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.

Factorisation Definition: Factorisation is the process of finding factors of a number or a polynomial or an algebraic expression. Generally, the term “factor” is used to express the number as the product of the numbers. The factorisation of an algebraic expression defines the process of finding out the factors of the given algebraic expression.

For example, the algebraic expression 5ab is factorised as 5×a×b.

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

Factorisation Questions with Solutions

The factorisation formula for any number can be given as follows:

N = Ax × By × Cz

where,

  • N be any number
  • A, B and C are the factors of number N
  • x, y, and z are the exponents of factors A, B, and C, respectively.

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.

 

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