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Factorisation using identities · 5 marks

Factor completely:

  1. (i) 9x2 + 24xy + 16y2
  2. (ii) 4s2 + 20st + 25t2
  3. (iii) 49x2 + 28xy + 4y2
  4. (iv) 64p2 + 323pq + 49q2
  5. *(v) 3a2 + 4ab + 43b2
  6. *(vi) 95s2 + 6sv + 5v2
Answer: (i) (3x + 4y)2 (ii) (2s + 5t)2 (iii) (7x + 2y)2 (iv) (8p + 23q)2 (v) 3(a + 23b)2 (vi) 5(35s + v)2

Step-by-step solution

Idea: Each expression has the shape a2 + 2ab + b2, which equals (a + b)2. Find a and b from the two square terms, then check that the middle term really is 2ab. For (v) and (vi) the printed hint applies: first take a common number out of all three terms, just as 50p2 + 60pq + 18q2 = 2(25p2 + 30pq + 9q2) = 2(5p + 3q)2. Taking out 3 in (v) and 5 in (vi) leaves a perfect square inside the bracket.

(i) 9x2 + 24xy + 16y2

  1. 9x2 = (3x)2 and 16y2 = (4y)2, so try a = 3x, b = 4y.
  2. Check the middle term: 2ab = 2(3x)(4y) = 24xy ✓
  3. So 9x2 + 24xy + 16y2 = (3x)2 + 2(3x)(4y) + (4y)2 = (3x + 4y)2½ mark
(3x + 4y)2 = (3x + 4y)(3x + 4y)

(ii) 4s2 + 20st + 25t2

  1. 4s2 = (2s)2 and 25t2 = (5t)2, so a = 2s, b = 5t.
  2. Middle term: 2(2s)(5t) = 20st ✓
  3. = (2s + 5t)2½ mark
(2s + 5t)2

(iii) 49x2 + 28xy + 4y2

  1. 49x2 = (7x)2 and 4y2 = (2y)2, so a = 7x, b = 2y.
  2. Middle term: 2(7x)(2y) = 28xy ✓
  3. = (7x + 2y)2½ mark
(7x + 2y)2

(iv) 64p2 + 323pq + 49q2

  1. 64p2 = (8p)2 and 49q2 = (23q)2, so a = 8p, b = 23q.
  2. Middle term: 2 × 8p × 23q = 323pq ✓
  3. = (8p + 23q)2½ mark
  4. (The same answer can also be written 19(24p + 2q)2 = 49(12p + q)2; all three forms are equal.)
(8p + 23q)2

*(v) 3a2 + 4ab + 43b2

  1. Hint used: 3a2 is not a perfect square (3 is not a square number). So first take a common factor out of all three terms. Taking out 3 makes the first term a2, which is a perfect square.
  2. 3a2 + 4ab + 43b2 = 3(a2 + 43ab + 49b2)
    (4 ÷ 3 = 43 and 43 ÷ 3 = 49)½ mark
  3. Inside the bracket: a2 = (a)2 and 49b2 = (23b)2. Middle term: 2 × a × 23b = 43ab ✓½ mark
  4. = 3(a + 23b)2½ mark
  5. (Equivalent form without fractions inside: 13(3a + 2b)2.)
3(a + 23b)2 = 13(3a + 2b)2

*(vi) 95s2 + 6sv + 5v2

  1. Hint used: 5v2 is not a perfect square, and neither is 95s2. Take 5 out as a common factor so the last term becomes v2.
  2. 95s2 + 6sv + 5v2 = 5(925s2 + 65sv + v2)
    (95 ÷ 5 = 925 and 6 ÷ 5 = 65)½ mark
  3. Inside the bracket: 925s2 = (35s)2 and v2 = (v)2. Middle term: 2 × 35s × v = 65sv ✓½ mark
  4. = 5(35s + v)2½ mark
  5. (Equivalent form: 15(3s + 5v)2.)
5(35s + v)2 = 15(3s + 5v)2
(i) (3x + 4y)² (ii) (2s + 5t)² (iii) (7x + 2y)² (iv) (8p + (2/3)q)² (v) 3(a + (2/3)b)² (vi) 5((3/5)s + v)²

Check: Put the letters equal to 1. (i) 9 + 24 + 16 = 49 = 72 = (3 + 4)2 ✓. (v) 3 + 4 + 43 = 253 and 3 × (53)2 = 3 × 259 = 253 ✓. (vi) 95 + 6 + 5 = 645 and 5 × (85)2 = 5 × 6425 = 645 ✓.

Answer to write in the exam

(i)

9x2 + 24xy + 16y2 = (3x)2 + 2(3x)(4y) + (4y)2 [a2 + 2ab + b2 = (a + b)2]

∴ 9x2 + 24xy + 16y2 = (3x + 4y)2

(ii)

4s2 + 20st + 25t2 = (2s)2 + 2(2s)(5t) + (5t)2 [a2 + 2ab + b2 = (a + b)2]

∴ 4s2 + 20st + 25t2 = (2s + 5t)2

(iii)

49x2 + 28xy + 4y2 = (7x)2 + 2(7x)(2y) + (2y)2 [a2 + 2ab + b2 = (a + b)2]

∴ 49x2 + 28xy + 4y2 = (7x + 2y)2

(iv)

64p2 + 323pq + 49q2 = (8p)2 + 2(8p)(23q) + (23q)2 [a2 + 2ab + b2 = (a + b)2]

∴ 64p2 + 323pq + 49q2 = (8p + 23q)2

*(v)

3a2 + 4ab + 43b2 = 3(a2 + 43ab + 49b2)

= 3[a2 + 2(a)(23b) + (23b)2] [a2 + 2ab + b2 = (a + b)2]

∴ 3a2 + 4ab + 43b2 = 3(a + 23b)2

*(vi)

95s2 + 6sv + 5v2 = 5(925s2 + 65sv + v2)

= 5[(35s)2 + 2(35s)(v) + v2] [a2 + 2ab + b2 = (a + b)2]

∴ 95s2 + 6sv + 5v2 = 5(35s + v)2

Common mistakes that cost marks

  • Not checking the middle term. 9x2 + 20xy + 16y2 has square end terms too, but 2(3x)(4y) = 24xy, not 20xy, so it is not (3x + 4y)2.
  • In (v), dividing only some terms by 3 when taking 3 out. Every term must be divided: 4ab becomes 43ab and 43b2 becomes 49b2.
  • Dropping the common factor at the end: writing (a + 23b)2 as the answer to (v). The 3 taken out must stay in front: 3(a + 23b)2.

How this can come in the exam

MCQ (1 mark)

25x2 + 30x + 9 can be factored as

  1. (5x + 3)(5x − 3)
  2. (5x + 3)2
  3. (25x + 3)(x + 3)
  4. (5x + 9)2
Show answer

(B) (5x + 3)2
25x2 = (5x)2, 9 = 32 and 2(5x)(3) = 30x, so it is (5x + 3)2.

Short answer (2 marks)

Factor completely: 2x2 + 12xy + 18y2.

Show answerTake out the common factor 2: 2(x2 + 6xy + 9y2) (½ mark). Inside, x2 + 2(x)(3y) + (3y)2 = (x + 3y)2 (1 mark). Answer: 2(x + 3y)2 (½ mark).

Try one yourself

Factor completely: 12m2 + 12mn + 3n2.

Show answer

Take out 3: 3(4m2 + 4mn + n2) = 3(2m + n)2, since 2(2m)(n) = 4mn.

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