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Factorisation using identities · 2 marks

Let us try to find factors of another algebraic expression: 36x2 + 12x + 1.

Answer: 36x2 + 12x + 1 = (6x + 1)2, so (6x + 1) is a factor.

Step-by-step solution

Given: 36x2 + 12x + 1
To find: Its factors

Idea: Try to write it in the form a2 + 2ab + b2, which equals (a + b)2. 36x2 and 1 are both perfect squares, which is the clue.

  1. 36x2 = (6x)2 and 1 = 12. So try a = 6x, b = 1.½ mark
  2. Middle term check: 2ab = 2(6x)(1) = 12x ✓.½ mark
  3. 36x2 + 12x + 1 = (6x)2 + 2(6x)(1) + 12½ mark
  4. = (6x + 1)2. Thus (6x + 1) is a factor of 36x2 + 12x + 1.½ mark
36x² + 12x + 1 = (6x + 1)² = (6x + 1)(6x + 1).

Check: Put x = 1: 36 + 12 + 1 = 49 and (6 + 1)2 = 49 ✓.

Answer to write in the exam

36x2 + 12x + 1 = (6x)2 + 2(6x)(1) + 12 [a2 + 2ab + b2 = (a + b)2]

∴ 36x2 + 12x + 1 = (6x + 1)2 = (6x + 1)(6x + 1)

Common mistakes that cost marks

  • Writing (36x + 1)2 or (18x + 1)2. The square root of 36x2 is 6x, not 36x or half of it.
  • Forgetting the coefficient check: 2 × 6x × 1 must equal the middle term 12x.
  • Writing (6x − 1)2: the middle term +12x is positive, so both terms in the bracket are positive.

How this can come in the exam

MCQ (1 mark)

49y2 + 14y + 1 equals

  1. (49y + 1)2
  2. (7y + 1)2
  3. (7y + 14)2
  4. (7y − 1)2
Show answer

(B) (7y + 1)2
(7y)2 + 2(7y)(1) + 12 = (7y + 1)2.

Short answer (2 marks)

Factorise 25p2 + 20p + 4.

Show answer25p2 = (5p)2, 4 = 22, 20p = 2(5p)(2) (1 mark). So it equals (5p + 2)2 (1 mark).

Try one yourself

Factorise 64a2 + 16a + 1.

Show answer

(8a)2 + 2(8a)(1) + 12 = (8a + 1)2.

More questions like this

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← The identity (a + b)² = a² + 2ab + b² can also be used to find factorsLet us try to factor 50p² + 60pq + 18q². What will a and b be in this →