Factorisation Questions

The factorisation questions and answers are provided here to help students grasp the concept fast. One of the most important topics taught in primary and secondary schools is factorisation. These questions can be used by students to get a fast overview of the topics and to practice them so that they are more familiar with the concept. Examine your answers with the answers provided on our page. Check this page for further information about factorisation.

Go through the below provided factorisation questions and answer and practice the questions as well.

Factorisation Questions with Solutions

1. Factorise the number 57

Solution:

As we know that the number 57 is a composite number. Thus, the factors of 57 are 1, 3, 19 and 57.

Thus, 57 can be factorised as 3 × 19, also known as the prime factorisation of 57.

Here, both the numbers 3 and 19 are prime numbers.

2. Prime factorise the number 150.

Solution:

As the number 150 is a composite number, the factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, and 150.

To factorise the number 150, we have to find the prime factorisation of a number.

Now, let us find the prime factors of the number 150, using the division method.

First, divide the number 150 by the smallest prime number. That is 2.

⇒ 150/2 = 75

So, again divide 75 by 2,

⇒75/2 = 37.5, which cannot be a factor.

Now, move to the next prime number 3.

⇒ 75/3 = 25

Again divide 25/3 = 8.333, which cannot be a factor.

So, the next prime number is 5 and now divide 25 by 5.

⇒ 25/5 = 5

⇒ 5/5 = 1

Finally, we get the quotient 1, and we cannot further continue with the division operation.

Hence, the factorisation of the number 150 is 2 × 3 × 5 × 5 (or) 2 × 3 × 5² where 2, 3, and 5 are the prime numbers.

Also, read: Prime Numbers.

3. Factorise the expression 5a³ – 10a².

Solution:

The factors of 5a³ are 5 × a × a × a.

The factors of 10a² are 2 × 5 × a × a.

Thus, the common factor for both the terms is 5a².

Now, take the common factor outside, we get

⇒ 5a² (a – 2)

Therefore, the factors of 5a³ – 10a² is 5a², (a-2)

4. Factorise the expression m² + 4m.

Solution:

The factors of m² are m × m.

The factors of 4m are 4 × m.

The common factor for both the terms is m.

Take the common factor outside,

⇒ m(m+4)

Thus, the given expression m²+4m is factorised as m(m+4).

5. Factorise the expression m² – 10m + 25.

Solution:

Given expression: m² – 10m + 25.

There is no common term in all the given three terms, such as m², 10m, 25.

So, use the algebraic identity that matches the given algebraic expression.

Here, we can use the algebraic identity (a-b)² = a² – 2ab + b².

Compare the given expression and the identity, we get a = m, b = 5

Hence, the factors of m² – 10m + 25 = (m – 5)² = (m – 5)(m-5).

Therefore, the given expression m² – 10m + 25 is factorised as (m – 5)² or (m – 5)(m – 5).

6. Factorise 49y² – 1.

Solution:

Given expression 49y² – 1.

This can be written as (7y)² – (1)^2.

Now, this expression is similar to the algebraic identity, a² – b², which is equal to (a-b)(a+b)

Here a = 7y and b = 1.

Therefore,

(7y)² – (1)² = (7y – 1)(7y + 1)

Therefore, the factorisation of 49y² – 1 is (7y -1 )(7y + 1).

7. Factorise 16x⁴ – y⁴.

Solution:

Given expression: 16x⁴ – y⁴

This can be written as (4x²)² – (y²)^2.

Now, this expression is similar to the algebraic identity, a² – b² = (a-b)(a+b)

Here a = 4x² and b = y².

Therefore,

(4x²)² – (y²)² = (4x² – y²)(4x² + y²)

Again (4x² – y²) = (2x)² – (y)², and apply the same identity.

(4x² – y²) = (2x + y)(2x – y)

So, (4x²)² – (y²)² = (2x + y) (2x – y) (4x² + y²)

Hence, the factorisation of 16x⁴ – y⁴ is (2x + y) (2x – y) (4x² + y²)

8. Factorise 54a³b + 81a⁴b².

Solution:

Given expression: 54a³b + 81a⁴b²

It can also be written as 54a³b + 81 a³abb.

Now, 54 = 2 × 27

81 = 3 × 27

⇒ 54a³b + 81a⁴b² = (2 × 27×a³b ) + (3 × 27 × a³abb)

Factor out the common terms, we get

⇒ 27a³b (2 + 3ab)

Therefore, the factorisation of 54a³b + 81a⁴b² is 27a³b (2 + 3ab).

9. Factorise 16a² – 25y².

Solution:

Given expression: 16a² – 25b²

16a² – 25b² = (4a)² – (5b)²

The expression is of the form a² – b².

⇒ (4a)² – (5b)² = (4a + 5b) (4a – 5b)

Hence, the factorisation of 16a² – 25b² is (4a + 5b) (4a – 5b)

10. Factorise a² + bc + ab + ac.

Solution:

Given expression: a² + bc + ab + ac.

Now, rearrange the terms, we get

⇒ a² + ab + bc + ac

⇒ a (a + b) + c( a + b)

⇒ (a + c) (a + b)

Therefore, the factorisation of a² + bc + ab + ac is (a + c) (a + b).

Practice Questions

1. Factorise the number 21.
2. Factorise the expression 4x² + 2x.
3. Factorise the expression a² + 6a + 9.