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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 2x+ 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, 30xyis  ___________. 

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)

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 in mathematics?
Ans.Factorisation is the process of breaking down an expression into a product of simpler factors. This can help simplify equations and make them easier to solve.
2. How do I factorise a quadratic equation?
Ans.To factorise a quadratic equation of the form \( ax^2 + bx + c \), you need to find two numbers that multiply to \( ac \) and add to \( b \). Then, rewrite the equation using these numbers to factor it into the form \( (px + q)(rx + s) \).
3. What are the common methods of factorisation?
Ans.Common methods of factorisation include grouping, using the difference of squares, factoring by trinomial, and using the quadratic formula. Each method is suitable for different types of expressions.
4. Can all algebraic expressions be factorised?
Ans.No, not all algebraic expressions can be factorised into simpler forms. Some expressions, like prime polynomials, do not have factors other than 1 and themselves.
5. Why is factorisation important in solving equations?
Ans.Factorisation is important because it allows us to simplify equations, making it easier to find solutions. By expressing an equation in its factored form, we can apply the zero product property to solve for the variable efficiently.
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