MSBSHSE Solutions For Class 8 Maths Part 1 Chapter 6- Factorisation of Algebraic Expressions

MSBSHSE Solutions For Class 8 Maths Part 1 Chapter 6 – Factorisation of Algebraic Expressions are provided here to help students come out with flying colours in their board examinations. As students are familiar with numbers, their problem-solving and analytical skills have improved considerably. We all know that algebra plays an essential role in maths. This chapter discusses the concept factors of a quadratic trinomial, rational algebraic expressions. Here in Maharashtra State Board Class 8 Textbooks Part 1 such problems are solved. Our expert faculty team has solved the questions in a step by step format, which helps students understand the concepts clearly. These solutions will also help them in obtaining knowledge and strong command over the subject. If the students wish to secure an excellent score then solving MSBSHSE Solutions Class 8 is a must. Students can refer to and download the PDF of Maharashtra Board Solutions for Class 8 Maths Chapter 6 Factorisation of Algebraic Expressions, which is readily available from the links.

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Practice set 6.1 PAGE NO: 30

1. Factorize:

(1) x² + 9x + 18

(2) x² – 10x + 9

(3) y² + 24y + 144

(4) 5y² + 5y – 10

(5) p² – 2p – 35

(6) p² – 7p – 44

(7) m² – 23m + 120

(8) m² – 25m + 100

(9) 3x² + 14x + 15

(10) 2x² + x – 45

(11) 20x² – 26x + 8

(12) 44x² – x – 3

Solution:

(1) x² + 9x + 18

On comparing with standard quadratic equation that is ax² + bx + c.

Let us simplify the given expression, we get

x² + 9x + 18 = x² + 6x + 3x + 18

= x (x + 6) + 3(x + 6)

= (x + 6) (x + 3)

(2) x² – 10x + 9

On comparing with standard quadratic equation that is ax² + bx + c.

Let us simplify the given expression, we get

x² – 10x + 9 = x² – 9x – x + 9

= x (x – 9) – 1(x – 9)

= (x – 9) (x – 1)

(3) y² + 24y + 144

On comparing with standard quadratic equation that is ax² + bx + c.

Let us simplify the given expression, we get

y² + 24y + 144 = y² + 12y + 12y + 144

= y(y + 12) + 12(y + 12)

= (y + 12) (y + 12)

(4) 5y² + 5y – 10

On comparing with standard quadratic equation that is ax² + bx + c.

Let us simplify the given expression, we get

5y² + 5y – 10 = 5(y² + y – 2) [By taking out the common factor 5]

= 5(y² + 2y – y – 2)

= 5[y(y + 2) – 1(y + 2)]

= 5 (p + 2) (y- 1)

(5) p² – 2p – 35

On comparing with standard quadratic equation that is ax² + bx + c.

Let us simplify the given expression, we get

p² – 2p – 35 = p² – 7p + 5p – 35

= p(p – 7) + 5(p – 7)

= (p – 7) (p + 5)

(6) p² – 7p – 44

On comparing with standard quadratic equation that is ax² + bx + c.

Let us simplify the given expression, we get

p² – 7p – 44 = p² – 11p + 4p – 44

= p(p – 11) + 4(p – 11)

= (p – 11) (p + 4)

(7) m² – 23m + 120

On comparing with standard quadratic equation that is ax² + bx + c.

Let us simplify the given expression, we get

m² – 23m + 120 = m² – 15m – 8m + 120

= m (m – 15) – 8 (m – 15)

= (m – 15) (m – 8)

(8) m² – 25m + 100

On comparing with standard quadratic equation that is ax² + bx + c.

Let us simplify the given expression, we get

m² – 25m + 100 = m² – 20m – 5m + 100

= m(m – 20) – 5(m – 20)

= (m – 20) (m – 5)

(9) 3x² + 14x + 15

On comparing with standard quadratic equation that is ax² + bx + c.

Let us simplify the given expression, we get

3x² + 14x + 15 = 3x² + 9x + 5x + 15

= 3x(x + 3) + 5(x + 3)

= (x + 3) (3x + 5)

(10) 2x² + x – 45

On comparing with standard quadratic equation that is ax² + bx + c.

Let us simplify the given expression, we get

2x² + x – 45 = 2x² + 10x – 9x – 45

= 2x(x + 5) – 9 (x + 5)

= (x + 5) (2x – 9)

(11) 20x² – 26x + 8

On comparing with standard quadratic equation that is ax² + bx + c.

Let us simplify the given expression, we get

20x² – 26x + 8 = 2(10x² – 13x + 4) 10 × 4 = 40 [By taking out the common factor 2]

= 2(10x² – 8x – 5x + 4)

= 2[2x (5x – 4) – 1(5x – 4)]

= 2 (5x – 4) (2x – 1)

(12) 44x² – x – 3

On comparing with standard quadratic equation that is ax² + bx + c.

Let us simplify the given expression, we get

44x² – x – 3 = 44x² – 12x + 11x – 3

= 4x (11x – 3) + 1(11x – 3)

= (11x – 3) (4x + 1)

Practice set 6.2 PAGE NO: 31

1. Factorize:

(1) x³ + 64y³

(2) 125p³ + q³

(3) 125k³ + 27m³

(4) 2l³ + 432m³

(5) 24a³ + 81b³

Solution:

(1) x³ + 64y³

We know that,

a³ + b³ = (a + b) (a² – ab + b²)

x³ + 64y³ = (x)³ + (4y)³

Here, a = x and b = 4y

Now, substituting in the above formula, we get

x³ + (4y)³ = (x + 4y) [x² – x(4y) + (4y)²]

= (x + 4y) (x² – 4xy + 16y²)

(2) 125p³ + q³

We know that,

a³ + b³ = (a + b) (a² – ab + b²)

125p³ + q³ = (5p)³ + q³

Here, a = 5p and b = q

Now, substituting in the above formula, we get

(5p)³ + q³ = (5p + q) [(5p)² – (5p)(q) + q²]

= (5p + q) (25p² – 5pq + q²)

(3) 125k³ + 27m³

We know that,

a³ + b³ = (a + b) (a² – ab + b²)

125k³ + 27m³ = (5k)³ + (3m)³

Here, a = 5k and b = 3m

Now, substituting in the above formula, we get

(5k)³ + (3m)³ = (5k + 3m) [(5k)² – (5k)(3m) + (3m)²]

= (5k + 3m) (25k² – 15km + 9m²)

(4) 2l³ + 432m³

We know that,

a³ + b³ = (a + b) (a² – ab + b²)

2l³ + 432m³ = 2 (l³ + 216m³) [By taking out the common factor 2]

= 2 (l³ + (6m)³)

Here, a = l and b = 6m

Now, substituting in the above formula, we get

2 (l³ + (6m)³) = 2 {(l + 6m) [l² – l(6m) + (6m)²]}

= 2 (l + 6m) (l² – 6lm + 36m²)

(5) 24a³ + 81b³

We know that,

a³ + b³ = (a + b) (a² – ab + b²)

24a³ + 81b³ = 3 [(2a)³ + (3b)³] [By taking out the common factor 3]

Here, a = 2a and b = 3b

Now, substituting in the above formula, we get

3 [(2a)³ + (3b)³] = 3 {(2a + 3b) [(2a)² – (2a)(3b) + (3b)²]}

= 3(2a + 3b) (4a² – 6ab + 9b²)

Practice set 6.3 PAGE NO: 32

1. Factorize:

(1) y³ – 27

(2) x³ – 64y³

(3) 27m³ – 216n³

(4) 125y³ – 1

(6) 343a³ – 512b³

(7) 64x² – 729y²

Solution:

(1) y³ – 27

We know that,

a³ – b³ = (a – b) (a² + b² + ab)

y³ – 27 = y³ – (3)³

Here, a = y and b = 3

Now, substituting in the above formula, we get

y³ – (3)³ = (y – 3) [y² + y(3) + (3)2]

= (y – 3) (y² + 3y + 9)

(2) x³ – 64y³

We know that,

a³ – b³ = (a – b) (a² + b² + ab)

x³ – 64y³ = x³ – (4y)³

Here, a = x and b = 4y

Now, substituting in the above formula, we get

x³ – (4y)³ = (x – 4y) [x² + x(4y) + (4y)²]

= (x – 4y) (x² + 4xy + 16y²)

(3) 27m³ – 216n³

We know that,

a³ – b³ = (a – b) (a² + b² + ab)

27m³ – 216n³ = 27 (m³ – 8n³) [By taking out the common factor 27]

= 27 [m³ – (2n)³]

Here, a = m and b = 2n

Now, substituting in the above formula, we get

27 [m³ – (2n)³] = 27 {(m – 2n) [m² + m(2n) + (2n)²]}

= 27 (m – 2n) (m² + 2mn + 4n²)

(4) 125y³ – 1

We know that,

a³ – b³ = (a – b) (a² + b² + ab)

125y³ – 1= (5y)³ – 1³

Here, a = 5y and b = 1

Now, substituting in the above formula, we get

(5y)³ – 1³ = (5y – 1) [(5y)² + (5y)(1) + (1)²]

= (5y – 1) (25y² + 5y + 1)

(6) 343a³ – 512b³

We know that,

a³ – b³ = (a – b) (a² + b² + ab)

343a³ – 512b³ = (7a)³ – (8b)³

Here, a = 7a and b = 8b

Now, substituting in the above formula, we get

(7a)³ – (8b)³ = (7a – 8b) [(7a)² + (7a)(8b) + (8b)²]

= (7a – 8b) (49a² + 56ab + 64b²)

(7) 64x² – 729y²

We know that,

a³ – b³ = (a – b) (a² + b² + ab)

64x² – 729y² = (4x)³ – (9y)³

Here, a = 4x and b = 9y

Now, substituting in the above formula, we get

(4x)³ – (9y)³ = (4x – 9y) [(4x)² + (4x) (9y) + (9y)²]

= (4x – 9y) (16x² + 36xy + 81y²)

2. Simplify:

(1) (x + y)³ – (x – y)³

(2) (3a + 5b)³ – (3a – 5b)³

(3) (a + b)³ – a³ – b³

(4) p³ – (p + 1)³

(5) (3xy – 2ab)³ – (3xy + 2ab)³

Solution:

(1) (x + y)³ – (x – y)³

Let us consider,

Here, a = x + y and b = x – y

By using the formula,

Let us simplify the given expression, we get

(x + y)³ – (x – y)³ = [(x + y) – (x – y)] [(x + y)² + (x + y) (x – y) + (x – y)]

= (x + y – x + y) [(x² + 2xy + y²) + (x² – y²) + (x² – 2xy + y²)]

= 2y(x² + x² + x² + 2xy – 2xy + y² – y² + y²)

= 2y (3x² + y²)

= 6x²y + 2y³

(2) (3a + 5b)³ – (3a – 5b)³

Let us consider,

Here, a = 3a + 5b and b = 3a – 5b

By using the formula,

Let us simplify the given expression, we get

(3a + 5b)³ – (3a – 5b)³ = [(3a + 5b) – (3a – 5b)] [(3a + 5b)² + (3a + 5b) (3a – 5b) + (3a – 5b)²]

= (3a + 5b – 3a + 5b) [(9a² + 30ab + 25b²) + (9a² – 25b²) + (9a² – 30ab + 25b²)]

= 10b (9a² + 9a² + 9a² + 30ab – 30ab + 25b² – 25b² + 25b²)

= 10b (27a² + 25b²)

= 270a²b + 250b³

(3) (a + b)³ – a³ – b³

By using the formula,

By substituting in the above equation, we get

(a + b)³ – a³ – b³ = a³ + 3a²b + 3ab² + b³ – a³ – b³

= 3a²b + 3ab²

(4) p³ – (p + 1)³

By using the formula,

By substituting in the above equation, we get

p³ – (p + 1)³ = p³ – (p³ + 3p² + 3p + 1)

= p³ – p³ – 3p² – 3p – 1

= – 3p² – 3p – 1

(5) (3xy – 2ab)³ – (3xy + 2ab)³

Let us consider,

Here, a = 3xy – 2ab and b = 3xy + 2ab

By using the formula,

Let us simplify the given expression, we get

(3xy – 2ab)³ – (3xy + 2ab)³ = [(3xy – 2ab) – (3xy + 2ab)] [(3xy – 2ab)² + (3xy – 2ab) (3xy + 2ab) + (3xy + 2ab)²]

= (3xy – 2ab – 3xy – 2ab) [(9x²y² – 12xyab + 4a²b²) + (9x²y² – 4a²b²) + (9x²y² + 12xyab + 4a²b²)]

= (- 4ab) (9x²y² + 9x²y² + 9x²y² – 12xyab + 12xyab + 4a²b² – 4a²b² + 4a²b²)

= (- 4ab) (27 xy² + 4a²b²)

= -108x²y²ab – 16a³b³

Practice set 6.4 PAGE NO: 33

1. Simplify:

Solution:

Chapter 6 Factorisation of Algebraic Expressions provides useful resources for students as it helps them in scoring full marks in the examination.

Our experts have solved the difficult problems into simpler steps, which can be easily solved by students. Regular revision of important concepts and formulas over time is the best way to strengthen the concepts.

Many such exercise problems are given in the MSBSHSE Solutions book. Students can refer to them and can secure good marks in their examination.