IGCSE Maths Algebraic manipulation notes

A letter is a container for a number. Substituting means replacing each letter with its value and then doing ordinary arithmetic — with one big trap: square before you multiply, and square negatives whole. If a = −2, then a² = (−2) × (−2) = +4, while 2a² = 2 × 4 = 8 — not −8. Brackets make it safe: write 2(−2)² each time.

The same skill runs formulas: with m = 3 and n = −2, the expression 4m + 3n = 12 + (−6) = 6, and 2m² − n = 18 − (−2) = 20 — subtracting a negative adds. You will reuse substitution everywhere: in formulas for area, interest, and later in functions.

A number in front of a bracket multiplies each term inside: 3(2x − 5) = 6x − 15. When the multiplier is negative, every sign flips: −2(x − 4) = −2x + 8. In a string like 2(x + 4) − 3(x − 1), expand both brackets first (2x + 8 and −3x + 3), then collect: 2x + 8 − 3x + 3 = 11 − x.

Two brackets multiply every term by every term. A grid keeps it honest: for (x + 3)(x + 5), the four products are x², 5x, 3x and 15, so the answer is x² + 8x + 15. For (2x − 1)(x + 4): 2x² + 8x − x − 4 = 2x² + 7x − 4. Always collect the middle terms — the answer is not finished until it has one term per power.

Extended. Three brackets are just two in disguise: pair two first, expand fully, then multiply the result by the third. The middle variable can be y or z too — (x + 2)(x − 3)(x + 1) is examined as readily as (x + a)(x + b)(x + c).

To factorise, look first for a common factor in every term — both numbers and letters. 6x² + 9x has 3x in both terms, so 6x² + 9x = 3x(2x + 3). Check by expanding: 3x × 2x = 6x² and 3x × 3 = 9x ✓. “Factorise fully” means keep going until nothing else comes out: 3x³ + 6x² = 3x²(x + 2), because 3x² divides both terms.

The difference of two squares is a pattern to memorise: x² − 49 = (x − 7)(x + 7), since (x − 7)(x + 7) = x² − 7x + 7x − 49. Spot it whenever two perfect squares are subtracted — including with coefficients: 4x² − 25 = (2x − 5)(2x + 5). A sum of squares, x² + 49, does not factorise.

For x² + bx + c, find two numbers that multiply to c and add to b: x² + 5x + 6 needs a pair with product 6 and sum 5 — 2 and 3 — so x² + 5x + 6 = (x + 2)(x + 3). When the constant is negative the pair has opposite signs: x² − 3x − 10 needs product −10, sum −3 → −5 and +2 → (x − 5)(x + 2).

Extended. When a ≠ 0, split the middle term using a factor pair of a × c. For 2x² + 7x + 3: a × c = 6, and the pair with sum 7 is 1 and 6, so 2x² + 7x + 3 = 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3). For 6x² + 7x − 3: a × c = −18, pair 9 and −2 → 6x² + 9x − 2x − 3 = 3x(2x + 3) − 1(2x + 3) = (3x − 1)(2x + 3). Expand your answer to check — every factorisation can be verified for free.

Extended. Any quadratic x² + bx + c can be rewritten with the first two terms as a perfect square: take half of b, square it, and build the square around it. For x² + 8x + 3: half of 8 is 4, and 4² = 16, so x² + 8x + 3 = (x + 4)² − 16 + 3 = (x + 4)² − 13. Check by expanding: (x + 4)² − 13 = x² + 8x + 16 − 13 = x² + 8x + 3 ✓.

The payoff is the turning point: a squared bracket is never negative, so (x + 4)² − 13 is smallest when x + 4 = 0, i.e. at x = −4, where it equals −13. The curve y = x² + 6x + 2 = (x + 3)² − 7 therefore has its minimum at (−3, −7). This form also feeds straight into the quadratic formula and into sketching — a later topic reuses it constantly.

Extended. Algebraic fractions obey the same laws as numerical ones. To add or subtract, build a common denominator: 2⁄(x + 1) + 3⁄(x + 2) = [2(x + 2) + 3(x + 1)] ⁄ [(x + 1)(x + 2)] = (5x + 7) ⁄ (x + 1)(x + 2). Multiply tops and bottoms straight across; divide by multiplying by the flipped fraction.

To simplify a rational expression, factorise top and bottom completely and cancel a whole common factor — never individual terms. (x² + 7x + 10) ⁄ (x² − 25) = (x + 5)(x + 2) ⁄ (x − 5)(x + 5) = (x + 2) ⁄ (x − 5), because x + 5 is a factor of both. Writing (x² + x) ⁄ x as “x + 1” by cancelling the two x terms is the classic crime — factorise first: x(x + 1) ⁄ x = x + 1.