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Factorisation using identities · 5 marks

Factor using suitable identities:

  1. (i) 16y2 − 24y + 9
  2. (ii) 94s2 + 6st + 4t2
  3. (iii) m29 + mk3 + k24 + 3nk + 2mn + 9n2
  4. (iv) p216 − 2 + 16p2
  5. (v) 9a2 + 4b2 + c2 − 12ab + 6ac − 4bc
Answer: (i) (4y − 3)2 (ii) (32s + 2t)2 (iii) (m3 + k2 + 3n)2 (iv) (p4 − 4p)2 (v) (3a − 2b + c)2

Step-by-step solution

Idea: Count the terms first. Three terms with two perfect squares → (a + b)2 or (a − b)2 (a minus middle term means (a − b)2). Six terms with three perfect squares → (a + b + c)2. Take the square roots of the square terms, then use the signs of the cross terms (2ab, 2bc, 2ca) to decide which of a, b, c carry a minus sign.

(i) 16y2 − 24y + 9

  1. 16y2 = (4y)2 and 9 = 32. The middle term is negative, so use (a − b)2 = a2 − 2ab + b2 with a = 4y, b = 3.
  2. Check: 2ab = 2(4y)(3) = 24y ✓½ mark
  3. 16y2 − 24y + 9 = (4y − 3)2½ mark
(4y − 3)2

(ii) 94s2 + 6st + 4t2

  1. 94s2 = (32s)2 and 4t2 = (2t)2. Use (a + b)2 with a = 32s, b = 2t.
  2. Check: 2 × 32s × 2t = 6st ✓½ mark
  3. = (32s + 2t)2 (which can also be written 14(3s + 4t)2)½ mark
(32s + 2t)2

(iii) m29 + mk3 + k24 + 3nk + 2mn + 9n2

  1. Six terms, so try (a + b + c)2. The square terms are m29 = (m3)2, k24 = (k2)2, 9n2 = (3n)2. All cross terms are positive, so take a = m3, b = k2, c = 3n.
  2. Check the cross terms:
    2ab = 2 × m3 × k2 = mk3 ✓
    2bc = 2 × k2 × 3n = 3nk ✓
    2ca = 2 × 3n × m3 = 2mn ✓½ mark
  3. = (m3 + k2 + 3n)2½ mark
(m3 + k2 + 3n)2

(iv) p216 − 2 + 16p2

  1. p216 = (p4)2 and 16p2 = (4p)2. The middle term is negative, so try (a − b)2 with a = p4, b = 4p.
  2. Check: 2ab = 2 × p4 × 4p = 2 (the p and the 4 both cancel) ✓. So the −2 is exactly −2ab.½ mark
  3. = (p4 − 4p)2½ mark
(p4 − 4p)2

(v) 9a2 + 4b2 + c2 − 12ab + 6ac − 4bc

  1. Six terms: use (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx (new letters, because a, b, c are already used). Square roots: 3a, 2b, c.
  2. Signs: the ab and bc terms are negative but the ac term is positive. The term common to both negative products is b, so b gets the minus sign: x = 3a, y = −2b, z = c.
  3. Check: 2(3a)(−2b) = −12ab ✓; 2(−2b)(c) = −4bc ✓; 2(c)(3a) = 6ac ✓½ mark
  4. = (3a − 2b + c)2½ mark
(3a − 2b + c)2
(i) (4y − 3)² (ii) ((3/2)s + 2t)² (iii) (m/3 + k/2 + 3n)² (iv) (p/4 − 4/p)² (v) (3a − 2b + c)²

Check: (v) with a = b = c = 1: 9 + 4 + 1 − 12 + 6 − 4 = 4 and (3 − 2 + 1)2 = 4 ✓. (iv) with p = 4: 1 − 2 + 1 = 0 and (1 − 1)2 = 0 ✓. (i) with y = 1: 16 − 24 + 9 = 1 and (4 − 3)2 = 1 ✓.

Answer to write in the exam

(i)

16y2 − 24y + 9 = (4y)2 − 2(4y)(3) + 32 [a2 − 2ab + b2 = (a − b)2]

∴ 16y2 − 24y + 9 = (4y − 3)2

(ii)

94s2 + 6st + 4t2 = (32s)2 + 2(32s)(2t) + (2t)2 [a2 + 2ab + b2 = (a + b)2]

∴ 94s2 + 6st + 4t2 = (32s + 2t)2

(iii)

m29 + mk3 + k24 + 3nk + 2mn + 9n2 = (m3)2 + (k2)2 + (3n)2 + 2(m3)(k2) + 2(k2)(3n) + 2(3n)(m3) [(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]

∴ m29 + mk3 + k24 + 3nk + 2mn + 9n2 = (m3 + k2 + 3n)2

(iv)

p216 − 2 + 16p2 = (p4)2 − 2(p4)(4p) + (4p)2 [a2 − 2ab + b2 = (a − b)2]

∴ p216 − 2 + 16p2 = (p4 − 4p)2

(v)

9a2 + 4b2 + c2 − 12ab + 6ac − 4bc = (3a)2 + (−2b)2 + c2 + 2(3a)(−2b) + 2(−2b)(c) + 2(c)(3a) [(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]

∴ 9a2 + 4b2 + c2 − 12ab + 6ac − 4bc = (3a − 2b + c)2

Common mistakes that cost marks

  • In (v), giving the minus sign to the wrong letter, e.g. (3a + 2b − c)2. That would make the ac and bc terms negative. Always check all three cross terms.
  • In (iv), not noticing that the constant −2 is the middle term. Since p4 × 4p = 1, the term 2ab is just the number 2.
  • In (i), writing (4y + 3)2. A negative middle term with positive square terms always means (a − b)2.

How this can come in the exam

MCQ (1 mark)

x2 + 4y2 + 9z2 + 4xy + 12yz + 6xz is equal to

  1. (x − 2y + 3z)2
  2. (x + 4y + 9z)2
  3. (x + 2y − 3z)2
  4. (x + 2y + 3z)2
Show answer

(D) (x + 2y + 3z)2
Square roots x, 2y, 3z; all cross terms positive, and 2(x)(2y) = 4xy, 2(2y)(3z) = 12yz, 2(3z)(x) = 6xz.

Short answer (2 marks)

Factor: 4a2 + b2 + c2 − 4ab + 2bc − 4ca.

Show answerSquare roots 2a, b, c. The ab and ca terms are negative and share a, so a takes the minus sign: (−2a + b + c)2, which is the same as (2a − b − c)2. Check: 2(2a)(−b) = −4ab, 2(−b)(−c) = 2bc, 2(−c)(2a) = −4ca ✓.

Try one yourself

Factor: x24 + 2 + 4x2.

Show answer

(x2 + 2x)2, since 2 × x2 × 2x = 2.

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