Class 8 Maths - Factorisation CBSE Worksheets Solutions

Multiple Choice Questions

Q1: Factorize 18a²b + 27ab².
(a) 3ab(6a + 9b)
(b) 9ab(2a + 3b)
(c) 6ab(3a + 5b)
(d) 12ab(3a + 6b)
Ans: (b)
We have:
18a²b = 2 × 3² × a × a × b
27ab² = 3³ × a × b × b
The two terms have 3², a, and b as common factors. Therefore:
18a²b + 27ab² = 9ab[(2 × a) + (3 × b)] = 9ab(2a + 3b).
Hence, the required factor form is: 9ab(2a + 3b).

Q2: Which of the following is the greatest common factor (GCF) of 8xy and 4x²?
(a) 4x
(b) 8x
(c) 2xy
(d) 4xy
Ans: (a)
Factors of 8xy are 2³ × x × y.
Factors of 4x² are 2² × x × x.
Common prime factors: 2² and x (Since 4x² has no y, y is not part of the common factor).
Thus, the GCF is 2² × x = 4x.
So Answer is: 4x

Q3: Factorize 49p² - 36.
(a) (49p - 36)(p + 1)
(b) (7p - 6)(7p + 6)
(c) (49p - 6)(p + 6)
(d) (7p + 6)(p - 1)
Ans: (b)
Both 49p² and 36 are perfect squares, and the expression is in the form a² - b².
Using the difference of squares identity:
a² - b² = (a - b)(a + b)
49p² - 36 = (7p)² - 6² = (7p - 6)(7p + 6).
So Answer is: (7p - 6)(7p + 6)

Q4: Which of the following statements is most correct regarding the factorization of 5xy?
(Remember: 5xy = 1 × 5 × x × y.)
(a) 1 is not a factor of 5xy.
(b) 1 is always a factor of any term, including 5xy.
(c) We must always show 1 as a factor when writing 5xy.
(d) 1 cannot be a factor unless there is a special requirement.
Ans: (b)
1 is, by definition, a factor of every number or term.
In factorization, we typically do not list 1 separately unless there is a special reason.
So Answer is: B. 1 is always a factor of any term, including 5xy.

Q5: Factorize the quadratic expression x² - 11x + 30.
(a) (x - 1)(x - 10)
(b) (x - 5)(x - 6)
(c) (x + 5)(x + 6)
(d) (x - 3)(x - 8)
Ans: (b)
We need two numbers that multiply to 30 and add up to -11.
The pair -5 and -6 works because:
-5 + -6 = -11, and -5 × -6 = 30.
Therefore:
x² - 11x + 30 = (x - 5)(x - 6).
So Answer is:  (x - 5)(x - 6)

True and False

Q1: Factorization of 3x²+15x−72 is 3(x+8)(x−3).
Ans:True

Sol: Factorizing3x²+15x−72,
=3(x²+5x−24)
=3(x²+8x−3x−24)
=3[x(x+8)−3(x+8)]
=3(x+8)(x−3)

Q2: Factorization of 6x³−8x² is 2x²(3x−4).
Ans: True 

Sol: Factorising 6x³- 8x²
=2x²(3−4x)

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

Sol: Given statement is false, because the expression does not contain h as the one of the terms in products.
The two factors are 2π and (h + r)

Q4: Factorization of 2a²−8a−64 is 2(a−8)(a+4).
Ans: True

Sol: Factorizing 2a²−8a−64,
=2(a²−4a−32)
=2(a²−8a+4a−32)
=2[a(a−8)+4(a−8)]
=2(a−8)(a+4)

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

Sol: 

Express each term in factored form:

17abc = 17 × a × b × c

34ab² = 17 × 2 × a × b²

51a²b = 17 × 3 × a² × b

Identify common numerical factors:

The numerical coefficients are 17, 34, and 51.

Their greatest common factor (GCF) is 17.

Identify common variable factors:

For a: The powers are a¹, a¹, and a². The minimum power is a¹.

For b: The powers are b¹, b², and b¹. The minimum power is b¹.

For c: c appears only in the first term, so it is not common.

Combine the common factors:

The common factors are 17 × a × b = 17ab.

Fill in the Blanks

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

Sol: 

Take -x as a common factor:
-x + x³ = -x(1 - x²).

Recognize that 1 - x² is a difference of squares:
1² - x² = (1 - x)(1 + x).

Substitute back:
-x + x³ = -x(1 - x)(1 + x).

Q2: Simplified form of  x² + 5xy - 24y²x + 8y is __________
Ans:(x- 3y)

Sol: 

Step 1: Rewriting the middle term:

x² + 5xy - 24y² = x² + 8xy - 3xy - 24y²

Group terms:

(x² + 8xy) - (3xy + 24y²)

Factorize each group:

x(x + 8y) - 3y(x + 8y)

Take out the common factor (x + 8y):

x² + 5xy - 24y² = (x + 8y)(x - 3y)

Step 2: Substitute the factorized form into the original expression:

(x + 8y)(x - 3y)(x + 8y)

Cancel x + 8y (as long as x + 8y ≠ 0):

= x - 3y

Final Answer: x - 3y

Q3: x²+7x+12 is correctly factorised as . _______
Ans: (x+3)(x+4)

Sol: 

Factorizing x² + 7x + 12:

To factorize x² + 7x + 12, we look for two numbers that multiply to 12 and add up to 7. These numbers are 3 and 4. Rewrite the middle term 7x as 3x + 4x, so the expression becomes:

x² + 3x + 4x + 12

Group the terms:

(x² + 3x) + (4x + 12)

Factorize each group:

x(x + 3) + 4(x + 3)

Take out the common binomial factor:

(x + 3)(x + 4)

Final Result: x² + 7x + 12 = (x + 3)(x + 4)

Q4: Factorised form 2x² + 4x is _____
Ans: 2x(x+2) 

Sol: 
To factorize 2x² + 4x, take out the common factor, which is 2x.

2x² + 4x = 2x(x + 2)

Thus, 2x² + 4x can be factorized as 2x(x + 2).

Q5:  Common factor 25x²y, 30xy² is  ___________. 

Ans: Common factor is 5 xy

Sol: 25x²y, 30xy²
25x²y = 5 × 5 × x × x × y
30xy² = 2 × 3 × 5 × x × y × y
Common factors are 5× x × y = 5 xy

Answer the following Questions

Q1: Factorize 15x² - 20x³ + 25x⁴.
Ans:

1. Express each term in its prime factors:

15x² = 3 × 5 × x × x

20x³ = 2 × 2 × 5 × x × x × x

25x⁴ = 5 × 5 × x × x × x × x

2. Find the common factors of the three terms:

The common factors are 5, x, and x = 5x².

3. Factor out the common terms:

15x² - 20x³ + 25x⁴ = 5x² [(3) - (2 × 2 × x) + (5 × x)].

4. Simplify the terms inside the brackets:

= 5x² [3 - 4x + 5x²].

Final answer = 5x² (5x² - 4x + 3).

Q2: Factorize x² - 2xy + y² - z².
Ans: The first three terms of the given expression form (x - y)². The fourth term is a square. So, the expression can be reduced to a difference of two squares.
Thus,
x² - 2xy + y² - z² = (x - y)² - z²
Using the difference of squares identity:
= [(x - y) - z] [(x - y) + z]
Final factorization:
= (x - y - z)(x - y + z).

Q3: Find the factors of 4n² + 8n + 4.
Ans: We notice that 4 is a common factor of all the terms.
Therefore,
4n² + 8n + 4 = 4(n² + 2n + 1)
Now,
n² + 2n + 1 = n² + n + n + 1 (as 2n = n + n)
= n(n + 1) + 1(n + 1)
= (n + 1)(n + 1)
Therefore,
4n² + 8n + 4 = 4(n + 1)(n + 1)

Q4: Divide n(4n² - 36) by 4n(n + 3).
Ans: 
Dividend = n(4n² - 36)
= n[(4 × n²) - (4 × 9)]
= n × 4 × (n² - 9)
= 4n × (n + 3)(n - 3) [using the identity a² - b² = (a + b)(a - b)]
Thus,
n(4n² - 36) / 4n(n + 3)
= [4n(n + 3)(n - 3)] / [4n(n + 3)]
= (n - 3)

Q5: Divide as directed: 7(3x + 2)(x - 4) ÷ (3x + 2)
Ans: The given expression is:
7(3x + 2)(x - 4) / (3x + 2)
Cancel the common factor (3x + 2) from the numerator and denominator:
= 7(x - 4)
Therefore, the result is:
7(x - 4)

Q6: Factorize 81q² - 16.
Ans: There are two terms; both are squares, and the second is negative. The expression is of the form a² - b². 
Using the Identity [(a² - b²) = (a + b)(a - b)]:
81q² - 16 = (9q)² - (4)²
= (9q - 4)(9q + 4) (required factorization)

Q7: Find the factors of x² - 5x + 6.
Ans: We note that 6 = 2 × 3, and 2 + 3 = 5.
Therefore,
x² - 5x + 6 = x² - 2x - 3x + 6
= x(x - 2) - 3(x - 2)
= (x - 2)(x - 3)

Q8: Factorize the expressions and divide them as directed:
x² + 6x + 8 ÷ (x + 4)
Ans: First, factorize the expression x² + 6x + 8.
We note that 8 = 4 × 2, and 4 + 2 = 6.
Therefore,
x² + 6x + 8 = (x + 4)(x + 2)
Now, divide by (x + 4):
[(x + 4)(x + 2)] / (x + 4) = x + 2
Thus, the result is:
x + 2

Q9: Simplify the expression: 12xy(9x² - 16y²) ÷ 4xy(3x + 4y)
Ans: 
First, factorize the term 9x² - 16y² as it is a difference of squares:
9x² - 16y² = (3x - 4y)(3x + 4y) [Using the Identity (a² - b²) = (a - b)(a + b)]
Substitute the factorized form into the numerator:
12xy(9x² - 16y²) = 12xy · (3x - 4y)(3x + 4y)
The given expression becomes:
[12xy(3x - 4y)(3x + 4y)] / [4xy(3x + 4y)]
Cancel the common factors 4xy and (3x + 4y) from the numerator and denominator:
= 3(3x - 4y)
Final Result:
3(3x - 4y)

Q10: Factorize the expression x⁴ - (y + z)⁴ using the difference of squares formula.
Ans: We observe that the given expression is in the form of a⁴ - b⁴, which can be expressed using the difference of squares formula twice:
a⁴ - b⁴ = (a² - b²)(a² + b²)
Here, a = x² and b = (y + z)².
Step 1: Apply the difference of squares formula:
x⁴ - (y + z)⁴ = (x² - (y + z)²)(x² + (y + z)²)
Step 2: Factorize x² - (y + z)² further using the difference of squares formula again:
x² - (y + z)² = (x - (y + z))(x + (y + z))
Step 3: Combine all factors:
x⁴ - (y + z)⁴ = (x - (y + z))(x + (y + z))(x² + (y + z)²)