Class 8 Exam  >  Class 8 Notes  >  Mathematics (Maths) Class 8  >  Worksheet Solutions: Factorisation

Class 8 Maths - Factorisation CBSE Worksheets Solutions

Multiple Choice Questions

Q1:The radius of a circle is 7ab – 7bc – 14a cm, then circumference of the circle is (Given π = 22/7)
(a)
22(ab -bc-2a)
(b) 44(ab -bc-2a)
(c) (ab -bc-2a)
(d) None of these
Ans: (b)

Q2: Factorised form of x2 − (p − 5) x − 5p is
(a)
(x-5)(x-p)
(b) (x+5)(x+p)
(c) (x-p)(x+5)
(d) (x+p)(x-5)
Ans: (a)

Q3: Factorised form of r– 10 r + 21 is
(a)
(r - 7) (r - 3)
(b) (r -7) (r +3)
(c) (r +1) (r - 4)
(d) (r -1) (r - 4)
Ans:
(a)

Q4: The expressionClass 8 Maths - Factorisation CBSE Worksheets Solutions is
(a)
(2x+3)
(b) (3x+2)
(c) (2x -3)
(d) (3x-2)
Ans: (c)

True and False

Q1: x2 + 4x + 3 is factorized as (x+1) (x+3)
Ans: 
True

Q2: x2 + (a + b) x + ab = (a + b) (x + ab)
Ans: 
False

Q3: h is a factor of 2π (h + r).
Ans:
False

Q4: Factors of (96 − 4 x − x2) is (x+12)(8-x)
Ans:
True

Q5: Common factor of 17 abc, 34 ab2, 51a2b is 17ab
Ans:
True

Fill in the Blanks

Q1: −x + x3 is factorised as ______
Ans:
x(x-1)(x+1)

Q2: Class 8 Maths - Factorisation CBSE Worksheets Solutions __________
Ans: 
(x- 3y)

Q3: Factorised form of (x-10)(x+7) + 16 is _______
Ans:
(x-9)(x+6)

Q4: Factorised form of 23xy – 46x + 54y – 108 is _____
Ans:
(23x + 54) (y – 2)

Q5: Factorized form of a12 x4 − ax12 is ____________
Ans:
a4x4(a+ x4)(a2+x2) (a+x) (a−x)

Answer the following Questions

Q1: Find the common factors of the following terms.
(a) 25x2y, 30xy2
(b) 63m3n, 54mn4
Ans:
(a) 25x2y, 30xy2
25x2y = 5 × 5 × x × x × y
30xy2 = 2 × 3 × 5 × x × y × y
Common factors are 5× x × y = 5 xy
(b) 63m3n, 54mn4
63m3n = 3 × 3 × 7 × m × m × m × n
54mn4 = 2 × 3 × 3 × 3 × m × n × n × n × n
Common factors are 3 × 3 × m × n = 9mn

Q2: Factorise the following expressions.
(a) 54m3n + 81m4n2
(b) 15x2y3z + 25x3y2z + 35x2y2z2
Ans:

(a) 54m3n + 81m4n2
= 2 × 3 × 3 × 3 × m × m × m × n + 3 × 3 × 3 × 3 × m × m × m × m × n × n
= 3 × 3 × 3 × m × m × m × n × (2 + 3 mn)
= 27m3n (2 + 3mn)
(b) 15x2y3z + 25 x3y2z + 35x2y2z2 = 5x2y2z ( 3y + 5x + 7)

Q3: Factorise the following polynomials.
(a) 6p(p – 3) + 1 (p – 3)
(b) 14(3y – 5z)3 + 7(3y – 5z)2

Ans:
(a) 6p(p – 3) + 1 (p – 3) = (p – 3) (6p + 1)
(b) 14(3y – 5z)3 + 7(3y – 5z)2
= 7(3y – 5z)2 [2(3y – 5z) +1]
= 7(3y – 5z)2 (6y – 10z + 1)

Q4: Factorise the following:
(a) p2q – pr2 – pq + r2
(b) x2 + yz + xy + xz

Ans:
(a) p2q – pr2 – pq + r2
= (p2q – pq) + (-pr2 + r2)
= pq(p – 1) – r2(p – 1)
= (p – 1) (pq – r2)
(b) x2 + yz + xy + xz
= x2 + xy +xz + yz
= x(x + y) + z(x + y)
= (x + y) (x + z)

Q5: Factorise the following polynomials.
(a) xy(z2 + 1) + z(x2 + y2)
(b) 2axy2 + 10x + 3ay2 + 15
Ans:
(a) xy(z2 + 1) + z(x2 + y2)
= xyz2 + xy + 2x2 + zy2
= (xyz2 + zx2) + (xy + zy2)
= zx(yz + x) + y(x + yz)
= zx(x + yz) + y(x + yz)
= (x + yz) (zx + y)
(b) 2axy2 + 10x + 3ay2 + 15
= (2axy2 + 3ay2) + (10x + 15)
= ay2(2x + 3) +5(2x + 3)
= (2x + 3) (ay2 + 5)

Q6: Factorise the following expressions.
(а) x2 + 4x + 8y + 4xy + 4y2
(b) 4p2 + 2q2 + p2q2 + 8
Ans:

(a) x2 + 4x + 8y + 4xy + 4y2
= (x+ 4xy + 4y2) + (4x + 8y)
= (x + 2y)2 + 4(x + 2y)
= (x + 2y)(x + 2y + 4)
(b) 4p2 + 2q2 + p2q2 + 8
= (4p2 + 8) + (p2q+ 2q2)
= 4(p2 + 2) + q2(p2 + 2)
= (p2 + 2)(4 + q2)

Q7: Factorise:
(a) a2 + 14a + 48
(b) m2 – 10m – 56
Ans:

(a) a2 + 14a + 48
= a2 + 6a + 8a + 48
[6 + 8 = 14 ; 6 × 8 = 48]
= a(a + 6) + 8(a + 6)
= (a + 6) (a + 8)
(b) m2 – 10m – 56
= m2 – 14m + 4m – 56
[14 – 4 = 10; 4 × 4 = 56]
= m(m – 14) + 6(m – 14)
= (m – 14) (m + 6)

Q8: Factorise:
(a) x4 – (x – y)4
(b) 4x2 + 9 – 12x – a2 – b2 + 2ab

Ans:
(a) x4 – (x – y)4
= (x2)2 – [(x – y)2]2
= [x2 – (x – y)2] [x2 + (x – y)2]
= [x + (x – y] [x – (x – y)] [x2 + x2 – 2xy + y2]
= (x + x – y) (x – x + y)[2x2 – 2xy + y2]
= (2x – y) y(2x2 – 2xy + y2)
= y(2x – y) (2x2 – 2xy + y2)
(b) 4x2 + 9 – 12x – a2 – b2 + 2ab
= (4x2 – 12x + 9) – (a2 + b2 – 2ab)
= (2x – 3)2 – (a – b)2
= [(2x – 3) + (a – b)] [(2x – 3) – (a – b)]
= (2x – 3 + a – b)(2x – 3 – a + b)

Q9: Factorise the following polynomials.
(a) 16x4 – 81
(b) (a – b)2 + 4ab
Ans:
 (a) 16x4 – 81
= (4x2)– (9)2
= (4x2 + 9)(4x2 – 9)
= (4x2 + 9)[(2x)2 – (3)2]
= (4x2 + 9)(2x + 3) (2x – 3)
(b) (a – b)2 + 4ab
= a2 – 2ab + b2 + 4ab
= a2 + 2ab + b2
= (a + b)2

Q10: Factorise:
(а) 14m5n4p2 – 42m7n3p7 – 70m6n4p3
(b) 2a2(b2 – c2) + b2(2c2 – 2a2) + 2c2(a2 – b2)

Ans:
(a) 14m5n4p– 42m7n3p7 – 70m6n4p3
= 14m5n3p2(n – 3m2p5 – 5mnp)
(b) 2a2(b2 – c2) + b2(2c2 – 2a2) + 2c2(a2 – b2)
= 2a2(b2 – c2) + 2b2(c2 – a2) + 2c2(a2 – b2)
= 2[a2(b2 – c2) + b2(c2 – a2) + c2(a2 – b2)]
= 2 × 0
= 0

Q11: Factorise:
(a) (x + y)2 – 4xy – 9z2
(b) 25x– 4y2 + 28yz – 49z2
Ans:
(a) (x + y)2 – 4xy – 9z2
= x2 + 2xy + y2 – 4xy – 9z2
= (x2 – 2xy + y2) – 9z2
= (x – y)2 – (3z)2
= (x – y + 3z) (x – y – 3z)
(b) 25x2 – 4y2 + 28yz – 49z2
= 25x2 – (4y2 – 28yz + 49z2)
= (5x)– (2y – 7)2
= (5x + 2y – 7) [5x – (2y – 7)]
= (5x + 2y – 7) (5x – 2y + 7)

The document Class 8 Maths - Factorisation CBSE Worksheets Solutions is a part of the Class 8 Course Mathematics (Maths) Class 8.
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FAQs on Class 8 Maths - Factorisation CBSE Worksheets Solutions

1. What is factorisation?
Ans. Factorisation is the process of breaking down an algebraic expression or number into its factors. It helps us simplify expressions and solve equations by identifying common factors.
2. How can I factorise a quadratic expression?
Ans. To factorise a quadratic expression, we use the technique of "splitting the middle term." We find two numbers that multiply to give the product of the coefficient of the quadratic term and the constant term, and at the same time, add up to give the coefficient of the linear term. These numbers are then used to write the quadratic expression as a product of two binomial expressions.
3. What are the common methods of factorisation?
Ans. Some common methods of factorisation include: - Factorising by taking out the greatest common factor (GCF). - Factorising quadratic expressions using the splitting the middle term method. - Factorising the difference of squares. - Factorising by grouping terms.
4. Can factorisation help in solving equations?
Ans. Yes, factorisation can be extremely helpful in solving equations. By factorising an equation, we can identify the values of the variable that make the equation true. These values are known as the solutions or roots of the equation.
5. What are the applications of factorisation in real life?
Ans. Factorisation has various applications in real life, particularly in fields such as cryptography, engineering, and computer science. It is used in encryption algorithms, signal processing, and data compression techniques. Factorisation also plays a significant role in prime number factorization, which is crucial for security in modern communication systems.
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