Important Questions for CBSE Class 8 Maths Chapter 14 Factorisation

Important Questions Class 8 Maths Chapter 14 – Factorisation

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Chapter 14 of Class 8 Maths is about Factorisation. The essential topics covered in this chapter are:  

• Introduction
• What is Factorisation?
• Division of Algebraic Expressions
• Division of Algebraic Expressions
• Can you Find the Error?

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Our faculty experts acknowledge the significance of solving problems to attain mastery in Maths. After careful research and study of CBSE past years’ question papers and NCERT textbook and exemplar, our team has compiled the most critical questions in our question bank of class 8 maths chapter 14 extra questions. This set of important questions and solutions will help students to furnish themselves with the proper concepts of all the topics relevant to Factorisation. Our step-by-step solutions with formulas and explanations make it easy for students to understand the answers without much difficulty. Extramarks believes in incorporating joyful learning experiences through its own repository of resources. Besides providing Important Questions Class 8 Maths Chapter 14, Extramarks also gives many other exam-oriented study materials. All the solutions provided by us are strictly in accordance with the NCERT books following the latest CBSE syllabus and guidance, making them completely reliable for the students to enhance their learning experience and achieve excellent results.

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Factorisation Class 8 Extra Questions with Solutions

Given below are a few questions and their solutions from factorization class 8 important questions. Students can get the question bank by registering on our Extramarks website.

Question 1:  Find the common factors of the given term:

• 6abc, 24ab², 12a²b

• 10pq, 20qr, 30rp

• 3x²y³, 10x³y², 6x²y²z

Answer 1: (a)  On factorising 6abc, 24ab² and 12a²b, we get

6abc = 2 × 3 × a × b × c

24ab² = 2 × 2 × 2 × 3 × a × b × b

12a²b = 2 × 2 × 3 × a × a × b

Hence, the common factors of 6abc, 24ab² and 12a²b are 2, 3, a and b

Therefore, multiplying the common factors we get 

2 × 3 × a × b = 6ab

(b) On factorising 10pq, 20qr and 30rp, we get

10pq = 2 × 5 × p × q

20qr = 2 × 2 × 5 × q × r

30rp = 2 × 3 × 5 × r × p

Hence, the common factors are 2 and 5

Therefore, multiplying the common factors we get

2 × 5 = 10

(c ) On factorising 3x²y³, 10x³y², 6x²y²z, we get

3x²y³ = 3 × x × x × y × y × y

10x³y² = 2 × 5 × x × x × x × y × y

6x²y²z = 2 × 3 × x × x × y × y × z

Hence, the common factors are x, x, y and y

Therefore, multiplying the common factors we get

x × x × y × y = x²y²

Question 2: Factorise the following expressions

• ax²y + bxy² + cxyz

• z – 7 + 7xy – xyz

Answer 2: (a) On factorising ax²y,  bxy² and cxyz, we get

ax²y = a + x + x + y

bxy² = b × x × y × y

cxyz = c × x × y × z

Hence, the common factors are x and y

Therefore, ax²y + bxy² + cxyz = xy ( ax + by + cz )

(b) z – 7 + 7xy – xyz

=⟩ z -7 – z (xy) + 7 (xy)

=⟩ ( z – 7) – xy ( z – 7 )

=⟩ ( 1 – xy ) ( z – 7 )

Question 3: Factorise the following expressions.

• (l + m)² – 4lm (Hindi: Expand (l + m)² first)

• 25m² + 30m + 9

• 16x5 – 144x³ 

• (l + m)² – (l – m)²

Answer 3: (a) (l + m)² – 4lm

=⟩ l² + m² + 2lm – 4lm  

[Using (x + y)² = x² + 2xy + y²]

=⟩ l² + m² – 2lm

=⟩ (l – m)²  

[Using (x – y)² = x² – 2xy + y²]

(b) 25m² + 30m + 9

=⟩ (5m)² + 2 × 5m × 3 + 3²

=⟩ (5m + 3)²    

[Using (x + y)² = x² + 2xy + y² ]

(c ) 16x5 – 144x³

=⟩ 16x³ (x² – 9)

=⟩ 16x³ (x – 3) (x + 3).  [Using (x² – y²) = (x + y)(x – y)

(d) (l + m)² – (l – m)²

=⟩ {(l + m) – (l – m)} {(l + m) + (l – m)}.   

 [Using x² – y² = (x + y) (x – y)]

=⟩ (l + m – l + m) (l + m + l – m) 

=⟩ (2m) (2l)

= 4ml

Question 3: Factorise the following expressions:

• 10ab + 4a + 5b + 2

• a⁴ – 2a²b² + b⁴

• q² – 10q + 21

Answer 3: (a) 10ab + 4a + 5b + 2 

=⟩ 5b (2a + 1) + 2(2a + 1)

=⟩ (5b +2) (2a + 1)

(b) a⁴ – 2a²b² + b⁴

=⟩ (a²)² – 2a²b² + (b²)²

=⟩ (a² – b²)²

=⟩ {(a – b) (a + b)}²

=⟩ (a – b) ² (a + b)²

(c ) q² – 10q + 21

Here we observe that,

21 = -7 × -3 and -7 + (-3) = -10

=⟩ q² – 10q + 21 = q² – 3q – 7q + 21

=⟩ q(q – 3) – 7(q – 3)

=⟩ (q – 7) (q- 3)

Question 4: Carry out the following divisions.

• 34x³y³z³ ÷ 51xy²z³

• (x³ + 2x² + 3x) ÷ 2x

• 9x²y²(3z – 24) ÷ 27xy(z – 8)

Answer 4: (a) 34x³y³z³ / 51xy²z³

= 2 × 17 × x × x × x × y × y × y × z × z × z / 3 × 17 × x × y × y × z × z × z 

= 2 x²y /3

(b) (x³ + 2x² + 3x) = x(x² + 2x + 3)

Therefore, x(x² + 2x + 3) / 2x 

= (x² + 2x + 3) / 2

(c ) 9x²y²(3z – 24) / 27xy(z – 8)

= 9x²y² × 3(z – 8) / 27xy (z – 8)

= xy

Question 5: Divide the following as directed

• 20(y + 4) (y² + 5y + 3) ÷ 5(y + 4)

• 39y³ (50y² – 98) ÷ 26y²(5y + 7)

Answer 5: (a) 20(y + 4) (y² + 5y + 3) / 5(y + 4)

= 4(y² + 5y + 3)

(b) In this case, first we have to factorise 50y² – 98

50y² – 98 = 2(25y² – 49) = 2(5y + 7) (5y – 7)

Therefore, 39y³(50y² – 98) / 26y² (5y + 7)

= 2 × 3 × 13 × y³ (5y + 7) (5y – 7) / 2 × 13 × y²(5y + 7)

= 3y (5y – 7) 

Question 6:  Find and correct the errors in the statement

(3x + 2)² = 3x² + 6x + 4

Answer 6: L. H. S. = (3x + 2)² 

= (3x)² + 2² + 2 × 2 × 3x 

= 9x² + 4 + 12x 

1. H. S. = 3x² + 6x + 4

Therefore, L. H. S. ≠ R. H. S.

Hence, correct statement is (3x + 2)² = 9x² + 4 + 12x

Question 7:Find and correct the errors in the statement

 (2a + 3b) (a – b) = 2a² – 3b²

Answer 7: L. H. S. = (2a + 3b) (a – b) 

= 2a(a – b) + 3b(a – b)

= 2a² – 2ab + 3ab – 3b²

= 2a² + ab – 3b²

1. H. S. = 2a² – 3b²

Therefore, L. H. S. ≠ R. H. S.

Hence, the correct statement is (2a + 3b) (a – b) = 2a² + ab – 3b²

Question 8:Find and correct the errors in the statement

 (z + 5)² = z² + 25

Answer 8: L. H. S. = (z + 5)²

(z + 5)² = z² + 10z + 25 

[Using identity (a + b)² =  a² + 2ab + b²]

1. H. S. = z² + 25

Hence, L. H. S. ≠  R. H. S.

Therefore, the correct statement is (z + 5)² = z² + 10z + 25

Question 9: Find and correct the errors in the statement

 3x / (3x + 2) = 1 / 2

Answer 9: L. H. S. = 3x / (3x + 2) 

1. H. S. = 1 / 2

Therefore, L. H. S. ≠  R. H. S. 

Hence, 3x / (3x + 2) = 3x / (3x + 2)

Question 10:Find and correct the errors in the statement

 (7x + 5) / 5 = 7x

Answer 10: L. H. S. = (7x + 5) / 5 

= 7x/5 + 5/5 

= 7x/5 + 1

1. H. S. = 7x

Therefore, L. H. S. ≠ R. H. S.

Hence, the correct statement is (7x + 5) / 5 = (7x/5) + 1

Question 11: Factorise 4x² – 20x + 25.

Answer 11: 4x² – 20x + 25

= (2x)² – 2 × 2x × 5 + (5)²

= (2x – 5)²  

[Using the identity a² – 2ab + b² = (a – b)²]

Question 12: Verify that

(3x + 5y)² – 30xy = 9x² + 25y² 

Answer 12: L. H. S. = (3x + 5y)² – 30xy 

9x² + 30xy + 25y² – 30xy = 9x² + 25y²

1. H. S. = 9x² + 25y²

Therefore, L. H. S. = R. H. S. (verified)

Question 13: Verify that

(11pq + 4q)² – (11pq – 4q)² = 176pq²

Answer 13: L. H. S. = (11pq + 4q)² – (11pq – 4q)²

= 121p²q² + 88pq² + 16q² – (121p²q² – 88pq² + 16q²)

[ Using identities (a + b)² = (a² + 2ab + b²) 

And (a – b)² = (a² – 2ab + b²) ]

= 121p²q² + 88pq² + 16q² – 121p²q² + 88pq² – 16q²

= 88pq² + 88pq²

= 176pq²

1. H. S. = 176pq²

Therefore, L. H. S. = R. H. S. (verified)

Question 14: The area of a rectangle is x² + 12xy + 27y² and its  length is (x + 9y). Find the breadth of the rectangle.

Answer 14: Area / Length

= (x² + 12xy + 27y²) / (x + 9y)

= x(x + 9y) + 3y(x + 9y) / (x + 9y)

= (x + 3y) (x + 9y) / (x + 9y)

= (x + 3y)

Hence, the breadth of the rectangle is (x + 3y)

Question 15: Divide 15(y + 3)(y² – 16) by 5(y² – y – 12).

Answer 15: On factorising 15(y + 3)(y² – 16), we get 5 × 3 × (y + 3)(y – 4)(y + 4).

On factorising 5(y² – 4y + 3y – 12)

= 5(y – 4)(y + 3)

Therefore, on dividing the first expression by second expression, we get

15(y + 3) (y² – 16) / 5 (y +3) (y – 4)

= 3(y + 4)

Question 16: Factorise 2ax² + 4axy + 3bx² + 2ay² + 6bxy + 3by².

Answer 16: 2ax² + 4axy + 3bx² + 2ay² + 6bxy + 3by²

= 2ax² + 4axy + 3bx² + 6bxy + 2ay² + 3by²

= 2ax(x + 2y) + 3bx(x + 2y) + 2y²(2a + 3b) 

= x(2a + 3b)(x + 2y) + 2y²(2a + 3b)

= (2a + 3b) [x(x + 2y) + 2y²]

= (2a + 3b) ( x² + 2y² + 2xy]

Question 17: Factorise 4a² – 4ab + b²

Answer 17: 4a² – 4ab + b² 

= (2a)² – 2(2a)(b) + b²

= (2a – b)²

[Using the identity a² – 2ab + b² = (a – b)² ]

Question 18: Factorise 3a²b³ – 27a⁴b

Answer 18: 3a²b³ – 27a⁴b

= 3a²b(b² – 9a²)

= 3a²b(b² – (3a)²)

= 3a²b(b + 3a)(b – 3a)

[Using the identity (a² – b²) = (a + b)(a – b)

Question 19: Factorise (4x² / 9) – (9y² / 16)

Answer 19: (4x² / 9) – (9y² / 16)

= (2x / 3)² – (3y / 4)²

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

[Using the identity (a² – b²) = (a + b)(a – b)]

Question 20: Factorise 1331x³y – 11y³x

Answer 20: 1331x³y – 11y³x

= 11xy (121x² – y²)

= 11xy [(11x)² – y²]

= 11xy (11x – y)(11x + y)

[Using the identity (a² – b²) = (a + b)(a – b)]

Question 21: The area of a rectangle is x² + 19x – 20. Find the possible length and the breadth of the rectangle.

Answer 21: Area of Rectangle = length × breadth

= x² + 19x – 20 

= x² + 20x – x – 20

= x(x + 20) – 1(x + 20)

= (x – 1) (x + 20)

Thus, the length and the breadth are (x – 1) and (x + 20) 

Question 22: Perform the following division:

(3pqr – 6p²q²r²) ÷ 3pq

Answer (3pqr – 6p²q²r²) ÷ 3pq

= (3pqr – 6p²q²r²) / 3pq

= 3pqr (1 – 2pqr) / 3pq

= r(1 – 2pqr)

Question 23: Perform the following division:

(x³y)/9 – (xy³)/16

Answer 23: (x³y)/9 – (xy³)/16 

= xy(x²/9 – y²/16)

= xy [(x/3)² – (y/4)²]

= xy (x/3 – y/4)(x/3 + y/4)

[Using the identity (a² – b²) = (a + b)(a – b) ]

Question 24: The area of a rectangle is x² + 7x + 12. If the breadth is (x + 3), find its length.

Answer 24: Area of Rectangle = Length × Breadth

=⟩ x² + 7x + 12 = Length × (x + 3)

=⟩ Length = (x² + 7x + 12) / (x +3)

=⟩ Length = (x² + 3x + 4x + 12) / (x + 3)

=⟩ Length = x(x + 3) + 4(x + 3) / (x + 3)

=⟩ Length = (x + 3)(x + 4) / (x + 3)

=⟩ Length = (x + 4)

Question 25: The area of a circle is given by the expression πx² + 6πx + 9π. Find the radius of the circle.

Answer 25: Area of a circle = πr²

Where radius = r

Then, πx² + 6πx + 9π = πr²

=⟩ π(x² + 6x + 9) = πr²

=⟩ (x² + 6x + 9) = r²

=⟩ r² = (x² + 2.x.3 + 3²)

=⟩ r² = (x + 3)²

Therefore, r = x + 3

Question 26: The sum of the first n natural numbers is given by the expression n²/2 + n/2. Factorise this expression.

Answer 26: Given that the sum of the first n natural number = n²/2 + n/2 = n/2 (n + 1)

Question 27: The sum of (x + 5) observations is x⁴ – 625. Find the mean of the observations.

Answer 27: Mean = (x⁴ – 625) / (x + 5)

Mean = [( x²)² – (25)²] / (x + 5)

Mean = [(x² + 25)(x² – 5²)] / (x + 5)

Mean = [(x² + 25)(x – 5)(x + 5)] / (x + 5)

Mean = (x² + 25)(x – 5)

Question 28: The height of a triangle is x⁴ + y⁴ and its base is 14xy. Find the area of the triangle.

Answer 28: Area of the triangle = 1 /2 × height × base

=⟩ Area = 1 /2 × (x⁴ + y⁴) × (14xy)

=⟩ Area = 7xy (x⁴ + y⁴)

Question 29: The cost of a chocolate is Rs (x + y) and Rohit bought (x + y) chocolates. Find the total amount paid by him in terms of x. If x = 10, find the amount paid by him.

Answer 29: The cost of chocolate = Rs (x + y)

No. of chocolates Rohit bought = (x + y)

Therefore, total amount he paid = Rs (x + y)(x + y)

= Rs (x + y)²

If x = 10, then Rs (10 + y)²

Question 30: The base of a parallelogram is (2x + 3 units) and the corresponding height is (2x – 3 units). Find the area of the parallelogram in terms of x. What will be the area of the parallelogram of x = 30 units?

Answer: Area of Parallelogram = Base × Height

Therefore, Area = (2x + 3)(2x – 3)

Area = (2x)² – (3)² = 4x² – 9

Putting x = 30 units, we get

Area = 4 × (30)² – 9 = 4 × 900 – 9 = 3600 – 9 = 3591sq. units.

Question 31: The radius of a circle is 7ab – 7bc – 14ac. Find the circumference of the circle. (π = 22/7)

Answer 31: The circumference of the circle = 2πr

Therefore, Circumference = 2π (7ab – 7bc – 14ac)

Circumference = 2 × 22/7 (7ab – 7bc – 14ac)

= 2 × 22 (ab – bc – 2ac)

= 44(ab – bc – 2ac)

Question 32: Factorise p⁴ + q⁴ + p²q²

Answer 32: p⁴ + q⁴ + p²q²

= (p²)² + (q²)² + 2p²q² – p²q²

= (p²+ q²)² – (pq)²

[Using the identity a² + b² + 2ab = (a + b)²]

= (p² + q² + pq)(p² + q² – pq)

[Using the identity a² – b² = (a + b)(a – b)]

Question 33: Factorise the expression and divide them as directed:

(2x³ – 12x² + 16x) ÷ (x – 2)(x – 4)

Answer 33: (2x³ – 12x² + 16x) / [(x – 2)(x – 4)]

= [2x (x² – 6x + 8)] / [(x – 2)(x – 4)]

= [2x (x² – 2x – 4x + 8)] / [(x – 2)(x – 4)]

= [2x {x (x – 2) – 4 (x – 2)}] / [(x – 2) (x – 4)]

= [2x (x – 4)(x – 2)] / [(x – 2) (x – 4)]

= 2x

Question 34: Factorise x² + 1/x² + 2 – 3x – 3/x

Answer 34: x² + 1/x² + 2 – 3x – 3/x 

=⟩ x² + 1/x² + 2 – 3 (x + 1/x²)

=⟩ (x + 1/x)² – 3 (x + 1/x²)

[Using the identity a² + b² + 2ab = (a + b)² ]

=⟩ (x + 1/x) (x + 1/x – 3)

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