Factorisation: Identities

1Summary 重點摘要

Learn a few handy formulas. When you see a certain pattern, you can factorise immediately, no trial and error.

2Vocabulary cards Vocabulary

3Step by step Steps

1. Step 1: Why learn identitiesFactorisation rewrites a polynomial as a product of factors, and identities supply the shortcut of 'recognise the pattern, split at once'. Some quadratics are highly regular: x² + 6x + 9 is a perfect square the moment you know the formula, written straight off as (x + 3)² with no trial and error. Compared with cross-method guessing, identities are fast, exact and confident: if the pattern matches, the answer is right; if it does not, you know at once to try another route. Three identities must be memorised in secondary school — two perfect squares and the difference of squares — all standard tools for solving quadratics and simplifying algebraic fractions.
2. Step 2: Perfect square 1: (a+b)²The first identity: (a + b)² = a² + 2ab + b². Used in reverse, any trinomial with 'squares at the ends and twice the product in the middle' collapses into a perfect square. Test x² + 6x + 9: the head x² = (x)², the tail 9 = (3)², the middle 6x = 2 × x × 3 — all three conditions hold, so x² + 6x + 9 = (x + 3)². A coefficient version: 4x² + 12x + 9 has head (2x)², tail 3², and middle 12x = 2 × (2x) × 3, hence (2x + 3)². The chant: check head and tail are squares, then check the middle is twice the product; only after passing both tests should you write the answer.
3. Step 3: Perfect square 2: (a−b)²The second identity: (a − b)² = a² − 2ab + b². It differs from the first only in the middle sign: a minus inside the bracket makes the middle −2ab, while the tail stays +b². Test x² − 6x + 9: head and tail are again x² and 3², the middle is −6x = −2 × x × 3, so it equals (x − 3)². Two errors recur: writing the tail negative too, imagining (x − 3)² to end in −9, when in fact (−3)² = +9; and flipping the middle sign, factorising x² − 6x + 9 as (x + 3)². The rule: the middle term copies the bracket's sign exactly, and the end term is always positive.
4. Step 4: Difference of squares: a² − b²The third identity: a² − b² = (a + b)(a − b) — two terms, both squares, joined by a minus, splitting into sum times difference. Examples: x² − 9 = x² − 3² = (x + 3)(x − 3); x² − 25 = (x + 5)(x − 5); 4x² − 49 = (2x)² − 7² = (2x + 7)(2x − 7). This identity appears most often in school work: simplifying (x − 3)(x + 3), solving x² = 16, and mental arithmetic such as 101 × 99 = (100 + 1)(100 − 1) = 10000 − 1 = 9999 all rest on it. Recognition checklist: exactly two terms, each a perfect square, joined by subtraction — all three required; x² + 9 does not qualify.
5. Step 5: Choosing the right identityA decision path for choosing the identity. Step one, count the terms. Two terms: if both are squares joined by a minus, use the difference of squares; if joined by a plus (such as x² + 9), none of the three identities applies. Three terms: check that head and tail are squares and the middle is twice the product — a positive middle gives (a + b)², a negative one (a − b)²; if the middle is not twice the product (as in x² + 5x + 9), no identity fits and another method such as cross-multiplication is needed. Step two, extract any common factor first: 2x² − 8 = 2(x² − 4) = 2(x + 2)(x − 2); pull out the 2 before applying the difference of squares, or the answer is incomplete.
6. Step 6: Verify by multiplying backAfter factorising, multiply back to verify — a habit that must not be skipped. For x² − 16 answered as (x + 4)(x − 4), expand to check: x × x = x², x × (−4) = −4x, 4 × x = 4x, 4 × (−4) = −16; the middle terms −4x + 4x cancel, leaving x² − 16, identical to the original, so the factorisation stands. If the product came out as x² + 16 or x² − 8x + 16, the pattern was misread and the working must be reconsidered. Verification also catches dropped minus signs and missing brackets. Under exam pressure, at least expand mentally once; in practice, write it out until checking becomes a reflex.
7. Step 7: Common mistakes summaryFour errors recur with identities. First, the difference of squares misapplied to addition: x² + 9 is not (x + 3)(x − 3), which expands to x² − 9; over the reals x² + 9 does not factorise at all. Second, dropping the middle term of a perfect square: when expanding (x + 3)(x + 3), confirm the middle 6x really appears — losing it means confusion with x² + 9. Third, squaring coefficients wrongly: 9x² − 16 factorises as (3x + 4)(3x − 4); writing (9x + 4)(9x − 4) is wrong because (9x)² = 81x², not 9x². Fourth, forgetting the common factor: 3x² − 12 must first become 3(x² − 4). Multiplying back after each question exposes all four on the spot.

4Diagrams Diagrams

5Worked example Example

6Common mistakes Mistakes

7Quick check Quick check

8Recap Recap

• Three handy formulas, apply when you see the pattern
• (a+b)² = a² + 2ab + b²
• (a−b)² = a² − 2ab + b²
• a² − b² = (a+b)(a−b)
• Always verify after using!