Algebra Fundamentals: Factorisation, Simultaneous Equations, Indices, Logarithms & Progressions
What it is
Algebra turns arithmetic into a set of reusable rules: a letter stands in for a number so that one relationship, once established, works for every value that fits it. This skill bundles five tools built on that idea — the standard identities that expand or factorise an expression in one step, the methods for solving two equations in two unknowns together, the laws of indices that compress repeated multiplication, logarithms (the operation that undoes an index), and progressions, which are sequences generated by a fixed rule of formation. Every question here reduces to picking the one correct standard rule and applying it exactly; the safety net is always to substitute the result back into the original statement and confirm it holds.
Core concepts
Standard identities.
| Identity | Statement |
|---|---|
| Square of a sum | (a + b)² = a² + 2ab + b² |
| Square of a difference | (a − b)² = a² − 2ab + b² |
| Difference of squares | a² − b² = (a + b)(a − b) |
| Cube of a sum | (a + b)³ = a³ + 3a²b + 3ab² + b³ |
| Sum of cubes | a³ + b³ = (a + b)(a² − ab + b²) |
| Difference of cubes | a³ − b³ = (a − b)(a² + ab + b²) |
Factorising an expression
four techniques, roughly in the order to try them:
- Common factor — pull out whatever every term shares before anything else: 6x² + 9x = 3x(2x + 3).
- Grouping — split four terms into two pairs that yield the same bracket, then factor that bracket out: ax + ay + bx + by = a(x + y) + b(x + y) = (x + y)(a + b).
- Quadratic trinomial — for x² + bx + c, find two numbers that multiply to c and add to b, then split the middle term into those two parts.
- Identity-based — recognise a difference of squares, a perfect-square trinomial, or a sum/difference of cubes, and apply the matching identity above directly.
Simultaneous equations.
- Two linear equations — elimination: scale one or both equations so that a variable's coefficients match or are opposite, then add or subtract to remove it. Substitution: solve one equation for one variable, then substitute that expression into the other equation.
- One linear, one quadratic — always solve the LINEAR equation for one variable first, substitute that expression into the quadratic equation, and solve the resulting single-variable quadratic. It usually produces two roots, each pairing with its own value from the linear equation — report both pairs, and verify every pair in both original equations.
Laws of indices.
| Law | Statement |
|---|---|
| Product of powers | a^m × a^n = a^(m + n) |
| Power of a power | (a^m)^n = a^(m × n) |
| Zero power | a⁰ = 1 |
| Negative power | a^(−n) = 1 / a^n |
| Fractional power | a^(m/n) = the n-th root of a^m |
Laws of logarithms. A logarithm undoes an index: log_a(x) = y means the same thing as a^y = x, for a > 0, a ≠ 1, x > 0.
| Law | Statement |
|---|---|
| Product rule | log_a(xy) = log_a(x) + log_a(y) |
| Quotient rule | log_a(x/y) = log_a(x) − log_a(y) |
| Power rule | log_a(x^n) = n × log_a(x) |
| Change of base | log_a(x) = log_b(x) / log_b(a) |
| Log of its own base | log_a(a) = 1 |
| Log of 1 | log_a(1) = 0 |
Progressions.
| Progression | nth term | Sum of n terms |
|---|---|---|
| AP (first term a, common difference d) | a + (n − 1)d | n/2 × (2a + (n − 1)d) |
| GP (first term a, common ratio r ≠ 1) | a × r^(n − 1) | a × (r^n − 1)/(r − 1) |
| HP | reciprocal of the matching AP's nth term | no direct formula — solve through the reciprocal AP |
An HP is defined by its RECIPROCALS forming an AP — there is no standalone HP formula, so every HP question is really an AP question worked on the reciprocals and converted back at the end.
Worked example
Factorising with an identity. Factorise 4x² − 25y² and x² − 5x − 24.
4x² − 25y² = (2x)² − (5y)² = (2x − 5y)(2x + 5y). Check: (2x − 5y)(2x + 5y) = 4x² + 10xy − 10xy − 25y² = 4x² − 25y².
For x² − 5x − 24, find two numbers multiplying to −24 and adding to −5: −8 and 3. So x² − 5x − 24 = (x − 8)(x + 3). Check: (x − 8)(x + 3) = x² + 3x − 8x − 24 = x² − 5x − 24.
Two linear equations. Solve 2x + 3y = 13 and 3x − y = 3.
From the second equation, y = 3x − 3. Substitute into the first: 2x + 3(3x − 3) = 13 → 2x + 9x − 9 = 13 → 11x = 22 → x = 2. Then y = 3(2) − 3 = 3. Check in both originals: 2(2) + 3(3) = 4 + 9 = 13, and 3(2) − 3 = 3 — both hold.
One linear, one quadratic. Solve x + y = 6 and xy = 8.
From the linear equation, y = 6 − x. Substitute into the second: x(6 − x) = 8 → 6x − x² = 8 → x² − 6x + 8 = 0 → (x − 2)(x − 4) = 0 → x = 2 or x = 4. Pairing each root with y = 6 − x gives (x, y) = (2, 4) or (4, 2). Check both pairs in both originals: 2 + 4 = 6 and 2 × 4 = 8; 4 + 2 = 6 and 4 × 2 = 8 — both hold.
Indices. Simplify (2³ × 2⁴) ÷ 2⁵, and evaluate 27^(2/3).
(2³ × 2⁴) ÷ 2⁵ = 2^(3 + 4 − 5) = 2² = 4.
27^(2/3) = (the cube root of 27)² = 3² = 9. Check: 3³ = 27, so the cube root of 27 is 3, and 3² = 9.
Logarithms. Evaluate log₃81 + log₃3 − log₃9 two ways, then evaluate log₄8 by change of base.
Directly: log₃81 = 4 (3⁴ = 81), log₃3 = 1, log₃9 = 2 (3² = 9); total = 4 + 1 − 2 = 3.
Via the laws first: log₃81 + log₃3 − log₃9 = log₃(81 × 3 ÷ 9) = log₃27 = log₃3³ = 3 — the two methods agree.
For log₄8, switch to base 2: log₄8 = log₂8 / log₂4 = 3/2 (2³ = 8 and 2² = 4). Check: 4^(3/2) = (the square root of 4)³ = 2³ = 8.
Progressions. AP: find the 10th term and the sum of the first 10 terms of 3, 7, 11, 15, … (a = 3, d = 4).
10th term = a + 9d = 3 + 36 = 39. Sum = n/2 × (first + last) = 5 × (3 + 39) = 210. Cross-check with the other sum form: n/2 × (2a + (n − 1)d) = 5 × (6 + 36) = 210 — matches.
GP: find the 6th term and the sum of the first 6 terms of 2, 6, 18, … (a = 2, r = 3).
6th term = a × r⁵ = 2 × 243 = 486. Sum = a(r⁶ − 1)/(r − 1) = 2 × (729 − 1)/2 = 728. Cross-check by adding the terms directly: 2 + 6 + 18 + 54 + 162 + 486 = 728 — matches.
HP: the first three terms of an HP are 1/3, 1/5, 1/7. Find the 4th term.
The reciprocals 3, 5, 7 form an AP with a = 3, d = 2, so its 4th term is 3 + 3(2) = 9, and the HP's 4th term is the reciprocal, 1/9.
Common traps
- Expanding (a − b)² as a² − b² — dropping the middle term — instead of a² − 2ab + b².
- Mixing up the sum-of-cubes and difference-of-cubes patterns: the sign inside the linear factor matches the sign between a³ and b³, but the sign inside the trinomial factor is always the opposite one.
- In a linear-plus-quadratic system, solving the quadratic for both roots but pairing a root with the wrong partner value, or reporting only one of the two valid solution pairs.
- Accepting a value that satisfies only one of the two original equations — a genuine solution to the system must check out in BOTH.
- Writing a⁰ = 0 instead of 1, or a^(−n) = −a^n instead of 1/a^n.
- Treating (a^m)^n as if it were a^m × a^n — the first multiplies the exponents, the second adds them, and the two give different results whenever m ≠ n.
- Reading log_a(x) + log_a(y) as log_a(x + y) instead of log_a(xy) — the product rule turns addition OUTSIDE the log into multiplication INSIDE it, never addition inside.
- Using the GP sum formula a(r^n − 1)/(r − 1) when r = 1 — this divides by zero; a GP with r = 1 is just n copies of a, so its sum is simply n × a.
- Averaging HP terms directly instead of taking reciprocals, working the resulting AP, and only then taking the reciprocal of the answer.
Speed technique
- Any product of the form (round number − k)(round number + k) is a disguised difference of squares: 97 × 103 = (100 − 3)(100 + 3) = 100² − 3² = 9,991, without a single long multiplication.
- Choose elimination when the two equations already have matching or opposite coefficients on one variable; choose substitution when one equation already isolates a variable cleanly.
- Rewrite every index term to the SAME base before applying any law — a mix of 8s and 32s collapses instantly once both are seen as powers of 2 (2³ and 2⁵).
- Rewrite every logarithm to the SAME base, via the change-of-base rule, before combining them — terms in different bases cannot be added or subtracted directly.
- Identify the progression before reaching for a formula: a constant difference between terms is an AP, a constant ratio is a GP, and terms whose reciprocals share a constant difference are an HP.
- Once the last term of an AP is easy to find, n/2 × (first + last) is usually faster than n/2 × (2a + (n − 1)d).
Check yourself
- Factorise x² − 49. (
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x − 7)(x + 7)**. - Solve simultaneously: x + y = 9 and x − y = 3; find x and y.
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Adding the equations: 2x = 12, so x = 6; then y = 3. - Simplify (2³ × 2²) ÷ 2⁴.
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2^(3 + 2 − 4) = 2¹ = 2. - Evaluate log₅125.
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Since 5³ = 125, log₅125 = 3. - Find the sum of the first 8 terms of the AP 5, 8, 11, …
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a = 5, d = 3, so the 8th term = 5 + 7 × 3 = 26; sum = 8/2 × (5 + 26) = 4 × 31 = 124.
What the exam tests here
None of the 3 papers we hold has asked this. It is on the syllabus, so it can appear — but nothing in the paper record tells us how it would be framed. Treat it as insurance, not as a priority.
Try it: Algebra: Factorization, Logarithms, Progressions & Simultaneous Equations questions
Real questions from the CUET PG CS bank on exactly this skill. Pick an answer to see the full solution — the intuition, the worked steps, the faster methods and the traps.
- 81 is factorised as (x - a)(x + a). What is the value of a?
Show the answer and worked solution
Answer: option B
- 81 is a difference of squares, matching the pattern - = (p - q)(p + q) with p = x.
Write 81 as a square: 81 = , so q = 9.
Substituting gives - 81 = (x - 9)(x + 9), which matches the given form (x - a)(x + a) with a = 9, option B.
The equation - 2x - 35 = 0 has two roots. What is the larger of the two roots?
Show the answer and worked solution
Answer: option C
- 2x - 35 = 0 is solved using the quadratic formula x = .
The part under the root is - 4×1×(−35) = 4 + 140 = 144, and = 12.
So x = = 7 or x = = −5, and the larger root is 7, option C.
What is the greatest common numerical factor that can be taken out of both terms of 12x^3y - 18x^2y^2?
Show the answer and worked solution
Answer: option A
The numerical (coefficient) parts of the two terms are 12 and 18.
The greatest common factor of 12 and 18 is 6, since 6 divides both 12 (12 = 6×2) and 18 (18 = 6×3) exactly, and no larger number does.
So the greatest common numerical factor is 6, option A, giving 12x^3y - 18x^2y^2 = 6x^2y(2x - 3y).
A quadratic factorises as (x - a)(x - b), where a and b are positive with a > b. If a + b = 11 and a x b = 28, what is a - b?
Show the answer and worked solution
Answer: option A
Since a and b are the roots of - 11t + 28 = 0, the difference between them is .
Substituting gives = = .
= 3, so a - b = 3, option A (the roots themselves are a = 7 and b = 4).
+ 3x^2 − 4x - 12 factorises by grouping as (x + 3) − 4(x + 3) = (x + 3)(x - 2)(x + 2). What is the sum of all three roots of + 3x^2 − 4x - 12 = 0?
Show the answer and worked solution
Answer: option D
From the factorisation (x + 3)(x - 2)(x + 2) = 0, the three roots are x = −3, x = 2, and x = −2.
Adding them together gives −3 + 2 + (−2).
−3 + 2 − 2 = −3, option D.
Answer above — every one shows its working.