Basic Operations and Factorisation
~14 min read
- What: Polynomial operations — addition, subtraction, multiplication, division, the algebraic identities, factorisation of quadratics, cubics, and special forms (\(a^2 - b^2\), \(a^3 \pm b^3\), \(a^4 + 4b^4\), etc.).
- Why it matters: CDS papers from 2007 to 2023 average 4–7 questions per sitting on this chapter — the largest algebra head and the foundation for linear and quadratic equations.
- Key fact: The Sophie-Germain identity \(a^4 + 4b^4 = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2)\) is the most-tested non-standard factorisation in CDS — it appears at least once every two years.
Basic Operations and Factorisation is the foundation chapter for everything algebra-flavoured in CDS — equations, HCF/LCM of polynomials, simplification, even some geometry problems. Once you have the eight standard identities at fingertip recall, this chapter becomes 4–7 marks of nearly free arithmetic.
This page draws from CDS Previous Year Questions across 2007–2023 plus NCERT Class 8 Algebra Play, Class 9 Algebraic Identities, and Class 10 Polynomials. Pair with HCF and LCM (polynomial HCF/LCM uses these identities) and Quadratic Equations.
What This Topic Covers
CDS scope: (1) polynomial arithmetic — add, subtract, multiply, divide; (2) the eight identities — \((a+b)^2, (a-b)^2, a^2 - b^2, (a+b)(a-b), (a+b)^3, (a-b)^3, a^3 + b^3, a^3 - b^3\); (3) factorisation — common factor, grouping, splitting middle term, special forms; (4) the remainder and factor theorems — for polynomial divisibility; and (5) symmetric expressions — \(a^2 + b^2 = (a+b)^2 - 2ab\), \(a^3 + b^3 + c^3 - 3abc\), etc.
Why This Topic Matters
- 4–7 CDS questions per paper — the largest algebra head.
- Foundation for Quadratic Equations, Linear Equations, and HCF/LCM of polynomials.
- The factor theorem turns "is this polynomial divisible by \(x - a\)" into "compute \(p(a)\)" in one step.
Exam Pattern & Weightage
| Year / Paper | No. | Subtopics Tested |
|---|---|---|
| 2007-II | 4 | Identities, factor of cubic |
| 2009-II | 4 | Symmetric \(a^3 + b^3 + c^3\), middle-term split |
| 2010-I/II | 5 | Identities, remainder theorem |
| 2011-I/II | 5 | Factor theorem, cubic factorisation |
| 2012-I/II | 5 | Sophie-Germain, symmetric |
| 2013-I/II | 6 | Polynomial division, identities |
| 2014-I/II | 5 | Factor of polynomial, identities |
| 2015-I/II | 4 | Cubic identities, remainder |
| 2016-I/II | 4 | Factor theorem, symmetric |
| 2017-I/II | 5 | Mixed factorisation |
| 2018-I/II | 4 | Identities, polynomial value |
| 2019-II | 3 | Cubic factorisation |
| 2020-I/II | 4 | Identities, remainder theorem |
| 2021-I/II | 4 | Symmetric \(a^3+b^3+c^3-3abc\) |
| 2022-I / 2023-I | 5 | Mixed |
The identity \(a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\) is heavily tested in CDS. If \(a + b + c = 0\), then \(a^3 + b^3 + c^3 = 3abc\) — a one-line consequence that appears every other sitting.
Core Concepts
The Eight Standard Identities
Three-Variable Identities
Corollary: if \(a + b + c = 0\), then \(a^3 + b^3 + c^3 = 3abc\).
Symmetric Expressions
Remainder Theorem and Factor Theorem
Special Non-Standard Factorisations
Factoring a Quadratic by Middle-Term Split
For \(ax^2 + bx + c\): find two numbers \(p\) and \(q\) such that \(p + q = b\) and \(pq = ac\). Then \(ax^2 + bx + c = ax^2 + px + qx + c\), and group.
\((a + b)^2 \neq a^2 + b^2\) and \(\sqrt{a^2 + b^2} \neq a + b\). The middle term \(2ab\) must always appear. CDS plants this in simplification problems.
Worked Examples
Example 1 — Square Identity (2010-I)
Q: Find \((103)^2\) using identity.
- \((103)^2 = (100 + 3)^2 = 10000 + 600 + 9 = 10609\).
Example 2 — Sum-of-Cubes Identity (2011-I)
Q: Factorise \(8x^3 + 27y^3\).
- \(8x^3 + 27y^3 = (2x)^3 + (3y)^3 = (2x + 3y)(4x^2 - 6xy + 9y^2)\).
Example 3 — Three-Variable (2009-II)
Q: If \(a + b + c = 0\), find \(a^3 + b^3 + c^3\) in terms of \(abc\).
- From the identity: \(a^3 + b^3 + c^3 - 3abc = (a + b + c)(\ldots)\).
- Since \(a + b + c = 0\), the right side is 0, so \(a^3 + b^3 + c^3 = 3abc\).
Example 4 — Factor Theorem (2014-I)
Q: For what value of \(k\) is \(x - 2\) a factor of \(x^3 - 3x^2 + kx - 4\)?
- Apply factor theorem: \(p(2) = 0\).
- \(8 - 12 + 2k - 4 = 0 \implies 2k = 8 \implies k = 4\).
Example 5 — Middle-Term Split (2013-II)
Q: Factorise \(x^2 - 5x - 14\).
- Need two numbers with product \(-14\) and sum \(-5\). Try \(-7\) and \(2\). ✓
- \(x^2 - 7x + 2x - 14 = x(x - 7) + 2(x - 7) = (x + 2)(x - 7)\).
Example 6 — Sophie-Germain (2012-II)
Q: Factorise \(x^4 + 4\).
- Apply: \(x^4 + 4 \cdot 1^4 = (x^2 + 2x + 2)(x^2 - 2x + 2)\).
Example 7 — Symmetric (2018-I)
Q: If \(x + 1/x = 3\), find \(x^3 + 1/x^3\).
- Use identity: \(x^3 + 1/x^3 = (x + 1/x)^3 - 3(x)(1/x)(x + 1/x) = 3^3 - 3 \cdot 1 \cdot 3 = 27 - 9 = 18\).
How CDS Tests This Topic
Six archetypes: (1) apply a square or cube identity to a numerical expression, (2) factorise a polynomial via standard identity, (3) factor theorem to find \(k\) or verify a divisor, (4) middle-term split for quadratics, (5) Sophie-Germain or \(a^4 + a^2 + 1\) form, (6) symmetric \(x + 1/x \to x^n + 1/x^n\).
Exam Shortcuts (Pro-Tips)
Shortcut 1 — Identities Cold
Memorise the eight standard identities. They are the building blocks of every algebra question.
Shortcut 2 — \(x + 1/x\) Sequence
Shortcut 3 — \(a + b + c = 0\) Trick
Whenever \(a + b + c = 0\): \(a^3 + b^3 + c^3 = 3abc\). Recognise this in symmetric problems.
Shortcut 4 — Number Squaring via Identity
\((103)^2 = (100 + 3)^2 = 10000 + 600 + 9 = 10609\) is faster than long multiplication. Same for \((99)^2 = (100 - 1)^2 = 9801\).
Shortcut 5 — Factor Theorem in 5 Seconds
Test if \(x - a\) divides \(p(x)\) by computing \(p(a)\). If \(p(a) = 0\), yes. Saves long polynomial division.
Common Question Patterns
Pattern 1 — Apply an Identity
Compute a numerical expression like \((202)^2\) or \(99^3\) using the appropriate identity.
Pattern 2 — Factorise
Identify the form: sum/difference of squares or cubes, Sophie-Germain, middle-term split. Apply the matching identity or method.
Pattern 3 — Factor Theorem for \(k\)
"For what \(k\) is \(x - a\) a factor of \(p(x)\)?" Compute \(p(a) = 0\) and solve for \(k\).
Pattern 4 — Three-Variable Symmetric
Given \(a + b + c\), \(ab + bc + ca\), \(abc\), find symmetric expressions in \(a, b, c\). Use \((a + b + c)^2\) and \(a^3 + b^3 + c^3 - 3abc\) identities.
Pattern 5 — \(x + 1/x\) Chain
Given \(x + 1/x = k\), find \(x^n + 1/x^n\). Use the recursion \(x^{n+1} + 1/x^{n+1} = (x + 1/x)(x^n + 1/x^n) - (x^{n-1} + 1/x^{n-1})\).
Preparation Strategy
Week 1. Memorise all eight standard identities cold. Drill 30 problems applying them. Practice the middle-term split method on 15 quadratic factorisation problems.
Week 2. Three-variable identities (especially the \(a + b + c = 0 \implies a^3 + b^3 + c^3 = 3abc\) corollary). Factor theorem and remainder theorem. Special forms (Sophie-Germain, \(a^4 + a^2 + 1\)).
Mock testing. Take timed CDS papers. Tag every algebra question. Most slip-ups come from sign errors and forgetting the middle term in \((a + b)^2\). Use CDS mock tests for pace.
Cross-train with HCF and LCM (polynomial pairs), Quadratic Equations, and Linear Equations.
Drill Algebra in Real Time
CDS mocks with identity application, factorisation, factor theorem, and three-variable symmetric problems. Eight identities — six archetypes — reflex.
Start Free Mock TestFrequently Asked Questions
What are the eight standard algebraic identities?
\((a + b)^2\), \((a - b)^2\), \(a^2 - b^2 = (a + b)(a - b)\), \((a + b)^3\), \((a - b)^3\), \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\), \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\), and the three-variable \(a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\).
What happens if \(a + b + c = 0\)?
The identity \(a^3 + b^3 + c^3 - 3abc = (a + b + c)(\ldots)\) collapses to \(a^3 + b^3 + c^3 = 3abc\) — a one-line CDS-favourite. Recognise this whenever a problem gives \(a + b + c = 0\) explicitly or as a consequence.
What is the factor theorem?
\(x - a\) is a factor of polynomial \(p(x)\) if and only if \(p(a) = 0\). Instead of doing polynomial long division, just plug in \(x = a\). If the result is zero, \(x - a\) is a factor; the result itself is the remainder when \(p(x)\) is divided by \(x - a\).
How do I factor a quadratic by middle-term split?
For \(ax^2 + bx + c\): find two numbers \(p\) and \(q\) with \(p + q = b\) and \(pq = ac\). Then \(ax^2 + bx + c = ax^2 + px + qx + c\), and group to factor. Example: \(x^2 - 5x - 14\). Need product \(-14\), sum \(-5\). Try \(-7\) and \(2\). ✓
How do I find \(x^n + 1/x^n\) given \(x + 1/x\)?
Use the recursion: \(x^{n+1} + 1/x^{n+1} = (x + 1/x)(x^n + 1/x^n) - (x^{n-1} + 1/x^{n-1})\). Start from \(x + 1/x = k\) and \(x^2 + 1/x^2 = k^2 - 2\), then build up. For \(x^3 + 1/x^3 = k^3 - 3k\) directly.
What is the Sophie-Germain identity?
\(a^4 + 4b^4 = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2)\). It comes up when CDS gives an expression like \(x^4 + 4\) or \(x^4 + 64\) for factorisation. The trick is to recognise the 4 as \(4 \cdot 1^4\) or \(4 \cdot 2^4 = 64\).
Which CDS Maths topics build on Basic Operations and Factorisation?
Quadratic Equations — factorisation is one of the three solving methods. Linear Equations in two variables. HCF and LCM — polynomial pairs require these identities. Trigonometry identities rely on parallel algebraic manipulation.