β π2)(9π + π2)
9. π₯2
36β
π¦2
25
= (π₯
6)2 β (
π¦
5)2
= (π₯
6+
π¦
5) (
π₯
6β
π¦
5)
10. 49x2
144β
16x2
25
Refer que. 9
LESSON NOTES 2021 β 2022
Subject: Mathematics Chapter: 15 Factorisation
Grade: VIII
11. π₯2
8β
π¦2
32
Refer que. 9
12. 1 β (π₯ β π¦)2
= (1)2 β (π₯ β π¦)2
= (1 + (π₯ β π¦))(1 β (π₯ β π¦))
= (1 + π₯ β π¦)(1 β π₯ + π¦)
13. (2π₯ + 3π¦)2 β 16π§2
Refer que. 12
14. 4(π₯ β π¦)2 β (π₯ + π¦)2
= (2(π₯ β π¦))2 β (π₯ + π¦)2
= (2π₯ β 2π¦ + π₯ + π¦)(2π₯ β 2π¦ β π₯ β π¦)
= (3π₯ β π¦)(π₯ β 3π¦)
15. 25 β π2 β 2ππ β π2
= 25 β (π2 + 2ππ + π2)
= 52 β (π + π)2
= (5 + (π + π))(5 β (π + π))
= (5 + π + π)(5 β π β π)
16. π₯4 + π¦4 + π₯2π¦2
Add and subtract 2π₯2π¦2
= (π₯)22+ 2π₯2π¦2 + (π¦)22
+ π₯2π¦2 β 2π₯2π¦2
= (π₯2 + π¦2)2 β π₯2π¦2
= [(π₯2 + π¦2) + π₯π¦][(π₯2 + π¦2) β π₯π¦]
17. π₯4 + π₯2 + 1 [π»πππ‘: πππ πππ π π’ππ‘ππππ‘ π₯2]
Refer que 16.
18. π₯4 + 4 [π»πππ‘: πππ πππ π π’ππ‘ππππ‘ 4π₯2]
Refer que 16.
LESSON NOTES 2021 β 2022
Subject: Mathematics Chapter: 15 Factorisation
Grade: VIII
19. Evaluate
(203)2 β (197)2
= (203 + 197)(203 β 197)
= (400)(6)
= 2400
20. Evaluate
(6.6)2 β (3.4)2
Refer que 19.
21. Simplify
0.75 Γ 0.75 β 0.25 Γ 0.25
0.75 β 0.25
=(0.75)2 β (0.25)2
0.75 β 0.25
=(0.75 + 0.25)(0.75 β 0.25)
(0.75 β 0.25)
= 0.75 + 0.25
= 1
* Factorization of Quadratic Trinomial
In order to factories quadratic trinomial ππ + ππ + π , two numbers a&b have to be found such that a+b=p and ab=q
Then, ππ + ππ + π = ππ + (π + π)π + ππ[πππππππππππ π & π]
= (π + π)(π + π)
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)
LESSON NOTES 2021 β 2022
Subject: Mathematics Chapter: 15 Factorisation
Grade: VIII
2. π₯2 + 9π₯ + 20
= π₯2 + 5π₯ + 4π₯ + 20
= π₯(π₯ + 5) + 4(π₯ + 5)
= (π₯ + 5)(π₯ + 4)
3. π2 β 2π β 15
= π2 β 5π + 3π β 15
= π(π β 5) + 3(π β 5)
= π(π β 5) + 3(π β 5)
= (π β 5)(π + 3)
4. π₯2 β 18π₯ + 65
= π₯2 β 13π₯ β 5π₯ + 65
= π₯(π₯ β 13) β 5(π₯ β 13)
= (π₯ β 13)(π₯ β 5)
5. π2 β 11π β 80
Refer que 3
6. π2 β 11π β 102
= π2 β 17π + 6π β 102
= π(π β 17) + 6(π β 17)
= (π β 17) + 6(π β 17)
= (π β 17)(π + 6)
7. π2 β 26π + 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 β 33π₯ + 4π₯2
= 4π₯2 β 33π₯ + 35
= 4π₯2 β 28π₯ β 5π₯ + 35
= 4π₯(π₯ β 7) β 5(π₯ β 7)
= (4π₯ β 5)(π₯ β 7)
LESSON NOTES 2021 β 2022
Subject: Mathematics Chapter: 15 Factorisation
Grade: VIII
11. 12π₯2 β 29π₯ + 15
Refer que 10
12. 4π₯2 β 8π₯ + 3
Refer que 10
13. π₯2 + 10π₯π¦ + 21π¦2
= π₯2 + 7π¦π₯ + 3π₯π¦ + 21π¦2
= π₯(π₯ + 7π¦) + 3π¦(π₯ + 7π¦)
= (π₯ + 7π¦)(π₯ + 3π¦)
14. π2 β ππ β 20π2
= π2 β 5ππ + 4ππ β 20π2
= π(π β 5π) + 4π(π β 5π)
= (π β 5π)(π + 4π)
15. 6 β 2π₯2 β π₯
= β2π₯2 β π₯ + 6
= 2π₯2 + 4π₯ β 3π₯ β 6
= 2π₯(π₯ β 2) β 3(π₯ β 2)
= (π₯ β 2)(2π₯ β 3)
16. 14π2 + 11ππ β 15π2
Refer que 10.
17. 36π₯2 + 12π₯π¦π§ β 15π¦2π§2
= 3(12π₯2 + 4π₯π¦π§ β 5π¦2π§2)
= 3(12π₯2 + 10π₯π¦π§ β 6π₯π¦π§ β 5π¦2π§2)
= 3[2π₯(6π₯ + 5π¦π§) β π¦π§(6π₯ + 5π¦π§)]
= 3(6π₯ + 5π¦π§)(2π₯ β π¦π§) 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.
LESSON NOTES 2021 β 2022
Subject: Mathematics Chapter: 15 Factorisation
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) ππππππ ππ¦ 7ππππ
=35π₯6ππ
7ππππ
= (35
7) (
π₯6
π₯2) (π¦4
π¦2)
= 5π₯6β2π¦4β2 = 5π₯4π¦2
ii) 12π6π6π5 ππ¦ β 3π4π2π
=12π6π6π5
β3π4π2π
= (12
β3) (
π6
π4)(
π6
π2)(
π5
π)
= (β4)(π6β4)(π6β2)(π5β1)
= β4π2π4π4
LESSON NOTES 2021 β 2022
Subject: Mathematics Chapter: 15 Factorisation
Grade: VIII
iii) β63π7π8π3 ππ¦ β9π5π5π3
=β63π7π8π3
β9π5π5π3
= (β63
β9)(
π7
π5)(
π8
π5)(
π3
π3)
= 7π2π3
iv) 16π2π¦π₯2ππ¦ β 2π₯π¦ Refer que no. iii
v) 4(π₯4π¦4 β 2π₯3π¦2 + 3π₯π¦2)ππ¦ β 2π₯π¦ Refer que no. iii
vi) β3π₯2 + 9
2π₯π¦ β 6π₯π§ ππ¦
β3
2π₯
=β3π₯2 +
92 π₯π¦ β 6π₯π§
β32 π₯
=β3π₯2
β32 π₯
+
92 π₯π¦
β32 π₯
β6π₯π§
β32 π₯
= β3(2
β3) (
π₯2
π₯) +
9
2 (
β2
3) (
π₯π¦
π₯) β 6(
2
β3)(
π₯π§
π₯)
= 2π₯ β 3π¦ + 4π§
2. Divide the following
i) (7π₯ β 35)ππ¦7
=7π₯ β 35
7
=7π₯
7β
35
7
= π₯ β 5
LESSON NOTES 2021 β 2022
Subject: Mathematics Chapter: 15 Factorisation
Grade: VIII
ii) (6π₯ + 33) Γ· (2x + 11)
=(6π₯ + 33)
(2x + 11)
=3(2π₯ + 11)
(2x + 11)
= 3
iii) 72ππ(π+5) (πβ4)Γ· 36π(π β 4)
=72ab(a + 5)(b β 4)
36π(π β 4)
= 2π(π + 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 β 23π₯ + 132) Γ· (π₯ β 11)
Letβs factorise (π₯2 β 23π₯ + 132) (π₯2 β 23π₯ + 132) = (π₯2 β 12π₯ β 11π₯ + 132)
= π₯(π₯ β 12) β 11(π₯ β 12) = (π₯ β 12)(π₯ β 11)
(π₯2 β 23π₯ + 132) Γ· (π₯ β 11)
=(π₯2 β 23π₯ + 132)
(π₯ β 11)
=(π₯ β 12)(π₯ β 11)
(π₯ β 11)
= (π₯ β 12)
ii) 3π¦2 + 12π¦ β 63 Γ· (π¦ + 7)
Refer que no. i
LESSON NOTES 2021 β 2022
Subject: Mathematics Chapter: 15 Factorisation
Grade: VIII
iii) 14π¦π§(π§2 β 5π§ β 36) Γ· 7π§(π§ β 9)
Letβs factorise (π§2 β 5π§ β 36) (π§2 β 5π§ β 36) = (π§2 β 9π§ + 4π§ β 36) = π§(π§ β 9) + 4(π§ β 9) = (π§ β 9)(π§ + 4)
14π¦π§(π§2 β 5π§ β 36) Γ· 7π§(π§ β 9)
=14π¦π§(π§2 β 5π§ β 36)
7π§(π§ β 9)
=14π¦π§(π§ β 9)(π§ + 4)
7π§(π§ β 9)
= 2π¦(π§ + 4)
iv) 15ππ(25π2 β 16π2) Γ· 5ππ(5π + 4π)
Refer que. No. iii
4. Divide the following polynomials by binomials.
i) π₯2 β 11π₯ + 30 ππ¦ (π₯ β 5)
Letβs factorise π₯2 β 11π₯ + 30 = π₯2 β 6π₯ β 5π₯ + 30 = (π₯ β 6)(π₯ β 5)
π₯2 β 11π₯ + 30 ππ¦ (π₯ β 5)
=π₯2β11π₯+30
(π₯β5)
=(π₯β6)(π₯β5)
(π₯β5)
= (π₯ β 6)
ii) 2π¦2 + 17π¦ + 21 ππ¦ 2π¦ + 3
Letβs factorise 2π¦2 + 17π¦ + 21 = 2π¦2 + 14π¦ + 3π¦ + 21
= 2π¦(π¦ + 7) + 3(π¦ + 7) = (π¦ + 7)(2π¦ + 3)
LESSON NOTES 2021 β 2022
Subject: Mathematics Chapter: 15 Factorisation
Grade: VIII
2π¦2 + 17π¦ + 21 ππ¦ 2π¦ + 3
=2π¦2 + 17π¦ + 21
2π¦ + 3
=(π¦ + 7)(2π¦ + 3)
2π¦ + 3
= π¦ + 7
iii) 6π2 β 7π β 3 ππ¦ 2π β 3
Letβs factorise 6π2 β 7π β 3 = 6π2 β 9π + 2π β 3 = 3π(2π β 3) + (2π β 3) = (2π β 3)(3π + 1)
6π2 β 7π β 3 ππ¦ 2π β 3
=6π2 β 7π β 3
2π β 3
=(2π β 3)(3π + 1)
2π β 3
= 3π + 1
iv) 15 + 3π₯ β 7π₯2 β 4π₯3 ππ¦ (5 β 4π₯)
π₯2 + 3π₯ + 3 β4π₯ + 5 β4π₯3 β 7π₯2 + 3π₯ + 15 β4π₯3 + 5π₯2 + -
-12π₯2 + 3π₯ + 15 -12π₯2 + 15π₯ + - β12π₯ + 15 β12π₯ + 15 + - 0 0
LESSON NOTES 2021 β 2022
Subject: Mathematics Chapter: 15 Factorisation
Grade: VIII
5. Divide the following polynomials by trinomials.
i) π¦2 β 6π¦2 + 11π¦ β 6 ππ¦ π¦2 β 4π¦ + 3
π¦ β 2 π¦2 β 4π¦ + 3
π¦3 β 6π¦2 + 11π¦ β 6 π¦3 β 4π¦2 + 3
_ + _ -2π¦2 + 11π¦ β 9 -2π¦2 + 8π¦ β 6 + _ + 3π¦ β 3
ii) π₯3 β 14π₯2 + 37π₯ β 26 ππ¦ π₯2 β 12π₯ + 13 Refer que no. i
6. The product of two expressions isπ¦2 β π§2 + 2π₯π¦ β 2π₯π§. If one of them is 2π₯ + π¦ + π§ find the other.
π¦ β π§ 2π₯ + π¦ + π§ 2π₯π¦ β 2π₯π§ + π¦2 β π§2 2π₯π¦ + π¦2 + π¦π§
- - -
β2π₯π§ βπ§2 β π¦π§ β2π₯π§ βπ§2 β π¦π§ + + +
0 0 0
7. Find the value of a so that 1 β 7π₯ is a factor β14π₯3 β 47π₯2 β 14π₯ + π
Refer que 6.