Factorization

1. Introduction: What is Factorization?

In arithmetic, you have already written numbers as products of factors, for example 12=3×412=3×4 or 12=2×2×312=2×2×3. In algebra, factorisation means writing an algebraic expression as a product of simpler expressions called factors.

• Example: 3x+6=3(x+2)3x+6=3(x+2). Here, 3x+63x+6 is written as product of 33 and (x+2)(x+2).
• Example: x2−9=(x−3)(x+3)x2−9=(x−3)(x+3).

This chapter is important because:

• It simplifies algebraic expressions.
• It helps in solving equations, especially quadratic equations in Classes 9–10.
• It is used in many later topics like polynomials, algebraic fractions and coordinate geometry.

2. Basic Concepts and Vocabulary

Before learning methods, understand some terms used in Classes 8–10.

1. Term
   A term is a number, a variable, or their product, for example 5×5x, −3y2−3y2, 7ab7ab.
2. Like and unlike terms
   • Like terms: Same variables with same powers (e.g., 3×23x2 and −5×2−5x2).
   • Unlike terms: Different variables or powers (e.g., 2×2x and 3×23x2).
3. Monomial, binomial, trinomial, polynomial
   • Monomial: expression with one term, e.g., 7×37x3.
   • Binomial: two terms, e.g., x+5x+5.
   • Trinomial: three terms, e.g., x2+3x+2x2+3x+2.
   • Polynomial: expression with many terms, e.g., 2×3−x2+4x−72x3−x2+4x−7.​
4. Factor and factorisation
   If A×B=CA×B=C, then AA and BB are factors of CC. Writing CC as A×BA×B is factorisation.

3. Factorization by Taking Out Common Factor (Class 8–9 Core)

The simplest method is to look for a common factor in all the terms and take it out.

Steps

1. Split each term into its prime numerical factors and variable factors.
2. Identify the greatest common factor (GCF) of all terms.
3. Take this GCF outside a bracket.
4. Write the remaining expression inside the bracket.

Example 1

Factorise 12x2y+16xy212x2y+16xy2.

1. 12x2y=2×2×3×x×x×y12x2y=2×2×3×x×x×y
   16xy2=2×2×2×2×x×y×y16xy2=2×2×2×2×x×y×y
2. Common numerical factor: 2×2=42×2=4.
   Common variables: xx and yy.
   So GCF = 4xy4xy.
3. Take out 4xy4xy:
   12x2y+16xy2=4xy(3x+4y)12x2y+16xy2=4xy(3x+4y).

Example 2

Factorise 15a3b−20a2b2+5ab15a3b−20a2b2+5ab.

• Common numerical factor: 5.
• Common variables: aa and bb (each term has at least one aa and one bb).
  GCF = 5ab5ab.

So 15a3b−20a2b2+5ab=5ab(3a2−4ab+1)15a3b−20a2b2+5ab=5ab(3a2−4ab+1).

This method is widely used in Class 8 NCERT Factorization chapter.

4. Factorization by Grouping Terms

Sometimes all terms do not have a single common factor, but we can group terms to make pairs with common factors.

Steps

1. Arrange terms in a suitable order (sometimes rearrangement helps).
2. Group the terms in pairs so that each pair has a common factor.
3. Factorise each group by taking out common factor.
4. If you get a common binomial factor, factor it out.

Example 1

Factorise ax+ay+bx+byax+ay+bx+by.

1. Group as (ax+ay)+(bx+by)(ax+ay)+(bx+by).
2. Take common factor in each group:
   a(x+y)+b(x+y)a(x+y)+b(x+y).
3. Now (x+y)(x+y) is common:
   (x+y)(a+b)(x+y)(a+b).

Example 2

Factorise x2+5x+2x+10x2+5x+2x+10.

1. Group: (x2+5x)+(2x+10)(x2+5x)+(2x+10).
2. Common factors:
   x(x+5)+2(x+5)x(x+5)+2(x+5).
3. (x+5)(x+5) is common:
   (x+5)(x+2)(x+5)(x+2).

This method is a bridge between simple factorization and quadratic trinomials seen in Classes 9–10.

5. Factorization Using Standard Identities

NCERT repeatedly uses certain algebraic identities for factorisation.

Important identities:

1. a2−b2=(a−b)(a+b)a2−b2=(a−b)(a+b).
2. a2+2ab+b2=(a+b)2a2+2ab+b2=(a+b)2.
3. a2−2ab+b2=(a−b)2a2−2ab+b2=(a−b)2.
4. (x+a)(x+b)=x2+(a+b)x+ab(x+a)(x+b)=x2+(a+b)x+ab (used a lot in Class 9–10).

Example 1 (Difference of squares)

Factorise x2−16x2−16.

• Recognise x2−16=x2−42x2−16=x2−42.
• Apply a2−b2a2−b2:
  x2−42=(x−4)(x+4)x2−42=(x−4)(x+4).

Example 2 (Perfect square trinomial)

Factorise y2+6y+9y2+6y+9.

• Check: 9=329=32, and 6y=2×y×36y=2×y×3.
• So expression is y2+2⋅y⋅3+32y2+2⋅y⋅3+32.
• Using a2+2ab+b2a2+2ab+b2:
  y2+6y+9=(y+3)2y2+6y+9=(y+3)2.

Example 3 (Another perfect square)

Factorise 4×2−12x+94x2−12x+9.

• Recognise 4×2=(2x)24x2=(2x)2, 9=329=32, and −12x=−2⋅2x⋅3−12x=−2⋅2x⋅3.
• So it is (2x)2−2(2x)(3)+32(2x)2−2(2x)(3)+32.
• Apply a2−2ab+b2a2−2ab+b2:
  (2x−3)2(2x−3)2.

The Class 8 Factorization chapter has many problems based on these identities, and they are heavily used in Class 9 and 10 polynomials as well.

6. Factorization of Quadratic Expressions (Middle-Term Splitting)

From late Class 8 and especially in Class 9–10, you encounter quadratic expressions of the form ax2+bx+cax2+bx+c. One powerful method to factorise them is splitting the middle term.

General Idea

For ax2+bx+cax2+bx+c:

1. Find two numbers mm and nn such that:
   • m+n=bm+n=b
   • mn=acmn=ac
2. Rewrite bxbx as mx+nxmx+nx.
3. Group and factorise by grouping.
4. Write as product of two binomials.

Example 1

Factorise x2+7x+12x2+7x+12.

1. Here a=1a=1, b=7b=7, c=12c=12.
   We need two numbers whose sum is 7 and product is 12.
   3 and 4 work: 3+4=73+4=7, 3×4=123×4=12.
2. Rewrite middle term:
   x2+3x+4x+12x2+3x+4x+12.
3. Group: (x2+3x)+(4x+12)(x2+3x)+(4x+12).
4. Factorise:
   x(x+3)+4(x+3)=(x+3)(x+4)x(x+3)+4(x+3)=(x+3)(x+4).

Example 2

Factorise 2×2+5x+32x2+5x+3.

1. a=2a=2, b=5b=5, c=3c=3.
   ac=2×3=6ac=2×3=6.
   We need two numbers whose sum is 5 and product is 6: 2 and 3.
2. Split 5×5x as 2x+3×2x+3x:
   2×2+2x+3x+32x2+2x+3x+3.
3. Group: (2×2+2x)+(3x+3)(2x2+2x)+(3x+3).
4. Factorise:
   2x(x+1)+3(x+1)=(x+1)(2x+3)2x(x+1)+3(x+1)=(x+1)(2x+3).

This method is systematically taught in Class 8 notes and then extended further in Class 9 and 10 when solving quadratic equations.

7. Division of Algebraic Expressions via Factorization

NCERT also connects factorisation to division of algebraic expressions.

Key Idea

To divide one algebraic expression by another:

1. Factorise numerator and denominator.
2. Cancel common factors.
3. Write the simplified result.

Example 1 (Monomial by monomial)

12x3y24x2y4x2y12x3y2

• Factorise: numerator = 12x3y212x3y2, denominator = 4x2y4x2y.
• Simplify coefficients: 12÷4=312÷4=3.
• Simplify variables: x3÷x2=xx3÷x2=x, y2÷y=yy2÷y=y.
• Result: 3xy3xy.

Example 2 (Polynomial by monomial)

15×2−20x5x5x15x2−20x

• Factorise numerator: 15×2−20x=5x(3x−4)15x2−20x=5x(3x−4).
• So fraction = 5x(3x−4)5×5x5x(3x−4).
• Cancel 5×5x: result = 3x−43x−4.

Example 3 (Polynomial by binomial using factorization)

x2−9x−3x−3x2−9.

• Factorise numerator: x2−9=(x−3)(x+3)x2−9=(x−3)(x+3).
• So fraction = (x−3)(x+3)x−3x−3(x−3)(x+3).
• Cancel (x−3)(x−3): result = x+3x+3.

These types are emphasized in Class 8 Factorization exercises and reappear later in algebraic fractions in higher classes.

8. Typical NCERT-Style Examples (Classes 8–10)

Below are some representative examples you will see across Classes 8–10.

Example A: Mixed Methods

Factorise completely: 6x2y−3xy2+9x2y26x2y−3xy2+9x2y2.

1. First, take out common factor:
   Each term has 3xy3xy.
   6x2y−3xy2+9x2y2=3xy(2x−y+3xy)6x2y−3xy2+9x2y2=3xy(2x−y+3xy).
2. Check if the expression inside bracket can be further factorised; if not, this is final.

Example B: Using identity and common factor

Factorise: 4×2−25y24x2−25y2.

• Recognise it as difference of squares: (2x)2−(5y)2(2x)2−(5y)2.
• Apply identity:
  4×2−25y2=(2x−5y)(2x+5y)4x2−25y2=(2x−5y)(2x+5y).

Example C: Quadratic with coefficient

Factorise: 3×2−5x−23x2−5x−2.

1. a=3a=3, b=−5b=−5, c=−2c=−2.
   ac=−6ac=−6.
   Need two numbers with product −6−6 and sum −5−5: −6−6 and 11.
2. Split middle term:
   3×2−6x+x−23x2−6x+x−2.
3. Group: (3×2−6x)+(x−2)(3x2−6x)+(x−2).
4. Factorise:
   3x(x−2)+1(x−2)=(x−2)(3x+1)3x(x−2)+1(x−2)=(x−2)(3x+1).

These patterns are consistent with worked examples in Class 8 notes and Class 9–10 factorization related exercises.

9. Common Mistakes and How to Avoid Them

Students in Classes 8–10 often make similar errors while factorising.

1. Forgetting to factorise completely
   They stop after taking one common factor but do not check if the remaining bracket can be factorised further.
   Always look inside the bracket again.
2. Sign mistakes (+/−)
   While splitting middle term or grouping, students often mis-handle negative signs.
   Check carefully that sum and product conditions are correctly satisfied.
3. Wrong application of identities
   For example, treating a2+b2a2+b2 as (a+b)2(a+b)2 is wrong.
   Remember:
   • (a+b)2=a2+2ab+b2(a+b)2=a2+2ab+b2.
   • (a−b)2=a2−2ab+b2(a−b)2=a2−2ab+b2.
     There is always a middle term with 2ab2ab.
4. Cancelling wrongly in division
   While simplifying x2+5xxxx2+5x, cancelling only one xx from one term and not the other is incorrect.
   First split: x2x+5xx=x+5xx2+x5x=x+5, or factorise as x(x+5)/xx(x+5)/x.

10. Quick Summary Notes for Grade 8–10 Revision

Use this as a last-minute revision sheet.

Key Ideas

• Factorisation = writing expression as product of simpler factors.
• Main methods (Class 8 NCERT):
  • Taking out common factor.
  • Grouping of terms.
  • Using identities.
  • Splitting middle term (for quadratics).
  • Using factorisation to do division of algebraic expressions.

Important Identities

1. a2−b2=(a−b)(a+b)a2−b2=(a−b)(a+b).
2. a2+2ab+b2=(a+b)2a2+2ab+b2=(a+b)2.
3. a2−2ab+b2=(a−b)2a2−2ab+b2=(a−b)2.
4. (x+a)(x+b)=x2+(a+b)x+ab(x+a)(x+b)=x2+(a+b)x+ab.

Basic Procedure Checklist

1. Check if there is a common factor in all terms.
2. If not, try grouping terms in pairs.
3. Look for standard identities (difference of squares, perfect square trinomials).
4. For quadratics ax2+bx+cax2+bx+c, use middle-term splitting.
5. For division tasks, factorise numerator and denominator, then cancel common factors.

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