Selina Concise Class 9 Maths Chapter 5 Factorisation Exercise 5B Solutions

Selina Concise Class 9 Maths Chapter 5 Factorisation Exercise 5B Solutions

Selina Concise Class 9 Maths Chapter 5 Factorisation Exercise 5B Solutions

EXERCISE – 5B

 

(1) Factorize:

a2 + 10a + 24

Solution:

a2 + 10a + 24

=> a2 + 6a + 4a + 24

=> a (a + 6) + 4 (a + 6)

=> (a + 6) (a + 4)

 

(2) a2 – 3a – 40

Solution:

a2 – 3a – 40

=> a2 – 8a + 5a – 40

=> a (a – 8) + 5 (a – 8)

=> (a – 8) (a + 5)

 

(3) 1 – 2a – 3a2

Solution:

1 – 2a – 3a2

=> First rearrange the term –

– 3a2 – 2a + 1

=> -3a2 – 3a + 1a + 1

=> -3a (a + 1) + 1 (a + 1)

=> (a + 1) (-3a + 1)

 

(4) 4x2 – 3ax – 88a2

Solution:

x2 – 3ax – 88a2

=> x2 – 3ax – 88a2

=> x2 – 11ax + 8ax – 88a2

=> x (x – 11a) + 8a (x – 11a)

= (x – 11a) (x + 8a)

 

(5) 6a2 – a – 15

Solution:

6a2 – a – 15

= 6a2 – 10a + 9a – 15

= 2a (3a – 5) + 3 (3a – 5)

= (3a – 5) (2a + 3)

 

(6) 24a3 + 37a2 – 5a

Solution:

24a3 + 37a2 – 5a

=> Divide ‘a’

=> 24a3/a + 37a2/a – 5a/a

=> 24a2 + 37a – 5

= 24a2 + 40a – 3a – 5

= 8a (3a + 5) – 1 (3a + 5)

= (3a + 5) (8a – 1)

 

(7) a (3a – 2) – 1

Solution:

a (3a – 2) -1

=> 3a2 – 2a – 1

=> 3a2 – 3a + a – 1

=> 3a (a – 1) + 1 (a – 1)

=> (a – 1) (3a + 1)

 

(8) a2 b + 8ab – 9

Solution:

a2b2 + 8ab – 9

=> a2b2 + 9ab – 1ab – 9

=> ab (ab + 9) – 1 (ab + 9)

=> (ab + 9) (ab – 1)

 

(9) 3 – a (4 + 7a)

Solution:

3 – (4a + 7a2)

= 3  – 4a – 7a2

First we have rearrange the term.

= – 7a2 – 4a + 3

= -7a2 – 7a + 3a + 3

= -7a (a + 1) + 3 (a + 1)

= (a + 1) (-7a + 3)

 

(10) (2a + b)2 – 6a – 3b – 4

Solution:

(2a + b)2 – 6a – 3b – 4

= (2a + b)2 – 3 (2a + b) – 4

Let, 2a + b = y

= y2 – 3y – 4

= y2 – 4y + y – 4

= y (y – 4) + 1 (y – 4)

= (y – 4) (y + 1)

But y = 2a + b

∴ => (2a + b – 4) (2a + b + 1)

 

(11) 1 – 2 (a + b) – 3 (a + b)2

Solution:

1 – 2 (a + b) – 3 (a + b)2

=> -3 (a + b)2 – 2 (a + b) + 1

Let, a + b = y

=> -3y2 – 2y + 1

=> -3y2 – 3y + y + 1

=> -3y (y + 1) + 1 (y + 1)

=> (y + 1) (-3y + 1)

But a + b = y

=> (a + b + 1) (-3 (a + b) + 1)

= (a + b + 1) (-3a – 3b + 1)

 

(12) 3a2 – 1 – 2a

Solution:

3a2 – 1 – 2a

First we have to rearrange the term

3a2 – 2a – 1

=> 3a2 – 3a + a – 1

=> 3a (a – 1) + 1 (a – 1)

=> (a – 1) (3a + 1)

 

(13) x2 + 3x + 2 + ax + 2a

Solution:

x2+ 3x + 2 + ax + 2a

= x2 + 3x + ax + 2 + 2a

= x2 + x (3 + a) + 2 (1 + a)

= x2 + x [2 + 1 + a] + 2 (1 + a)

=> x2 + 2x + (1 + a) x + 2 (1 + a)

=> x (x + 2) + (1 + a) (x + 2)

=> (x + 2) (x + 1 +a)

 

(14) (3x – 2y)2 + 3 (3x – 2y) – 10

Solution:

(3x – 2y)2 + 3 (3x – 2y) – 10

Let, 3x – 2y = a

a2 + 3a – 10

=> a2 + 5a – 2a – 10

=> a (a + 5) – 2 (a + 5)

=> (a + 5) (a – 2)

But, a = 3x – 2y

=> (3x – 2y + 5) (3x – 2y – 2)

 

(15) 5 – (3a2 – 2a) (6 – 3a2 + 2a)

Solution:

 

5 – (3a2 – 2a) (6 – 3a2 + 2a)

Let, 3a2 – 2a = y

=> 5 – (3a2 – 2a), (6 – (3a2 – 2a))

=> 5 – y (6 – y)

=> 5 – 6y + y2

=> y2 – 6y + 5

=> y2 – 5y – y + 5

=> y (y – 5) – 1 (y – 5)

=> (y – 5) (y – 1)

But, y = 3a2 – 2a

=> (3a2 – 2a – 5) (3a2 – 2a – 1)

First we have to factorize

3a2 – 2a – 5

=> 3a2 – 5a + 3a – 5

= a (3a – 5) + 1 (3a – 5)

= (3a – 5) (a + 1)

Now, 3a2 – 2a –1

=> 3a2 – 3a + 1a – 1

=> 3a (a – 1) + 1 (a – 1)

= (a – 1) (3a + 1)

∴ (3a – 5) (a + 1) (a – 1) (3a + 1)

 

(16) 1/35 + 12a/35 + a2

Solution:

1/35 + 12/35 a + a2

=> Take L.C.M

1 + 12a + 35a2/35

=> 1/35 (35a2 + 12a + 1)

=> 1/35 (35a2 + 7a + 5a + 1)

=> 1/35 (7a (5a + 1) + 1 (5a+ 1))

=> 1/35 {(5a + 1) (7a + 1)}

= 1/35 (5a + 1) (7a + 1)

 

(17) (x2 – 3x) (x2 – 3x – 1) – 20

Solution:

(x2 – 3x) (x2 – 3x – 1) – 20

Let, x2 – 3x = y

=> y (y – 1) – 20

=> y2 – y – 20

=> y2 – 5y + 4y – 20

=> y (y – 5) + 4 (y – 5)

=> (y – 5) (y + 4)

But y = x2 – 3x

(x2 – 3x – 5) (x2 – 3x + 4)

 

(18) Find each trinomial (Quadratic expression) given below, find whether it is factorization or not. Factorize, if possible.

(i) x2 – 3x – 54

Solution:

x2 – 3x – 54

Comparing with general form of ax2 + bx + c

∴ a = 1, b = 3, c = -54, D = b2 – 4ac

= (-3)2 – 4 (1) (-54)

= 9 + 216

D = 225

If D is positive then it is possible

If D is negative then it is impossible.

x2 – 3x – 54

=> x2 – 9x + 6x – 54

=> x (x – 9) + 6 (x – 9)

=> (x – 9) (x + 6)

 

(ii) 2x2 – 7x – 15

Solution:

2x2 – 7x – 15

Comparing with general form ax2 + bx + c

a = 2, b = -7, c = -15

∴ D = b2 – 4ac

= (-7)2 – 4 (2) (-15)

= 49 + 120

D = 169

∴ D is positive

So, it is possible

∴ 2x2 – 7x – 15

=> 2x2 – 10x + 3x – 15

=> 2x (x – 5) + 3 (x – 5)

=> (x – 5) (2x + 3)

 

(iii) 2x2 + 2x – 75

Solution:

2x2 + 2x – 75

Comparing with general form of ax2 + bx + c

∴ a = 2, b = 2, c = -75

D = b2 – 4ac

= (2)2 – 4×2×(-75)

= 4 + 600

D = 604

∴ D is positive

So, it is possible.

But, 604 is not a perfect square

∴ It is not a factorisable

 

(iv) 3x2 + 4x – 10

=> Solution: 3x2 + 4x – 10

Comparing with the general form of ax2 + bx + c

∴ a = 3, b = 4, c = -10

∴ D = b2 – 4ac

= (4)2 – 4 (3) (-10)

= 16 + 120

D = 136

∴ D is positive.

But 136 is not a perfect square

∴ It is not factorisable.

 

(v) x (2x – 1) – 1

Solution:

x (2x – 1) – 1

=> 2x2 – x – 1

Comparing with the general form of ax2 + bx + c

∴ a = 2, b = -1, c = -1

∴ D = b2 – 4ac

=> (-1)2 – 4 × (2) (-1)

=> 1 + 8

D = 9

∵ D is positive and also 9 is a perfect square.

So, it is possible to factorise.

∴ 2x2 – x – 1

=> 2x2 – 2x + x – 1

=> 2x (x – 1) + 1 (x – 1)

=> (x – 1) (2x + 1)

 

(19) Factorise:

(i) 4√3 x2 + 5x – 2√3

Solution:

4√3 x2 + 5x – 2√3

=> 4√3 x2 + 8x – 3x – 2√3

=> 4x (√3x + 2) -√3 (√3 + 2)

= (√3x + 2) (4x – √3)

 

(ii)  7√2 x2 – 10x – 4√2

Solution:

 

7√2 x2 – 10x – 4√2

=> 7√2 x2 – 14x + 4x – 4√2

=> 7√2 x (x – √2) + 4 (x – √2)

=> (x – √2) (7√2 x + 4)

 

(20) Give possible expressions for the length and the breadth of the rectangle whose area is 12x2 – 35x + 25

Solution:

Given that,

12x2 – 35x + 25

=> 12x2 – 20x – 15x + 25

=> 2x (6x – 10) – 5 (3x – 5)

=> 2×2x (3 – 5) – 5 (3x – 5)

=> 4x (3x – 5) – 5 (3x – 5)

=> (3x – 5) (4x – 5)

Hence, Length of the rectangle is (3x – 5) and breadth of the rectangle is (4x – 5)

{Area of rectangle = length × breadth}

 

Here is your solution of Selina Concise Class 9 Maths Chapter 5 Factorisation Exercise 5B

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