Factorise quadratics: special cases

  • A quadratic expression is in the form ax² + bx + c.
  • Special cases involve patterns that make factorisation easier.
  • These take the form a² + 2ab + b² = (a + b)² or a² – 2ab + b² = (a – b)².
  • Example: x² + 6x + 9 = (x + 3)(x + 3) = (x + 3)².
  • This follows the identity a² – b² = (a – b)(a + b).
  • Example: x² – 9 = (x – 3)(x + 3).
  • Always check for common factors before applying special cases.
  • Example: 4x² – 36 = 4(x² – 9) = 4(x – 3)(x + 3).
  • If a quadratic expression is factorised and set to zero, solve by setting each factor to zero.
  • Example: (x – 5)(x + 2) = 0 gives x = 5 or x = -2.
  • Identify perfect squares and differences of squares quickly.
  • Look for missing middle terms (b = 0) in the difference of squares.
  • Example 1: x² + 16x + 64 = (x + 8)².
  • Example 2: 49y² – 25 = (7y – 5)(7y + 5).
  • Use factorisation to simplify algebraic expressions.
  • Apply in solving quadratic equations efficiently.

Factorising perfect square trinomials:

a2+2ab+b2=(a+b)2

a2–2ab+b2=(a–b)2

Learn with an example

c2–8c+16=——

Notice that c2–8c+16 is a perfect square trinomial because it can be written in the form a2–2ab+b2, where a is c and b is 4.

a2–2ab+b2

c2–2c4+42

c2–8c+16

Now use the formula for factorising perfect square trinomials.

a2–2ab+b2=(a–b)2

c2–2c4+42=(c–4)2

c2–8c+16=(c–4)2

The factorised form of c2–8c+16 is (c–4)2.

Finally, check your work.

(c–4)2

(c–4)(c–4)Expand

c2–4c–4c+16Apply the distributive property (FOIL)

c2–8c+16

Yes, c2–8c+16=(c–4)2.

9x2–6x+1=——

Notice that 9x2–6x+1 is a perfect square trinomial because it can be written in the form a2–2ab+b2, where a is 3x and b is 1.

a2–2ab+b2

(3x)2–2 . 3x . 1+12

9x2–6x+1

Now use the formula for factorising perfect square trinomials.

a2–2ab+b2=(a–b)2

(3x)2–2 . 3x . 1+12=(3x–1)2

9x2–6x+1=(3x–1)2

The factorised form of 9x2–6x+1 is (3x–1)2.

Finally, check your work.

(3x–1)2

(3x–1)(3x–1)Expand

9x2–3x–3x+1Apply the distributive property (FOIL)

9x2–6x+1

Yes, 9x2–6x+1=(3x–1)2.

z2–1=—–

Notice that z2–1 is a difference of squares, because it can be written in the form a2–b2, where a is z and b is 1.

a2–b2

z212

z2–1

Now use the formula for factorising a difference of squares.

a2–b2=(a+b)(a–b)

z212=(z+1)(z–1)

z2–1=(z+1)(z–1)

The factorised form of z2–1 is (z+1)(z–1).

Finally, check your work.

(z+1)(z–1)

z2+z–z–1Apply the distributive property (FOIL)

z2–1

Yes, z2–1=(z+1)(z–1).

let’s practice!