GCSE · Algebra

Double Brackets and Factorising Quadratics

Multiply two brackets together and you get a quadratic: an expression whose highest power is x². Going the other way, splitting a quadratic back into two brackets, is called factorising. One picture, a rectangle cut into four pieces, explains both directions.

GCSE subject content: simplify and manipulate algebraic expressions by expanding products of two binomials; factorise quadratic expressions of the form x² + bx + c, including the difference of two squares; Higher: factorise ax² + bx + c and expand products of more than two binomials

Quick recap (from KS3). To expand a single bracket, multiply every term inside by the term outside: 3(x + 4) = 3x + 12, x(x − 5) = x² − 5x. To factorise, take out the highest common factor: 6x + 9 = 3(2x + 3). Multiplying negatives: (−3) × (−4) = +12, (−3) × 4 = −12. If any of that feels shaky, start with Expanding and factorising.

1 · Expanding double brackets: the area grid

A binomial is an expression with two terms, like x + 3. To expand (x + 3)(x + 5) means to multiply it out so there are no brackets left. Think of it as the area of a rectangle with sides x + 3 and x + 5. Cut each side into its two parts and the rectangle splits into four smaller rectangles. Every term in the first bracket meets every term in the second: 2 × 2 = 4 products.

Change the numbers. The grid works for negatives too: a "length" of −2 just gives a negative area. Then press F, O, I, L in turn.

FOIL check:

FOIL is a memory aid for the same four products: First × first, Outer × outer, Inner × inner, Last × last. It is a good check that you haven't missed one, but the grid is safer, because FOIL only works for two brackets of two terms each, while the grid works for any size.

Common mistakes. Only multiplying the firsts and lasts: (x + 3)(x + 5) ≠ x² + 15. Forgetting to collect the two middle terms (3x + 5x = 8x). Sign slips: in (x − 4)(x + 2) the corner is (−4) × 2 = −8, and the middle is −4x + 2x = −2x.

2 · Squaring a bracket: the (a + b)² trap

(x + 3)² means (x + 3)(x + 3): a bracket multiplied by itself. The most common GCSE algebra mistake is to "square each bit" and write x² + 9. The picture shows what that misses. A square of side a + b is made of an a² square, a b² square and two a × b rectangles.

(a + b)²
a² + b² only
Missing piece 2ab

(x + k)² = x² + 2kx + k²

Double the number for the middle term, square it for the end. With a negative, (x − 4)² = x² − 8x + 16: the end is (−4)² = +16, always positive.

3 · The difference of two squares

Expand (x + 5)(x − 5): x² − 5x + 5x − 25. The middle terms cancel, leaving x² − 25. This happens whenever the brackets are the same apart from the sign, so

x² − a² = (x + a)(x − a)

It's called the difference of two squares ("difference" means subtraction). Here's why it works, with no algebra at all. Take an x by x square, cut away an a by a corner, then move the leftover strip.

Spot it: two terms, both perfect squares, with a minus between them. x² − 49 = (x + 7)(x − 7), 4x² − 25 = (2x + 5)(2x − 5) (because 4x² = (2x)²), 9 − y² = (3 + y)(3 − y).

Trap: x² + 25 is a sum of two squares and does not factorise. And x² − 25 ≠ (x − 5)²: that expands to x² − 10x + 25.

Mental-maths trick: 21 × 19 = (20 + 1)(20 − 1) = 400 − 1 = 399.

4 · Factorising x² + bx + c: two numbers

Look back at the grid. When both brackets start with x, expanding (x + m)(x + n) gives

x² + (m + n)x + mn

So to factorise x² + bx + c you need two numbers that multiply to make c (the product) and add to make b (the sum). Start with the product, because there are only a few factor pairs; then check which pair has the right sum. Click the pairs to test them; the grid is the expansion run in reverse.

Signs in x² + bx + cThe two numbers are…Example
c positive, b positiveboth positivex² + 7x + 12 = (x + 3)(x + 4)
c positive, b negativeboth negativex² − 9x + 20 = (x − 4)(x − 5)
c negativeone positive, one negativex² − 2x − 15 = (x − 5)(x + 3)

5 · Practice: factorise it yourself

Type the two numbers in the brackets. If it isn't right, the feedback tells you what went wrong (product, sum or sign), so you can fix it. Always expand your answer in your head to check.

6 · Factorising ax² + bx + c Higher

When x² has a coefficient other than 1, like 6x² + x − 12, guessing brackets gets slow. Splitting the middle term always works:

  1. Take out any common factor first.
  2. Multiply a × c. Find two numbers that multiply to a × c and add to b.
  3. Use them to split bx into two x terms.
  4. Put the four terms into the grid (or factorise in pairs). The common factors of the rows and columns are the brackets.

Why a × c? In (px + r)(qx + s) the two middle cells are psx and rqx. Multiply them: (ps)(rq) = (pq)(rs) = a × c. So the two middle numbers always multiply to ac and add to b.

Check by expanding. (2x + 3)(3x − 4) = 6x² − 8x + 9x − 12 = 6x² + x − 12 ✓. If the middle term comes out with the wrong sign, swap the signs in the brackets.

7 · Which method? Spot the type

In an exam the question just says "factorise fully". "Fully" means keep going until no bracket can be factorised any more. Work through this checklist every time:

  1. Common factor? Take it out first: 2x² − 50 = 2(x² − 25).
  2. Two square terms with a minus? Difference of two squares.
  3. x² + bx + c? Two numbers: product c, sum b.
  4. ax² + bx + c? Split the middle term (Higher).
  5. No pair works? It doesn't factorise (with whole numbers).

Worked examples

Expand and simplify (x + 4)(x − 7)
Grid: x × x = x²; x × (−7) = −7x; 4 × x = 4x; 4 × (−7) = −28.
Write all four: x² − 7x + 4x − 28.
Collect the x terms: −7x + 4x = −3x.
Answer: x² − 3x − 28. Check: the numbers 4 and −7 add to −3 and multiply to −28 ✓
Expand (2x − 3)²
Write it as two brackets: (2x − 3)(2x − 3). Do not write 4x² + 9 or 4x² − 9.
Four products: 2x × 2x = 4x²; 2x × (−3) = −6x; (−3) × 2x = −6x; (−3) × (−3) = +9.
Collect: 4x² − 6x − 6x + 9 = 4x² − 12x + 9.
Pattern check: middle term = 2 × (2x) × (−3) = −12x, end = (−3)² = 9 ✓
Factorise x² − 2x − 15
Product −15, sum −2. The product is negative, so the numbers have opposite signs.
Factor pairs of −15: 1 and −15 (sum −14), −1 and 15 (sum 14), 3 and −5 (sum −2) ✓, −3 and 5 (sum 2).
Answer: (x + 3)(x − 5).
Check: x² − 5x + 3x − 15 = x² − 2x − 15 ✓
Factorise fully 3x² − 48
Common factor first: 3 goes into both: 3(x² − 16).
The bracket is a difference of two squares: x² − 16 = x² − 4² = (x + 4)(x − 4).
Answer: 3(x + 4)(x − 4). Stopping at 3(x² − 16) would lose a mark: that isn't "fully" factorised.
Factorise 6x² + x − 12 Higher
Common factor? 6, 1 and 12 have none.
a × c = 6 × (−12) = −72. Need two numbers with product −72 and sum +1: 9 and −8.
Split the middle: 6x² + 9x − 8x − 12.
Factorise in pairs: 3x(2x + 3) − 4(2x + 3). Both pairs share the bracket (2x + 3) — if they don't, check your signs.
Answer: (2x + 3)(3x − 4). Check: 6x² − 8x + 9x − 12 = 6x² + x − 12 ✓
Expand (x + 1)(x + 2)(x − 3) Higher
Two brackets first: (x + 1)(x + 2) = x² + 2x + x + 2 = x² + 3x + 2.
Multiply by the third: (x² + 3x + 2)(x − 3). A grid with 3 rows and 2 columns: x³ − 3x² + 3x² − 9x + 2x − 6.
Collect: x³ + 0x² − 7x − 6 = x³ − 7x − 6.
Check with x = 1: (2)(3)(−2) = −12, and 1 − 7 − 6 = −12 ✓
Exam-style: show that (x + 5)² − (x + 1)² = 8(x + 3)
Expand each square: (x + 5)² = x² + 10x + 25; (x + 1)² = x² + 2x + 1.
Subtract the whole bracket (the minus applies to every term): x² + 10x + 25 − x² − 2x − 1.
Simplify: 8x + 24 = 8(x + 3), as required ✓
Shortcut: it is a difference of two squares: (A − B)(A + B) with A = x + 5, B = x + 1 gives (4)(2x + 6) = 8(x + 3).

Quick check

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