Algebra is the silent workhorse of the CDS Maths paper. Before you can solve equations, simplify expressions or crack mensuration formulas, you must be fluent in basic operations on polynomials and factorisation. This page rebuilds those skills from the ground up — the standard identities, polynomial division, the factor and remainder theorems — and shows you exactly how examiners test them.
Why algebra basics decide your score
Roughly one-fifth of every CDS Maths paper rests directly or indirectly on algebra. Linear equations, quadratics, simplification, even age and ratio word-problems all collapse into polynomial manipulation. If your basic operations and factorisation are shaky, every downstream chapter costs you extra time.
The good news: this topic is rule-based and predictable. There is no guesswork — only identities to memorise and procedures to drill. Master the dozen-or-so identities below and you will recognise the shortcut in almost every algebra question. Unlike data interpretation or geometry, algebra rewards mechanical accuracy over insight, which means a disciplined student can score full marks here with steady practice rather than flashes of cleverness.
A polynomial is an expression of the form anxn + … + a1x + a0 with whole-number powers of the variable. The highest power is its degree: degree 1 is linear, degree 2 quadratic, degree 3 cubic.
Key terminology you must not confuse
Examiners love to test definitions disguised as quick questions. Fix this vocabulary firmly.
- Term — a single part separated by + or − signs, e.g. 3x2.
- Coefficient — the numerical factor of a term; in 3x2 it is 3.
- Like terms — terms with the same variable raised to the same power; only like terms can be added or subtracted.
- Monomial, binomial, trinomial — expressions with 1, 2 and 3 terms respectively.
- Constant term — the term with no variable.
- Degree of a polynomial — the highest power of the variable present in any term.
Getting these labels right matters because many CDS questions are phrased entirely in this language — for example asking for the coefficient of a particular term or the degree of a product of two polynomials.
3x and 3x2 are not like terms — the powers differ. You may not combine them. Writing 3x + 3x2 = 6x3 is a classic trap answer.
Addition and subtraction of polynomials
To add or subtract polynomials, collect like terms and combine their coefficients. The variable parts stay unchanged.
Example: (4x2 + 3x − 5) + (2x2 − x + 7) = 6x2 + 2x + 2.
When subtracting, change the sign of every term inside the bracket being removed:
(5x2 + 2x − 1) − (3x2 − 4x + 6) = 5x2 + 2x − 1 − 3x2 + 4x − 6 = 2x2 + 6x − 7.
The single most common slip in this chapter is forgetting to flip the sign of the inner terms after a subtraction sign. Distribute the minus to every term before combining.
Multiplication of polynomials
Multiply every term of the first expression by every term of the second, then collect like terms. Add the exponents of the same base: xm × xn = xm+n.
Example: (x + 3)(x2 − 2x + 4) = x3 − 2x2 + 4x + 3x2 − 6x + 12 = x3 + x2 − 2x + 12.
For two binomials, use FOIL (First, Outer, Inner, Last) to avoid missing a cross-term. But where an identity fits — like (a+b)2 — using the identity is far faster than expanding term by term.
The standard algebraic identities
These identities are the heart of the chapter. Recognising them instantly is what separates a 30-second solve from a two-minute grind.
- (a + b)2 = a2 + 2ab + b2
- (a − b)2 = a2 − 2ab + b2
- (a + b)(a − b) = a2 − b2
- (x + a)(x + b) = x2 + (a + b)x + ab
- (a + b)3 = a3 + b3 + 3ab(a + b)
- (a − b)3 = a3 − b3 − 3ab(a − b)
- a3 + b3 = (a + b)(a2 − ab + b2)
- a3 − b3 = (a − b)(a2 + ab + b2)
- a2 + b2 + c2 + 2(ab + bc + ca) = (a + b + c)2
A powerful corollary worth memorising: when a + b + c = 0, then a3 + b3 + c3 = 3abc. This single fact answers many CDS questions in one line. Spend dedicated practice time writing out each identity from memory until reproducing it is automatic; in the exam hall you will not have a second to rederive them.
Using identities to evaluate quickly
Identities are not only for factorising — they let you compute awkward numbers mentally and avoid long multiplication under time pressure.
Find 103 × 97. Write it as (100 + 3)(100 − 3) = 1002 − 32 = 10000 − 9 = 9991. Find 992. Write it as (100 − 1)2 = 10000 − 200 + 1 = 9801. Find 1001 × 999 = (1000)2 − 12 = 999999.
Whenever you see two numbers symmetric about a round figure (like 103 and 97 about 100), reach for a2 − b2. It turns a tedious multiplication into a one-step subtraction that you can do without scratch work.
The four routes to factorisation
Factorisation is writing an expression as a product of simpler factors. CDS questions are solved by one of four standard routes:
- Common factor — take out the highest common factor (HCF) of all terms. 6x2 + 9x = 3x(2x + 3).
- Grouping — group terms in pairs that share a factor. ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y).
- Using identities — recognise a difference of squares, perfect square or sum/difference of cubes.
- Splitting the middle term — for a quadratic ax2 + bx + c, split b into two numbers whose product is ac and whose sum is b.
Always check first for a common factor before trying any other method. Pulling out the HCF often turns a messy expression into a familiar identity.
Splitting the middle term in detail
This is the workhorse method for quadratics. To factorise ax2 + bx + c, find two numbers p and q such that p × q = a × c and p + q = b.
Factorise x2 + 7x + 12. Here a·c = 12 and b = 7. The pair 3 and 4 works (3 × 4 = 12, 3 + 4 = 7). So x2 + 3x + 4x + 12 = x(x + 3) + 4(x + 3) = (x + 3)(x + 4).
Factorise 6x2 + 11x + 3. Here a·c = 18 and b = 11. The pair 9 and 2 works. So 6x2 + 9x + 2x + 3 = 3x(2x + 3) + 1(2x + 3) = (2x + 3)(3x + 1).
When a ≠ 1, students forget to multiply a by c — they look for factors of c alone. The product to factor is always a × c, not just c.
Remainder theorem and factor theorem
For higher-degree expressions, these two theorems are indispensable.
Remainder theorem: when a polynomial p(x) is divided by (x − a), the remainder is p(a).
Factor theorem: (x − a) is a factor of p(x) if and only if p(a) = 0.
So to test whether (x − 2) divides p(x) = x3 − 3x2 + 4x − 4, compute p(2) = 8 − 12 + 8 − 4 = 0. Since p(2) = 0, (x − 2) is a factor.
Once one factor is found, divide it out (by long division or synthetic division) to reduce a cubic to a quadratic, then factorise the quadratic by splitting the middle term. This two-step strategy — find one root by inspection, then reduce the degree — is the standard professional approach to any cubic in the CDS paper. The remainder theorem is also tested directly: a question may simply ask for the remainder when a polynomial is divided by (x − 3), which you answer instantly by substituting x = 3, with no division at all.
Worked example: factorising a cubic
Factorise completely: x3 − 6x2 + 11x − 6.
Verification: the three roots 1, 2, 3 multiply to 6, matching the constant term (with sign), and add to 6, matching the coefficient of x2. This quick check catches arithmetic slips.
Mistakes that quietly cost marks
Most algebra errors are not conceptual — they are careless. Watch for these.
- Sign errors after a minus sign — the leading cause of wrong answers in subtraction.
- Treating (a + b)2 as a2 + b2 — the middle term 2ab is missing. This is the most penalised mistake in the chapter.
- Forgetting to factorise completely — e.g. leaving 4x2 − 16 as 4(x2 − 4) instead of 4(x − 2)(x + 2).
- Wrong product in middle-term splitting — using c instead of ac when the leading coefficient is not 1.
(a + b)2 ≠ a2 + b2. The correct expansion always carries the cross term 2ab. Burn this into memory before exam day.
Previous-year style question
Q. If x + 1/x = 5, what is the value of x3 + 1/x3?
Answer: Use the identity a3 + b3 = (a + b)3 − 3ab(a + b) with a = x, b = 1/x, so ab = 1. Then x3 + 1/x3 = (x + 1/x)3 − 3(x + 1/x) = 53 − 3×5 = 125 − 15 = 110.
Whenever you see x + 1/x = k, the powers follow from identities: x2 + 1/x2 = k2 − 2, and x3 + 1/x3 = k3 − 3k. Memorise these two — they appear almost every cycle.
Quick revision
- Combine only like terms; flip every sign after a subtraction bracket.
- Memorise the nine standard identities — especially (a±b)2, a2−b2, and the cube identities.
- If a + b + c = 0, then a3 + b3 + c3 = 3abc.
- Factorise via: common factor → grouping → identity → splitting the middle term.
- For quadratics ax2+bx+c, split b into factors of a·c.
- Factor theorem: (x − a) divides p(x) when p(a) = 0; use it to crack cubics.
- For x + 1/x = k: x2+1/x2 = k2−2 and x3+1/x3 = k3−3k.
Frequently asked questions
What is the difference between an expression and an equation?
An expression like 3x + 2 has no equals sign and cannot be solved — it is only simplified or factorised. An equation like 3x + 2 = 11 has an equals sign and can be solved for the variable.
When should I split the middle term versus use an identity?
Use an identity first if the quadratic is a perfect square or difference of squares, since it is faster. Otherwise, split the middle term by finding two numbers whose product is a×c and sum is b.
How do I know which factor to test in the factor theorem?
Test the factors of the constant term divided by factors of the leading coefficient. For a monic cubic, simply try small integers like ±1, ±2, ±3 that divide the constant term until one gives p(a) = 0.
Is (a + b)² the same as a² + b²?
No. (a + b)² = a² + 2ab + b². The cross term 2ab is essential and dropping it is one of the most common and most penalised errors in CDS algebra.
How important is algebra for the CDS and OTA exam?
Very. Algebra underpins linear equations, quadratics, simplification and many word problems, accounting for a large slice of the Maths paper. Strong basic operations and factorisation skills save time across the entire section.
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