Lesson 6: Factorisation of Algebraic Expressions | Practice Set 6.1 Solutions

 

📘 Lesson 6: Factorisation of Algebraic Expressions

Maharashtra State Board – Mathematics

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✏️ Practice Set 6.1

Practice Set 6.1 – Factorise

1) x² + 9x + 18

We need two numbers whose:

• Product = 18
• Sum = 9

The numbers are 3 and 6.

x² + 9x + 18
= x² + 3x + 6x + 18
= x(x + 3) + 6(x + 3)
= (x + 3)(x + 6)

2) x² − 10x + 9

We need two numbers whose:

• Product = 9
• Sum = −10

The numbers are −1 and −9.

x² − 10x + 9
= x² − x − 9x + 9
= x(x − 1) − 9(x − 1)
= (x − 1)(x − 9)

3) y² + 24y + 144

We need two numbers whose:

• Product = 144
• Sum = 24

The numbers are 12 and 12.

y² + 24y + 144
= y² + 12y + 12y + 144
= y(y + 12) + 12(y + 12)
= (y + 12)²

4) 5y² + 5y − 10

First take 5 common:

5y² + 5y − 10
= 5(y² + y − 2)

We need two numbers whose:

• Product = −2
• Sum = 1

The numbers are 2 and −1.

= 5[y² + 2y − y − 2]
= 5[y(y + 2) − 1(y + 2)]
= 5(y + 2)(y − 1)

5) p² − 2p − 35

We need two numbers whose:

• Product = −35
• Sum = −2

The numbers are −7 and 5.

p² − 2p − 35
= p² − 7p + 5p − 35
= p(p − 7) + 5(p − 7)
= (p − 7)(p + 5)

6) p² − 7p − 44

We need two numbers whose:

• Product = −44
• Sum = −7

The numbers are −11 and 4.

p² − 7p − 44
= p² − 11p + 4p − 44
= p(p − 11) + 4(p − 11)
= (p − 11)(p + 4)

7) m² − 23m + 120

We need two numbers whose:

• Product = 120
• Sum = −23

The numbers are −8 and −15.

m² − 23m + 120
= m² − 8m − 15m + 120
= m(m − 8) − 15(m − 8)
= (m − 8)(m − 15)

8) m² − 25m + 100

We need two numbers whose:

• Product = 100
• Sum = −25

The numbers are −5 and −20.

m² − 25m + 100
= m² − 5m − 20m + 100
= m(m − 5) − 20(m − 5)
= (m − 5)(m − 20)

9) 3x² + 14x + 15

Here, a = 3, b = 14 and c = 15.

Therefore, a × c = 3 × 15 = 45.

We need two numbers whose:

• Product = 45
• Sum = 14

The numbers are 5 and 9.

3x² + 14x + 15
= 3x² + 5x + 9x + 15
= x(3x + 5) + 3(3x + 5)
= (3x + 5)(x + 3)

10) 2x² + x − 45

Here, a × c = 2 × (−45) = −90.

We need two numbers whose:

• Product = −90
• Sum = 1

The numbers are 10 and −9.

2x² + x − 45
= 2x² + 10x − 9x − 45
= 2x(x + 5) − 9(x + 5)
= (2x − 9)(x + 5)

11) 20x² − 26x + 8

First take 2 common:

20x² − 26x + 8
= 2(10x² − 13x + 4)

Now, 10 × 4 = 40.

We need two numbers whose:

• Product = 40
• Sum = −13

The numbers are −5 and −8.

= 2(10x² − 5x − 8x + 4)
= 2[5x(2x − 1) − 4(2x − 1)]
= 2(5x − 4)(2x − 1)

12) 44x² − x − 3

Here, a × c = 44 × (−3) = −132.

We need two numbers whose:

• Product = −132
• Sum = −1

The numbers are 11 and −12.

44x² − x − 3
= 44x² + 11x − 12x − 3
= 11x(4x + 1) − 3(4x + 1)
= (11x − 3)(4x + 1)

Final Answers

No.	Factorised Form
1	(x + 3)(x + 6)
2	(x − 1)(x − 9)
3	(y + 12)²
4	5(y + 2)(y − 1)
5	(p − 7)(p + 5)
6	(p − 11)(p + 4)
7	(m − 8)(m − 15)
8	(m − 5)(m − 20)
9	(3x + 5)(x + 3)
10	(2x − 9)(x + 5)
11	2(5x − 4)(2x − 1)
12	(11x − 3)(4x + 1)