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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