Further Expanding and Factorising [H]

Further Expanding and Factorising

At Higher tier, the expanding and factorising skills extend to three or more brackets, and to quadratics with a leading coefficient other than 1 (\( ax^2 + bx + c \) with \( a \neq 1 \)). The AC method provides a reliable algorithm for the latter and connects directly to the grouping technique used in expanding.

Expanding Three or More Brackets

Expand two brackets first, simplify, then multiply the result by the third bracket. Apply the distributive law to every pair of terms.

\[ (x+1)(x+2)(x+3) \]

Step 1: expand the first two: \( (x+1)(x+2) = x^2 + 3x + 2 \)

Step 2: multiply by the third:

\[ (x^2+3x+2)(x+3) = x^3+3x^2+3x^2+9x+2x+6 = x^3 + 6x^2 + 11x + 6 \]

Factorising \( ax^2 + bx + c \) — The AC Method

When the leading coefficient \( a \neq 1 \), the simple "find two numbers" approach no longer works directly. The AC method adapts it:

  1. Find \( ac \) (multiply \( a \) and \( c \)).
  2. Find \( m \) and \( n \) such that \( m \times n = ac \) and \( m + n = b \).
  3. Rewrite \( bx \) as \( mx + nx \).
  4. Factorise by grouping: find the common factor in each pair of terms.

Example: \( 2x^2 + 7x + 3 \)

\[ ac = 2 \times 3 = 6. \quad \text{Need } m \times n = 6, m + n = 7: \quad m = 1, n = 6. \] \[ 2x^2 + x + 6x + 3 = x(2x+1) + 3(2x+1) = (2x+1)(x+3) \]

Verify: \( (2x+1)(x+3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3 \checkmark \)

Worked Examples

Expand \( (x-1)(x+2)(x-3) \).

Step 1: expand first two brackets.

\[ (x-1)(x+2) = x^2 + 2x - x - 2 = x^2 + x - 2 \]

Step 2: multiply by the third bracket.

\[ (x^2+x-2)(x-3) = x^3 - 3x^2 + x^2 - 3x - 2x + 6 = x^3 - 2x^2 - 5x + 6 \]
Factorise \( 3x^2 - 10x - 8 \) and \( 6x^2 + 11x - 10 \).

First: \( ac = 3 \times (-8) = -24 \). Need \( m \times n = -24, m + n = -10 \): try \( m = 2, n = -12 \).

\[ 3x^2 + 2x - 12x - 8 = x(3x+2) - 4(3x+2) = (3x+2)(x-4) \]

Second: \( ac = 6 \times (-10) = -60 \). Need \( m \times n = -60, m + n = 11 \): try \( m = 15, n = -4 \).

\[ 6x^2 + 15x - 4x - 10 = 3x(2x+5) - 2(2x+5) = (2x+5)(3x-2) \]
Factorise \( 4x^2 - 12x + 9 \) and \( 9x^2 - 6x + 1 \).

Recognise perfect square trinomials: \( a^2 - 2ab + b^2 = (a-b)^2 \).

For \( 4x^2 - 12x + 9 \): \( a = 2x \), \( b = 3 \), check: \( 2ab = 2(2x)(3) = 12x checkmark \)

\[ 4x^2 - 12x + 9 = (2x-3)^2 \]

For \( 9x^2 - 6x + 1 \): \( a = 3x \), \( b = 1 \), check: \( 2ab = 2(3x)(1) = 6x checkmark \)

\[ 9x^2 - 6x + 1 = (3x-1)^2 \]

 Key Takeaways

  • To expand three brackets: expand the first two, simplify fully, then multiply by the third.
  • The AC method for \( ax^2+bx+c \): find \( m,n \) with \( mn = ac \) and \( m+n = b \); split \( bx \); factorise by grouping.
  • A perfect square trinomial has the form \( a^2 \pm 2ab + b^2 = (a \pm b)^2 \) — check that the middle term equals \( 2ab \).
  • A common factor should always be extracted first, before applying the AC method or recognising special forms.
  • Verify every factorisation by expanding — a single sign error can produce an incorrect result that is hard to spot without checking.