FORMULA CHAPTER

Chapter 08 – Factorisation

WBBSEClass 09Mathematics20 formulas

Complete WBBSE Madhyamik Class 9 Factorisation formula list for exam preparation. Revise square and cube identities, reciprocal identities, difference of squares, sum and difference of cubes, three-variable cubic identities and binomial products. Each formula includes notation, common mistakes and a clearly solved step-by-step example for confident algebra practice.

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(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab + b²

(a + b)² + (a − b)² = 2(a² + b²)

(a + b)² − (a − b)² = 4ab

(a + 1/a)² + (a − 1/a)² = 2(a² + 1/a²)

(a + 1/a)² − (a − 1/a)² = 4

a² − b² = (a + b)(a − b)

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

(a + b)³ = a³ + b³ + 3ab(a + b)

a³ + b³ = (a + b)³ − 3ab(a + b)

(a − b)³ = a³ − b³ − 3ab(a − b)

a³ − b³ = (a − b)³ + 3ab(a − b)

a³ + b³ = (a + b)(a² − ab + b²)

a³ − b³ = (a − b)(a² + ab + b²)

a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

a³ + b³ + c³ − 3abc = ½(a + b + c)[(a − b)² + (b − c)² + (c − a)²]

If a + b + c = 0, then a³ + b³ + c³ = 3abc

(x + a)(x + b) = x² + (a + b)x + ab

(x + a)(x − b) = x² + (a − b)x − ab

(x − a)(x − b) = x² − (a + b)x + ab

Square of a Sum

(a + b)² = a² + 2ab + b²

SymbolMeaningUnit

a, b, c, x, yAlgebraic numbers or expressions used in the identityNot applicable

Worked example

Question: Expand (x + 3)².

Calculation:

Step 1: Use (a + b)² = a² + 2ab + b².

Step 2: (x + 3)² = x² + 2(x)(3) + 3².

Step 3: Simplify: x² + 6x + 9.

Answer: (x + 3)² = x² + 6x + 9

Square of a Difference

(a − b)² = a² − 2ab + b²

SymbolMeaningUnit

a, b, c, x, yAlgebraic numbers or expressions used in the identityNot applicable

Worked example

Question: Expand (x − 4)².

Calculation:

Step 1: Use (a − b)² = a² − 2ab + b².

Step 2: (x − 4)² = x² − 2(x)(4) + 4².

Step 3: Simplify: x² − 8x + 16.

Answer: (x − 4)² = x² − 8x + 16

Sum of the Squares of a Sum and Difference

(a + b)² + (a − b)² = 2(a² + b²)

SymbolMeaningUnit

a, b, c, x, yAlgebraic numbers or expressions used in the identityNot applicable

Worked example

Question: Simplify (x + y)² + (x − y)².

Calculation:

Step 1: Expand both squares.

Step 2: x² + 2xy + y² + x² − 2xy + y².

Step 3: The middle terms cancel, giving 2x² + 2y².

Answer: 2(x² + y²)

Difference of the Squares of a Sum and Difference

(a + b)² − (a − b)² = 4ab

SymbolMeaningUnit

a, b, c, x, yAlgebraic numbers or expressions used in the identityNot applicable

Worked example

Question: Simplify (x + y)² − (x − y)².

Calculation:

Step 1: Expand both squares.

Step 2: x² + 2xy + y² − (x² − 2xy + y²).

Step 3: Combine like terms to obtain 4xy.

Answer: 4xy

Reciprocal Sum of Paired Squares

(a + 1/a)² + (a − 1/a)² = 2(a² + 1/a²)

SymbolMeaningUnit

a, b, c, x, yAlgebraic numbers or expressions used in the identityNot applicable

Worked example

Question: Simplify (x + 1/x)² + (x − 1/x)².

Calculation:

Step 1: Use the paired-square sum identity.

Step 2: Replace a by x.

Step 3: The middle terms cancel.

Answer: 2(x² + 1/x²)

Reciprocal Difference of Paired Squares

(a + 1/a)² − (a − 1/a)² = 4

SymbolMeaningUnit

a, b, c, x, yAlgebraic numbers or expressions used in the identityNot applicable

Worked example

Question: Simplify (x + 1/x)² − (x − 1/x)².

Calculation:

Step 1: Use the paired-square difference identity.

Step 2: Here a = x and b = 1/x.

Step 3: 4ab = 4(x)(1/x) = 4.

Answer: 4

Difference of Two Squares

a² − b² = (a + b)(a − b)

SymbolMeaningUnit

a, b, c, x, yAlgebraic numbers or expressions used in the identityNot applicable

Worked example

Question: Factorise x² − 25.

Calculation:

Step 1: Write 25 as 5².

Step 2: x² − 25 = x² − 5².

Step 3: Apply the difference-of-squares identity.

Answer: x² − 25 = (x + 5)(x − 5)

Square of Three Terms

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

SymbolMeaningUnit

a, b, c, x, yAlgebraic numbers or expressions used in the identityNot applicable

Worked example

Question: Expand (x + y + 2)².

Calculation:

Step 1: Use the square-of-three-terms identity.

Step 2: Substitute a = x, b = y and c = 2.

Step 3: Simplify x² + y² + 4 + 2xy + 4y + 4x.

Answer: x² + y² + 2xy + 4x + 4y + 4

Cube of a Sum

(a + b)³ = a³ + b³ + 3ab(a + b)

SymbolMeaningUnit

a, b, c, x, yAlgebraic numbers or expressions used in the identityNot applicable

Worked example

Question: Expand (x + 2)³.

Calculation:

Step 1: Use the cube-of-a-sum identity.

Step 2: (x + 2)³ = x³ + 2³ + 3(x)(2)(x + 2).

Step 3: Expand and combine terms.

Answer: (x + 2)³ = x³ + 6x² + 12x + 8

Sum of Cubes from Sum and Product

a³ + b³ = (a + b)³ − 3ab(a + b)

SymbolMeaningUnit

a, b, c, x, yAlgebraic numbers or expressions used in the identityNot applicable

Worked example

Question: Find a³ + b³ when a + b = 5 and ab = 6.

Calculation:

Step 1: Use a³ + b³ = (a + b)³ − 3ab(a + b).

Step 2: Substitute the given values: 5³ − 3(6)(5).

Step 3: 125 − 90 = 35.

Answer: a³ + b³ = 35

Cube of a Difference

(a − b)³ = a³ − b³ − 3ab(a − b)

SymbolMeaningUnit

a, b, c, x, yAlgebraic numbers or expressions used in the identityNot applicable

Worked example

Question: Expand (x − 2)³.

Calculation:

Step 1: Use the cube-of-a-difference identity.

Step 2: (x − 2)³ = x³ − 2³ − 3(x)(2)(x − 2).

Step 3: Expand and combine terms.

Answer: (x − 2)³ = x³ − 6x² + 12x − 8

Difference of Cubes from Difference and Product

a³ − b³ = (a − b)³ + 3ab(a − b)

SymbolMeaningUnit

a, b, c, x, yAlgebraic numbers or expressions used in the identityNot applicable

Worked example

Question: Find a³ − b³ when a − b = 3 and ab = 2.

Calculation:

Step 1: Use the identity.

Step 2: Substitute: 3³ + 3(2)(3).

Step 3: 27 + 18 = 45.

Answer: a³ − b³ = 45

Factorisation of a Sum of Cubes

a³ + b³ = (a + b)(a² − ab + b²)

SymbolMeaningUnit

a, b, c, x, yAlgebraic numbers or expressions used in the identityNot applicable

Worked example

Question: Factorise x³ + 8.

Calculation:

Step 1: Write 8 as 2³.

Step 2: Use the sum-of-cubes identity.

Step 3: Substitute a = x and b = 2.

Answer: x³ + 8 = (x + 2)(x² − 2x + 4)

Factorisation of a Difference of Cubes

a³ − b³ = (a − b)(a² + ab + b²)

SymbolMeaningUnit

a, b, c, x, yAlgebraic numbers or expressions used in the identityNot applicable

Worked example

Question: Factorise x³ − 27.

Calculation:

Step 1: Write 27 as 3³.

Step 2: Use the difference-of-cubes identity.

Step 3: Substitute a = x and b = 3.

Answer: x³ − 27 = (x − 3)(x² + 3x + 9)

Three-Variable Cubic Identity

a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

SymbolMeaningUnit

a, b, c, x, yAlgebraic numbers or expressions used in the identityNot applicable

Worked example

Question: Factorise a³ + b³ + c³ − 3abc.

Calculation:

Step 1: Identify the symmetric cubic expression.

Step 2: Apply the three-variable cubic identity.

Step 3: Write the linear factor followed by the quadratic factor.

Answer: (a + b + c)(a² + b² + c² − ab − bc − ca)

Three-Variable Cubic Identity in Squared-Difference Form

a³ + b³ + c³ − 3abc = ½(a + b + c)[(a − b)² + (b − c)² + (c − a)²]

SymbolMeaningUnit

a, b, c, x, yAlgebraic numbers or expressions used in the identityNot applicable

Worked example

Question: Rewrite a³ + b³ + c³ − 3abc using squared differences.

Calculation:

Step 1: Use the symmetric squared-difference form.

Step 2: Take ½(a + b + c) as the first factor.

Step 3: Add the three squared pairwise differences.

Answer: ½(a + b + c)[(a − b)² + (b − c)² + (c − a)²]

Cubic Identity When the Sum Is Zero

If a + b + c = 0, then a³ + b³ + c³ = 3abc

SymbolMeaningUnit

a, b, c, x, yAlgebraic numbers or expressions used in the identityNot applicable

Worked example

Question: If a = 2, b = 3 and c = −5, find a³ + b³ + c³.

Calculation:

Step 1: Check the condition: 2 + 3 − 5 = 0.

Step 2: Use a³ + b³ + c³ = 3abc.

Step 3: 3(2)(3)(−5) = −90.

Answer: a³ + b³ + c³ = −90

Product of Two Positive Binomials

(x + a)(x + b) = x² + (a + b)x + ab

SymbolMeaningUnit

a, b, c, x, yAlgebraic numbers or expressions used in the identityNot applicable

Worked example

Question: Expand (x + 3)(x + 5).

Calculation:

Step 1: Use (x + a)(x + b) = x² + (a + b)x + ab.

Step 2: Substitute a = 3 and b = 5.

Step 3: x² + (3 + 5)x + 3×5.

Answer: x² + 8x + 15

Product of Mixed-Sign Binomials

(x + a)(x − b) = x² + (a − b)x − ab

SymbolMeaningUnit

a, b, c, x, yAlgebraic numbers or expressions used in the identityNot applicable

Worked example

Question: Expand (x + 5)(x − 2).

Calculation:

Step 1: Use the mixed-sign binomial identity.

Step 2: Substitute a = 5 and b = 2.

Step 3: x² + (5 − 2)x − 5×2.

Answer: x² + 3x − 10

Product of Two Negative Binomials

(x − a)(x − b) = x² − (a + b)x + ab

SymbolMeaningUnit

a, b, c, x, yAlgebraic numbers or expressions used in the identityNot applicable

Worked example

Question: Expand (x − 3)(x − 4).

Calculation:

Step 1: Use the two-negative-binomial identity.

Step 2: Substitute a = 3 and b = 4.

Step 3: x² − (3 + 4)x + 3×4.

Answer: x² − 7x + 12

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