CBSE Class 8 Mathematics Chapter 12 ‘Factorisation’ Notes

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Welcome to Online Tutorial Classes! We are excited to present a valuable resource for CBSE students studying Class 8 Mathematics. In this post, we will delve into the CBSE class 8 Mathematics Chapter 12 ‘Factorisation’ from the CBSE curriculum and provide you with a comprehensive notes and worksheets to help you master this important topic.

Chapter 12 of CBSE Class 8 Maths, “Factorisation”, equips students with various methods to break down algebraic expressions into simpler forms by identifying their factors, covering techniques like common factors, regrouping, identities, and special cases like splitting the middle term, while also exploring division of polynomials.

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CBSE class 8 Mathematics Chapter 12 ‘Factorisation’ Overview

CBSE Class 8 Mathematics Chapter 12, titled ‘Factorisation’, provides a comprehensive introduction to the concept of factorisation, for algebraic expressions. The chapter begins by explaining the factors of natural numbers and how every natural number can be expressed as a product of other natural numbers. It then moves on to discuss the factors of algebraic expressions, explaining that terms in algebraic expressions are formed as products of factors. The chapter also introduces the concept of ‘regrouping’ in factorisation, which involves rearranging expressions to make factorisation easier.

The latter part of the chapter delves into more complex topics such as factorisation using identities, division of algebraic expressions, and division of a polynomial by a monomial and by polynomial.

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CBSE class 8 Maths Chapter 12 ‘Factorisation’ Notes

Factors of Natural Numbers

• Factors of natural numbers: You can express a natural number as a product of other natural numbers.
• Prime factors: Out of all the factors of a number, numbers that cannot be further expressed as a product of two or more numbers are called prime factors.
• Prime factor form: A number written as a product of its prime factors is said to be in prime factor form.
• Example:
  • The Prime factor form of 30 is 2 x 3 x 5
  • The prime factor form of 70 is 2 × 5 × 7.
  • The prime factor form of 90 is 2 × 3 × 3 × 5, and so on.

Factors of algebraic expressions

• An algebraic expression can be expressed as a product of its factors.
• expression 5xy + 3x the term 5xy has been formed by the factors 5, x and y, i.e., 5xy = 5 x × x y
• The factors 5, x and y of 5xy cannot further be expressed as a product of factors. We may say that 5, x and y are ‘prime’ factors of 5xy.
• Irreducible form: In algebraic expressions, factors that cannot be further expressed as a product of two or more factors are called irreducible factors.

What is Factorisation?

• Factorisation is expressing an algebraic expression as a product of factors.
• These factors can be numbers, algebraic variables, or algebraic expressions.
• The expression 3x(x + 2) is an example pf irreducable factors, where 3, x and (x + 2) are the irreducible factors.

• Examples of expressions in factor form:
  • 3xy
  • 5x2y
  • 2x(y+2)
  • 5(y+1)(x+2)
• Examples of expressions not in factor form:
  • 2x+4
  • 3x+3y
  • x2+5x
  • x2+5x+6

Method of common factors

• This method is used to factorise expressions that have common factors.
• We can factor out the common factor by using the distributive law.

Example:

Factorise 2x+4

• Write each term as a product of irreducible factors:
  • 2x=2×x
  • 4=2×2
• So the expression becomes: 2 (x) + (2 x 2)
• Notice that the factor 2 is common to both terms.
• Using the distributive law, we can rewrite the expression as: 2x + 4 = 2(x + 4)

• Therefore, 2x+4=2(x+2).

Factorisation by regrouping terms

• This method is used to factorise expressions that do not have common factors throughout.
• We can group the terms of the expression such that each group has a common factor.
• Then, we can factor out the common factors from each group and combine them.

Example 3: Factorise 6xy−4y+6−9x

Steps to solve:

1. Check for common factors: There are no common factors among all the terms.
2. Regrouping:
   • Notice that the first two terms, 6xy and −4y, have a common factor of 2y.
     • 6xy−4y=(2y×3x)+(2y×−2)
   • Observe that if we change the order of the last two terms, 6 −9x to −9x+6, they will have a common factor of 3.
     • −9x+6=−(3×3x)+(3×2)
3. Factoring the grouped expressions:
   • 2y(3x−2)−3(3x−2)
4. Combining the factored terms:
   • (3x−2)(2y−3)

Therefore, the factors of (6xy−4y+6−9x) are (3x−2) and (2y−3).

Factorisation Using Identities

• We can factorise expressions using certain algebraic identities.
• These identities are true for all values of the variables involved.

Commonly used identities for factorisation:

1. Square of a sum: (a+b)²=a²+2ab+b²
2. Square of a difference: (a−b)²=a²−2ab+b²
3. Difference of squares: a²−b²=(a+b)(a−b)

Steps to factorise using identities:

1. Observe the given expression.
2. If it matches the form on the right side of one of the identities, then the expression on the left side of the identity gives the desired factorisation.

Examples:

Example 1: Factorise x2+8x+16

• Observe that the expression is in the form of the square of a sum, where a=x and b=4.
• Therefore, using the identity (a+b)²=a²+2ab+b², we get:
  • x²+8x+4²=(x+4)²

Example 2: Factorise 49p2−36

• Observe that the expression is in the form of the difference of squares, where a=7p and b=6.
• Therefore, using the identity a²−b²=(a+b)(a−b), we get:
  • 49p²−36 = (7p)² – 6² = (7p+6)(7p−6)

Factors of the form (x + a)( x + b)

We know (x + a) (x + b) = x² + (a + b)x + ab

Steps to factorise expressions of the form (x+a)(x+b)

1. Compare the given expression with (x+a)(x+b):
   • Identify the coefficient of the x term and the constant term in the given expression.
   • Let these coefficients be p and q, respectively.
2. Find two numbers that satisfy the following conditions:
   • The product of the two numbers is equal to the constant term, q.
   • The sum of the two numbers is equal to the coefficient of the x term, p.
3. Write the expression as (x+the first number)(x+the second number).

Example 1: Factorise x²+5x+6

Steps to solve:

1. Compare the coefficients:
   • In x²+5x+6, the coefficient of the x term is 5 and the constant term is 6.
   • So, we need to find two numbers that multiply to 6 and add up to 5.
2. Find two suitable numbers:
   • Such two numbers are 2 and 3.
   • 2×3=6 and 2+3=5.
3. Factorise the expression:
   • Therefore, x²+5x+6=(x+2)(x+3).

Division of Algebraic Expressions

• Division is the inverse operation of multiplication.
• This section introduces the concept of dividing algebraic expressions.

Division of a monomial by another monomial

• The process includes factorization and cancellation of common factors.

Example 6x³ ÷ 2x

Division of a Polynomial by a Monomial

Steps to divide a polynomial by a monomial:

1. Express each term of the polynomial in the factored form, where the variable is raised to its highest power.
2. Divide each term of the factored polynomial by the monomial.
3. Simplify the resulting expression.

Example: Divide 24(x²yz + xy²z + xyz²) by 8xyz
Solution: 24 (x²yz + xy²z + xyz²)
= 2 × 2 × 2 × 3 × [(x × x × y × z) + (x × y × y × z) + (x × y × z × z)]
= 2 × 2 × 2 × 3 × x × y × z × (x + y + z) = 8 × 3 × xyz × (x + y + z)
Therefore, 24 (x²yz + xy²z + xyz2) ÷ 8xyz

Alternate Method:

24 (x²yz + xy²z + xyz2) ÷ 8xyz =

Division of a Polynomial by a Polynomial

• The solution involves factoring the dividend and divisor, cancelling out common factors, and arriving at the quotient

Example: Divide 44(x⁴ – 5x³ – 24x²) by 11x (x – 8)
Solution: Factorising 44(x⁴ – 5x³ – 24x²) we get
44(x⁴ – 5x³ – 24x²) = 2 × 2 × 11 × x² (x² – 5x – 24)
(taking the common factor x² out of the bracket)
= 2 × 2 × 11 × x²(x²– 8x + 3x – 24) = 2 × 2 × 11 × x² [x (x – 8) + 3(x – 8)] = 2 × 2 × 11 × x² (x + 3) (x – 8)

Therefore

44(x⁴ – 5x³ – 24x²) ÷ 11x (x – 8)

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CBSE class 8 Mathematics Notes

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