﻿ Factorisation - Class 8 Maths Chapter 6 Question Answer

# Factorisation

## Factorisation - Sub Topics

In this chapter, factorisation is a valuable tool in mathematics that simplifies complex expressions into simpler components and helps to solve equations. By understanding the basic methods of factorisation, you can make your math journey smoother and tackle a wide range of mathematical problems with ease.

• Factors
• Factorisation
• Standard Identities
• Division of Polynomial
• Methods of Factorisation
• Solved Questions on Factorisation
• ## Factors

A mathematical expression that is made up of two or more smaller expressions, each of those smaller expressions is called a factor.

Examples: Factors of an expression “3x + 7” is shown as:

Factors of an expression “3ab2 5a2b” is shown as:

## Factorisation

Factorisation is the process of writing a complex expression as a product of simpler expressions.

Example: Factorisation of an expression “3x2 + 9xy – 5x3 + 15x” is shown as:

### Standard Identities

These are important standard identities that can help us simplify expressions:

1. (a + b)² = a² + 2ab + b²
→ This tells us how to square the sum of two numbers.
2. (a b)² = a² 2ab + b²
→ This helps us understand how to square the difference between two numbers.
3. (a + b)(a b) = a²
→ It is a special formula for multiplying the sum and difference of two numbers.

### Some Useful Identities

Here are some useful identities that can help us simplify expressions:

### Division of Polynomial

If one polynomial (Dividend) is divided by another polynomial (Divisor), we get a result called a quotient. The polynomial we divide by is called the divisor and the extra part left over is called the remainder.

### Methods of Factorisation

Let's explore various methods to solve expression using factorization:

1) Method of Common Factors

Step 1: First, we look at all the parts of an expression and find the smallest parts (called irreducible factors) within each of them.

Step 2: Next, we check if there are any of these small parts that are the same in all the terms of the expression. These are called common factors.

Step 3: To factorize the expression, we take the common factor we found in Step 2 and multiply it by what is left after removing the common part.

Example: Factorisation of an expression “3x2 + 9xy – 2x3 + xy” using common factors is shown as:

2) Factorisation by Regrouping Terms

Step 1: If you don't see any common parts in the expressions, don't worry. First, group the terms together.

Step 2: Once you have grouped them, look inside each group and find any common pieces in those smaller groups.

Step 3: When you find common parts, put them together as a pair. This pair of common parts becomes one of the factors you're looking for.

Example: Factorisation of an expression “5x2 + 2x – 20x – 8” by regrouping terms is shown as:

3) Factorisation Using Identities

Identities help you to break down expressions easily. Here are a few of them:

1. (a + b)² = a² + b² + 2ab

2. (a b)² = a² + b² 2ab

3. b² = (a + b)(a b)

Now, when you have an expression, you can use the above identities. Here's how it works:

→ (x + 5)² = (x)² + (5)² + 2(x)(5) = x² + 25 + 10x = x² + 10x + 25
→ (3x − 7)² = (3x)² + (7)² − 2(3x)(7) = 9x² + 49 − 42x = 9x² − 42x + 49
→ (4x + 11)(4x − 11) = (4x)² − (11)² = 16x² − 121
→ 169x² − 144 = (13x)² − (12)² = (13x + 12)(13x − 12)

Example: Factorisation of an expression “9x2 21x + 10” by using identities is shown as:

Step 1: We start by looking for factors of the product of the first and last terms, which is 9 × 10 = 90.

Step 2: We want to find pairs of factors of 90 that add up to the middle term, which is −21. The pair is (−6, −15).

Step 3: Now, we need to split the middle term (−21x) into two parts using the pair we found in Step 2. So, we rewrite it as − 6x − 15x.

Step 4: After breaking the middle term, the expression looks like this: 9x2 − 6x − 15x + 10

Step 5: We group the terms into pairs: (9x2 − 6x) − (15x − 10)

Step 6: Now, we find the greatest common divisor (GCD) for each pair.

For the first pair, it is 3x such that:

(9x2 − 6x) = 3x(3x − 2)

Step 7: Now, we find the greatest common divisor (GCD) for each pair.

For the second pair, it is 5 such that: − (15x − 10) = −5(3x − 2)

Step 8: Combining steps 6 and 7.

(9x2 − 6x) − (15x − 10) = 3x(3x 2)  −5(3x 2)

Step 9: Finally, we notice that both pairs share the common factor (3x 2), so we factor that out from both terms: (3x 2)(3x − 5)

Hence, factorisation of an expression “9x2 21x + 10is (3x 2)(3x 5)”.