ML Aggarwal Solutions Class 9 Math Fourth Chapter Factorisation Exercise 4.4

ML Aggarwal Solutions Class 9 Math 4th Chapter Factorisation Exercise 4.4

ML Aggarwal Understanding ICSE Mathematics Class 9 Solutions Fourth Chapter Factorisation Exercise 4.4. APC Solution Class 9 Exercise 4.4.

 

(1) (i) x2 + 5x + 6

Solution:

x2 + 5x + 6

= x2 + 2x + 3x + 6

= x (x + 2) + 3 (x + 2)

= (x + 2) (x + 3)

 

(ii) x2 – 8x + 7

Solution:

x2 – 8x + 7

= x2 – 7x – x + 7

= x (x – 7) – 1 (x – 7)

= (x – 7) (x – 1)

 

(2) (i) x2 + 6x – 7

Solution:

x2 + 6x – 7

= x2 + 7x – x – 7

= x (x + 7) – 1 (x + 7)

= (x + 7) (x – 1)

 

(ii) y2 + 7y – 18

Solution:

y2 + 7y – 18

= y2 + 9y – 2y – 18

= y (y + 9) – 2 (y + 9)

= (y + 9) (y – 2)

 

(3) (i) y2 – 7y – 18

Solution:

y2 – 7y – 18

= y2 – 9y + 2y – 18

= y (y – 9) + 2 (y – 9)

= (y – 9) (y + 2)

 

(ii) a2 – 3a – 54

Solution:

a2 – 3a – 54

= a2 – 9a + 6a – 54

= a (a – 9) + 6 (a – 9)

= (a – 9) (a + 6)

 

(4) (i) 2x2 – 7x + 6

Solution:

2x2 – 7x + 6

= 2x2 – 3x – 4x + 6

= x (2x – 3) – 2 (2x – 3)

= (2x – 3) (x – 2)

 

(ii) 6x2 + 13x – 5

Solution:

6x2 + 13x – 5

= 622 – 2x + 15x – 5

= 2x (3x – 1) + 5 (3x – 1)

= (3x – 1) (2x + 5)

 

(5) (i) 6x2 + 11x – 10

Solution:

6x2 + 11x – 10

= 6x2 + 15x – 4x – 10

= 3x (2x + 5) – 2 (2x + 5)

= (2x + 5) (3x – 2)

 

(ii) 6x2 – 7x – 3

Solution:

6x2 – 7x – 3

= 6x2 – 9x + 2x – 3

= 3x (2x – 3) + 1 (2x – 3)

= (2x – 3) (3x + 1)

 

(6) (i) 2x2 – x – 6

Solution:

2x2 – x – 6

= 2x2 – 4x + 3x – 6

= 2x (x– 2) + 3 (x – 2)

= (x – 2) (2x + 3)

 

(ii) 1 – 18y – 63y2

Solution:

1 – 18y – 63y2

= 1 – 21y + 3y – 63y2

= 1 (1 – 21y) + 3y (1 – 21y)

= (1 – 21y) (1 + 3y)

 

(7) (i) 2y2 + y – 45

Solution:

2y2 + y – 45

= 2y2 + 10y – 9y – 45

= 2y (y + 5) – 9 (y + 5)

= (y + 5) (2y – 9)

 

(ii) 5 – 4x – 12x2

Solution:

5 – 4x – 12x2

= 5 – 10x + 6x – 12x2

= 5 (1 – 2x) + 6 (1 – 2x)

= (1 – 2x) (5 + 6x)

 

(8) (i) x (12x + 7) – 10

Solution:

x (12x + 7) – 10

= 12x2 + 7x – 10

= 12x2 + 15x – 8x – 10

= 3x (4x + 5) – 2 (4x + 5)

= (4x + 5) (3x – 2)

 

(ii) (4 – x)2 – 2x

Solution:

(4 – x)2 – 2x

= 42 – 8x + x2 – 2x

= x2 – 10x + 16

= x2 – 8x – 2x + 16

= x (x – 8) – 2 (x – 8)

= (x – 8) (x – 2)

 

(9) (i) 60x2 – 70x – 30

Solution:

60x2– 70x – 30

= 10 (6x2 – 7x – 3)

= 10 (6x2 – 9x + 2x – 3)

= 10 (3x (2x – 3) + 1 (2x – 3)}

= 10 (2x – 3) (3x + 1)

 

(ii) x2 – 6xy – 7y2

Solution:

x2 – 6xy – 7y2

= x2 – 7xy + xy – 7y2

= x (x – 7y) + y (x – 7y)

= (x – 7y) (x + y)

 

(10) (i) 2x2 + 13xy – 24y2

Solution:

2x2 + 13xy – 24y2

= 2x2 + 16xy – 3xy – 24y2

= 2x (x + 8y) – 3y (x + 8y)

= (x + 8y) (2x – 3y)

 

(ii) 6xy2 – 5xy – 6y2

Solution: 

6xy2 – 5xy – 6y2

= 6x2 – 9xy + 4xy – 6y2

= 3x (2x – 3y) + 2y (2x – 3y)

= (2y – 3y) (3x + 2y)

 

(11) (i) 5x2 + 17xy – 12y2

Solution:

5x2 + 17xy – 12y2

= 5x2 + 20xy – 3xy – 12y2

= 5x (x + 4y) – 3y (x + 4y)

= (x + 4y (5x – 3y)

 

(ii) x2y2 – 8xy – 48

Solution:

x2y2 – 8xy – 48

= x2y2 – 12xy + 4xy – 48

= xy (xy – 12) + 4 (xy – 12)

= (xy – 12) (xy + 4)

 

(12) (i) 2a2b2 – 7ab – 30

Solution:

2a2b2 – 7ab – 30

= 2a2b2 – 12ab + 5ab – 30

= 2ab (ab – 6) + 5 (ab – 6)

= (ab – 6) (2ab + 5)

 

(ii) a (2a – b) – b2

Solution:

a (2a – b) – b2

= 2a2 – ab – b2

= 2a2 – 2ab + ab – b2

= 2a (a – b) + b (a – b)

= (a – b) (2a + b)

 

(13) (i) (x – y)2 – 6 (x – y) + 5

Solution:

(x – y)2 – 6 (x – y) + 5

= (x – y)2 – 5 (x – y) – (x – y) + 5

= (x – y) (x – y – 5) – 1 (x – y – 5)

= (x – y – 5) (x – y – 1)

 

(ii) (2x – y)2 – 11 (2x – y) + 28

Solution:

(2x – y)2 – 11 (2x – y) + 28

= (2x – y)2 – 4 (2x – y) – 7 (2x – y) + 28

= (2x – y) (2x – y – 4) – 7 (2x – y – 4)

= (2x – y – 4) (2x – y – 7)

 

(14) (i) 4 (a – 1)2 – 4 (a – 1) – 3

Solution:

4 (a – 1)2 – 4 (a – 1) – 3

= 4 (a – 1)2 + 2 (a – 1) – 6 (a – 1) – 3

= 2 (a – 1) (2 (a – 1) + 1) – 3 (2 (a – 1) + 1)

= (2a – 2 + 1) (2 (a – 1) – 3)

= (2a – 2 + 1) (2a – 2 – 3)

= (2 – 1) (2a – 5)

 

(ii) 1 – 2a – 2b – 3 (a + b)2

Solution:

1 – 2a – 2b – 3 (a + b)2

= 1 – 2 (a + b) – 3 (a + b)2

= 1 – 3 (a + b) + (a + b) – 3 (a + b)2

= 1 (1 – 3 (a + b) + (a + b) (1 – 3 (a + b)

= (1 – 3a – 3b) (1 + a + b)

 

(15) (i) 3 – 5a – 5b – 12 (a + b)2

Solution:

3 – 5a – 5b – 12 (a + b)2

= 3 – 5 (a + b) – 12 (a + b)

= 3 – 9 (a + b) + 4 (a + b) – 12 (a + b)2

= 3 (1 – 3 (a + b) + 4 (a + b) (1 – 3 (a + b))

= (1 – 3a – 3b) (3 + 4a + 4b)

 

(ii) a4 – 11a2 + 10

Solution:

a4 – 11a2 + 10

= a4 – 10a2 – a2 + 10

= a2 (a2 – 10) – 1 (a2 – 10)

= (a2 – 10) (a2 – 1)

 

(16) (i) (x + 4)2 – 5xy – 20y – 6y2

Solution:

(x + 4)2 – 5xy – 20y – 6y2

= (x + 4)2 – 5xy (x + 4) – 6y2

= (x + 4)2 – 6y (x + 4) + y (x + y) – 6y2

= (x + y) (x + 4 – 6y) + y (x + 4 – 6y)

= (x + 4 – 6y) (x + 4 + 6y)

 

(ii) (x2 – 2x)2 – 23 (x2 – 2x) + 120

Solution:

(x2 – 2x)2 – 23 (x2 – 2x) + 120

= (x2 – 2x)2 – 15 (x2 – 28) – 8 (x2 – 2x) + 120

= (x2 – 2x) (x2 – 2x – 15) – 8 (x2 – 2x – 15)

= (x2 – 2x – 15) (x2 – 2x – 8)

= (x2 – 5x + 3x – 15) (x2 – 4x + 2x – 8)

= (x (x – 5) + 3 (x – 5)) (x (x – 4) + 2 (x – 4))

= (x – 5) (x + 3) (x – 4) (x + 2)

 

(17) 4 (2a – 3)2 – 3 (2a – 3) (a – 1) – 7 (a – 1)2

Solution:

4 (2a – 3)2 – 3 (2a – 3) (a – 1) – 7 (a – 1)2

Let, 2a – 3 = x & (a – 1) = y

∴ 4x2 – 3xy – 7y2

= 4x2 – 7xy + 4xy – 7y

= x (4x – 7y) + y (4x – 7y)

= (4x – 7y) (x + y)

Putting the value of x, y w get,

= (4 (2a – 3) – 7 (a – 1) (2a – 3 + a – 1)

= (8a – 12 – 7a + 7) (3a – 4)

= (a – 5) (3a – 4)

 

(18) (2x2 + 5x) (2x2 + 5x – 19) + 84

Solution:

(2x2 + 5x) (2x2 + 5x – 19) + 84

Let, 2x2 + 5x = a

∴ a (a – 19) + 84

= a2 – 19a + 84

= a2 – 7a – 12a + 84

= a (a – 7) – 12 (a – 7)

= (a – 7) (a – 12)

Putting the value of a we get,

= (2x2 + 5x – 7) (2x2 + 5x – 12)

= (2x2 + 7x – 2x – 7) (2x2 + 8x – 3x – 12)

= (2x (2x + 7) – 1(21 + 7)) (2x (x + 4) – 3 (x + 4))

= (2x + 7) (x – 1) (2x – 3) (x + 4)

Updated: June 17, 2022 — 4:20 pm

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