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ICSE • Class 9 • Mathematics • Ch 4
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Factorisation

In ICSE Class 9 Mathematics, "Factorisation" is the foundational algebraic discipline of decomposing a composite polynomial expression into an irreducible product of simpler algebraic factors. As the inverse operation of expansion, factorisation is the principal technique used throughout mathematics to solve polynomial equations, simplify rational algebraic expressions, and analyze functions. The ICSE curriculum methodically develops four major classes of factorisation: (1) Taking out common monomial and polynomial factors; (2) Factorisation by grouping terms into pairs or triplets; (3) Factorisation of the difference of two squares $a^2 - b^2 = (a - b)(a + b)$ and its nested iterations (e.g., $a^4 - b^4$); (4) Factorisation of quadratic trinomials of the general canonical form $ax^2 + bx + c$ by splitting the middle term (where two numbers $p$ and $q$ are chosen such that $p + q = b$ and $pq = ac$); (5) Factorisation by completing the square ($a^4 + a^2b^2 + b^4$); and (6) Factorisation of the sum and difference of two cubes using the foundational formulas $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ and $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. The chapter emphasizes strategic pattern recognition, multi-step nested factorisations, substitution of compound algebraic variables, and rigorous checking.

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The RSA Encryption Secret: How Modern Internet Security Depends Completely on the Difficulty of Factorisation

Every time you log into your email, make an online payment on UPI, or send an encrypted WhatsApp message, your data is protected by the mathematics of Factorisation! In the 1970s, mathematicians Rivest, Shamir, and Adleman created the RSA cryptosystem based on a fascinating mathematical asymmetry: It takes a computer a fraction of a microsecond to multiply two prime numbers together (expansion: $p \times q = N$), but if you give the computer the massive 300-digit number $N$, it would take all the supercomputers on planet Earth millions of years to break it back down into its factors $p$ and $q$ (factorisation)! In algebra, factorisation is your master key—it transforms an unruly, unsolvable polynomial into a clean product of factors that can be set to zero to reveal its roots. How do master mathematicians look at a tangled expression like $x^4 + 4y^4$ and instantly spot its hidden factors? Let us discover the art of algebraic decomposition!

Why This Chapter Matters

Factorisation is indispensable for solving quadratic and polynomial equations, simplifying rational expressions in calculus, finding limits, and computing roots in mechanical and electrical engineering.

Before You Begin (Prerequisites)

  • Fundamental identities from Expansions: $(a \pm b)^2 = a^2 \pm 2ab + b^2$ and $(a+b)(a-b) = a^2 - b^2$.
  • Finding the HCF of algebraic monomials.

What You Will Learn (Core Objectives)

  • Identify and factorise algebraic expressions by extracting common factors and grouping.
  • Factorise binomials using the difference of two squares $a^2 - b^2$ and iterative difference of squares.
  • Decompose quadratic trinomials $ax^2 + bx + c$ systematically by the middle-term splitting method.
  • Factorise expressions by completing the square into a difference of two squares.
  • Apply the cubic sum and difference formulas $a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$ to factorise complex polynomials.

Chapter Roadmap & Progression

1 1. Common Factors and Grouping Meth...
2 2. Difference of Two Squares: $a^2...
3 3. Quadratic Trinomials: Middle-Ter...
4 4. Sum and Difference of Two Cubes
5 5. Worked ICSE Problem Archetypes

Complete Concept Guide (100% Curriculum Coverage)

1. Common Factors and Grouping Method

Fundamental Techniques
A. Extracting Common Monomial/Binomial Factors:

Always inspect the entire expression first for the Highest Common Factor (HCF) of coefficients and variable powers:

$$12x^2y - 18xy^2 = 6xy(2x - 3y)$$
B. Factorisation by Grouping:

When an expression has 4 or more terms with no overall common factor, group terms in pairs sharing common factors:

$$ab + bc + ax + cx = b(a + c) + x(a + c) = (a + c)(b + x)$$

Warning on Sign Changes: When factoring out a negative sign from a group, ensure internal signs are inverted: $(ax - ay - bx + by) = a(x - y) - b(x - y) = (x - y)(a - b)$.

2. Difference of Two Squares: $a^2 - b^2$

The Difference of Squares Formula $$\mathbf{a^2 - b^2 = (a - b)(a + b)}$$
Iterative / Nested Difference of Squares:

Expressions of higher even degree must be factored completely until no further difference of squares can be applied:

$$x^4 - 16 = (x^2)^2 - 4^2 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)$$

Note: The sum of two squares $(x^2 + 4)$ cannot be factored further over real numbers ($\mathbb{R}$).

3. Quadratic Trinomials: Middle-Term Splitting ($ax^2 + bx + c$)

Middle-Term Splitting Algorithm
The Standard Splitting Algorithm:
  1. Write the quadratic trinomial in canonical descending order: $ax^2 + bx + c$.
  2. Compute the product of the leading coefficient and constant term: $P = a \times c$.
  3. Find two integers $p$ and $q$ such that: $$p + q = b \quad \text{and} \quad p \times q = a \times c$$
  4. Split the middle term $bx$ as $(px + qx)$ and factorise by grouping pairs: $$ax^2 + bx + c = ax^2 + px + qx + c$$
Sign Rules for $p$ and $q$:
  • If $ac > 0$: Both $p$ and $q$ must have the same sign (both positive if $b > 0$; both negative if $b < 0$).
  • If $ac < 0$: $p$ and $q$ must have opposite signs, with the numerically larger factor carrying the sign of $b$.

4. Sum and Difference of Two Cubes

Cubic Factorisation Formulas
Formulas:
$$\mathbf{a^3 + b^3 = (a + b)(a^2 - ab + b^2)}$$ $$\mathbf{a^3 - b^3 = (a - b)(a^2 + ab + b^2)}$$
Key Structural Observations:
  • The linear binomial factor $(a \pm b)$ carries the exact same sign as the original expression.
  • In the quadratic trinomial factor $(a^2 \mp ab + b^2)$, the middle term has the opposite sign, and there is no factor of 2 (it is $-ab$, NOT $-2ab$).

5. Worked ICSE Problem Archetypes

Exemplary Solutions
Problem 1: Factorise completely: $6x^2 - 17x + 12$.

Solution:

Here, $a = 6$, $b = -17$, $c = 12$.

Product $ac = 6 \times 12 = 72$.

We need two numbers $p$ and $q$ whose product is $72$ and sum is $-17$. The numbers are $-9$ and $-8$ ($(-9)(-8) = 72, (-9) + (-8) = -17$).

Splitting the middle term:

$$6x^2 - 17x + 12 = 6x^2 - 9x - 8x + 12$$ $$= 3x(2x - 3) - 4(2x - 3) = \mathbf{(2x - 3)(3x - 4)}$$
Problem 2: Factorise: $x^4 + x^2y^2 + y^4$.

Solution (Completing the Square):

Notice that $(x^2 + y^2)^2 = x^4 + 2x^2y^2 + y^4$.

Add and subtract $x^2y^2$ to complete the perfect square:

$$x^4 + x^2y^2 + y^4 = (x^4 + 2x^2y^2 + y^4) - x^2y^2$$ $$= (x^2 + y^2)^2 - (xy)^2$$

Applying the difference of two squares formula $A^2 - B^2 = (A - B)(A + B)$:

$$= \mathbf{(x^2 + y^2 - xy)(x^2 + y^2 + xy)}$$

Key Formulas, Identities & Theorems

Difference of Two Squares
$$a^2 - b^2 = (a - b)(a + b)$$
Applies to any binomial with two squared terms separated by minus.
Trinomial Splitting
$$ax^2 + bx + c = ax^2 + px + qx + c, \quad p+q=b, \; pq=ac$$
Decomposes quadratic trinomial into factorable pairs.
Sum of Two Cubes
$$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$
Middle term in trinomial has opposite sign and no coefficient 2.
Difference of Two Cubes
$$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$
Trinomial factor is strictly positive for real non-zero values.

Mathematics: Factorisation Decision Tree & Middle-Term Splitting

Master Factorisation Decision Flowchart for Polynomials Step 1: Extract Overall Common Factor (HCF) How many terms remaining? 2 Terms (Binomial) • Difference of Squares: a² - b² = (a - b)(a + b) • Sum / Diff of Cubes: a³ + b³ = (a+b)(a² - ab + b²) a³ - b³ = (a-b)(a² + ab + b²) 3 Terms (Trinomial) • Perfect Square Trinomial: a² ± 2ab + b² = (a ± b)² • Split Middle Term: ax² + bx + c Find p, q: p+q=b, pq=ac 4 or More Terms • Grouping by Pairs: ab + bc + ad + cd = b(a+c) + d(a+c) • 3 Terms to 1 Square: (a² + 2ab + b²) - c² = (a+b)² - c²

Chapter Summary & 10 Key Takeaways

Takeaway 1
Factorisation is the process of writing an algebraic expression as a product of irreducible factors.
Takeaway 2
Always check for and extract the greatest common factor (HCF) first.
Takeaway 3
When 4 terms are present, group them in pairs to find common binomial factors.
Takeaway 4
The difference of two squares identity is a^2 - b^2 = (a - b)(a + b).
Takeaway 5
Expressions like a^4 - b^4 factorise into (a^2 + b^2)(a + b)(a - b).
Takeaway 6
To factorise ax^2 + bx + c, find two numbers p and q such that p + q = b and pq = ac.
Takeaway 7
Sum of two cubes identity: a^3 + b^3 = (a + b)(a^2 - ab + b^2).
Takeaway 8
Difference of two cubes identity: a^3 - b^3 = (a - b)(a^2 + ab + b^2).
Takeaway 9
In a^3 ± b^3, the trinomial factor a^2 ∓ ab + b^2 does NOT contain a coefficient of 2.
Takeaway 10
Expressions of the form a^4 + a^2b^2 + b^4 are factorised by completing the square into (a^2 + b^2)^2 - (ab)^2.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
Factorise completely: $16x^4 - 81y^4$.
Reveal Answer & Explanation
Answer: • Write the terms as squares of squares:
$$16x^4 - 81y^4 = (4x^2)^2 - (9y^2)^2$$
• Apply the difference of two squares $A^2 - B^2 = (A - B)(A + B)$:
$$= (4x^2 - 9y^2)(4x^2 + 9y^2)$$
• Notice that the first factor $(4x^2 - 9y^2)$ is again a difference of two squares:
$$4x^2 - 9y^2 = (2x)^2 - (3y)^2 = (2x - 3y)(2x + 3y)$$
• The second factor $(4x^2 + 9y^2)$ is a sum of squares and cannot be factored over real numbers.
• Final complete factorisation:
$$\mathbf{(2x - 3y)(2x + 3y)(4x^2 + 9y^2)}$$
Apply difference of squares twice: (4x^2 - 9y^2)(4x^2 + 9y^2), then factor the first term again.
2
Factorise the quadratic trinomial: $12x^2 - 23x + 10$.
Reveal Answer & Explanation
Answer: • Identify $a = 12, b = -23, c = 10$.
• Compute product $ac = 12 \times 10 = 120$.
• We need two negative integers whose sum is $-23$ and product is $120$.
Factors of 120: $15 \times 8 = 120$ and $15 + 8 = 23$.
So $p = -15, q = -8$.
• Split the middle term:
$$12x^2 - 23x + 10 = 12x^2 - 15x - 8x + 10$$
$$= 3x(4x - 5) - 2(4x - 5) = \mathbf{(4x - 5)(3x - 2)}$$
Product ac = 120, sum = -23. Factors are -15 and -8.
3
Factorise completely: $a^3 - 8b^3 - 64c^3 - 24abc$.
Reveal Answer & Explanation
Answer: • Rewrite the terms in cubic form:
$$a^3 + (-2b)^3 + (-4c)^3 - 3(a)(-2b)(-4c)$$
• This matches the master cubic identity: $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$ with $x = a, y = -2b, z = -4c$.
• Linear factor: $(x + y + z) = (a - 2b - 4c)$
• Quadratic factor:
$$x^2 + y^2 + z^2 - xy - yz - zx = a^2 + (-2b)^2 + (-4c)^2 - a(-2b) - (-2b)(-4c) - (-4c)(a)$$
$$= a^2 + 4b^2 + 16c^2 + 2ab - 8bc + 4ca$$
• Therefore:
$$\mathbf{(a - 2b - 4c)(a^2 + 4b^2 + 16c^2 + 2ab - 8bc + 4ca)}$$
Identify x = a, y = -2b, z = -4c in x^3 + y^3 + z^3 - 3xyz.
4
Factorise: $x^2 - y^2 - 2y - 1$.
Reveal Answer & Explanation
Answer: • Group the last three terms and factor out the negative sign:
$$x^2 - (y^2 + 2y + 1)$$
• Notice that $(y^2 + 2y + 1)$ is a perfect square: $(y + 1)^2$.
$$= x^2 - (y + 1)^2$$
• Now apply the difference of two squares $A^2 - B^2 = (A - B)(A + B)$ with $A = x, B = (y + 1)$:
$$= [x - (y + 1)][x + (y + 1)] = \mathbf{(x - y - 1)(x + y + 1)}$$
Group as x^2 - (y^2 + 2y + 1) = x^2 - (y + 1)^2.
5
Factorise: $27x^3 + 64y^3$.
Reveal Answer & Explanation
Answer: • Write each term as a perfect cube:
$$27x^3 + 64y^3 = (3x)^3 + (4y)^3$$
• Apply the sum of cubes identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ with $a = 3x, b = 4y$:
$$= (3x + 4y)[(3x)^2 - (3x)(4y) + (4y)^2]$$
$$= \mathbf{(3x + 4y)(9x^2 - 12xy + 16y^2)}$$
Use a^3 + b^3 = (a + b)(a^2 - ab + b^2) with a = 3x, b = 4y.
6
Factorise by completing the square: $x^4 + 4$.
Reveal Answer & Explanation
Answer: • We have $x^4 + 4 = (x^2)^2 + 2^2$.
• To complete the square of $(x^2 + 2)^2$, we need the middle term $2(x^2)(2) = 4x^2$.
• Add and subtract $4x^2$:
$$x^4 + 4 = (x^4 + 4x^2 + 4) - 4x^2$$
$$= (x^2 + 2)^2 - (2x)^2$$
• Apply the difference of two squares $A^2 - B^2 = (A - B)(A + B)$:
$$= \mathbf{(x^2 + 2 - 2x)(x^2 + 2 + 2x)} = (x^2 - 2x + 2)(x^2 + 2x + 2)$$
Add and subtract 4x^2 to make (x^2 + 2)^2 - (2x)^2.
7
Factorise the expression: $2(x^2 + 5x)^2 - 5(x^2 + 5x) - 3$.
Reveal Answer & Explanation
Answer: • Let $y = x^2 + 5x$. The expression simplifies to a quadratic in $y$:
$$2y^2 - 5y - 3$$
• Split the middle term ($ac = 2 \times (-3) = -6$, sum = $-5 \implies -6$ and $+1$):
$$2y^2 - 6y + y - 3 = 2y(y - 3) + 1(y - 3) = (y - 3)(2y + 1)$$
• Substitute back $y = x^2 + 5x$:
$$= (x^2 + 5x - 3)(2(x^2 + 5x) + 1) = \mathbf{(x^2 + 5x - 3)(2x^2 + 10x + 1)}$$
Substitute y = x^2 + 5x to get 2y^2 - 5y - 3 = (y - 3)(2y + 1).
8
Factorise: $a^3 - \frac{1}{a^3} - 2a + \frac{2}{a}$.
Reveal Answer & Explanation
Answer: • Group the cubic terms and the linear terms separately:
$$\left(a^3 - \frac{1}{a^3}\right) - 2\left(a - \frac{1}{a}\right)$$
• Expand the difference of cubes $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$:
$$\left(a - \frac{1}{a}\right)\left(a^2 + 1 + \frac{1}{a^2}\right) - 2\left(a - \frac{1}{a}\right)$$
• Factor out the common binomial $\left(a - \frac{1}{a}\right)$:
$$= \left(a - \frac{1}{a}\right)\left[a^2 + 1 + \frac{1}{a^2} - 2\right] = \left(a - \frac{1}{a}\right)\left(a^2 - 1 + \frac{1}{a^2}\right)$$
Group as (a^3 - 1/a^3) - 2(a - 1/a), then factor out (a - 1/a).
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