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Factorisation using identities · 5 marks

Factor the following:

  1. (i) 9a2 + b2 + 4c2 − 6ab + 12ac − 4bc
  2. (ii) 16s2 + 25t2 − 40st
  3. (iii) r2 − r − 42
  4. (iv) 49g2 + 14gh + h2
  5. (v) 64u2 + 121v2 + 4w2 − 176uv − 32uw + 44vw
Answer: (i) (3a − b + 2c)2 (ii) (4s − 5t)2 (iii) (r − 7)(r + 6) (iv) (7g + h)2 (v) (8u − 11v − 2w)2

Step-by-step solution

Idea: First decide which identity fits. Six terms with three perfect squares → (a + b + c)2. Three terms with two perfect squares → (a ± b)2. A quadratic like r2 − r − 42 whose constant is not a perfect square → (x + a)(x + b). For the six-term ones, the signs of the cross terms tell you which square root takes a minus sign.

(i) 9a2 + b2 + 4c2 − 6ab + 12ac − 4bc

  1. Square roots of the square terms: 3a, b, 2c. Use (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx.
  2. The ab and bc terms are negative, the ac term is positive. The letter shared by the two negative terms is b, so take x = 3a, y = −b, z = 2c.
  3. Check: 2(3a)(−b) = −6ab ✓; 2(−b)(2c) = −4bc ✓; 2(2c)(3a) = 12ac ✓½ mark
  4. = (3a − b + 2c)2½ mark
(3a − b + 2c)2

(ii) 16s2 + 25t2 − 40st

  1. Rearrange in the usual order: 16s2 − 40st + 25t2. Here 16s2 = (4s)2, 25t2 = (5t)2, and the middle term is negative, so use (a − b)2.
  2. Check: 2(4s)(5t) = 40st ✓½ mark
  3. = (4s − 5t)2½ mark
(4s − 5t)2

(iii) r2 − r − 42

  1. 42 is not a perfect square, so this is not (a ± b)2. Use (r + a)(r + b) = r2 + (a + b)r + ab with a + b = −1 and ab = −42.
  2. The product is negative, so one number is positive and one negative; the sum is −1, so the negative one is bigger. 6 × 7 = 42 and −7 + 6 = −1 ✓. So a = −7, b = 6.½ mark
  3. = (r − 7)(r + 6)½ mark
(r − 7)(r + 6)

(iv) 49g2 + 14gh + h2

  1. 49g2 = (7g)2 and h2 = (h)2. Use (a + b)2 with a = 7g, b = h.
  2. Check: 2(7g)(h) = 14gh ✓½ mark
  3. = (7g + h)2½ mark
(7g + h)2

(v) 64u2 + 121v2 + 4w2 − 176uv − 32uw + 44vw

  1. Square roots: 64u2 = (8u)2, 121v2 = (11v)2, 4w2 = (2w)2.
  2. Signs: uv and uw are negative, vw is positive. So v and w have the same sign and u has the opposite sign. Take 8u, −11v, −2w.
  3. Check: 2(8u)(−11v) = −176uv ✓; 2(8u)(−2w) = −32uw ✓; 2(−11v)(−2w) = +44vw ✓½ mark
  4. = (8u − 11v − 2w)2 (the same as (11v + 2w − 8u)2, since a square does not change when every sign is flipped)½ mark
(8u − 11v − 2w)2
(i) (3a − b + 2c)² (ii) (4s − 5t)² (iii) (r − 7)(r + 6) (iv) (7g + h)² (v) (8u − 11v − 2w)²

Check: Put every letter equal to 1. (i) 9 + 1 + 4 − 6 + 12 − 4 = 16 and (3 − 1 + 2)2 = 16 ✓. (v) 64 + 121 + 4 − 176 − 32 + 44 = 25 and (8 − 11 − 2)2 = 25 ✓. (iii) 1 − 1 − 42 = −42 and (−6)(7) = −42 ✓.

Answer to write in the exam

(i)

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

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

(ii)

16s2 + 25t2 − 40st = (4s)2 − 2(4s)(5t) + (5t)2 [a2 − 2ab + b2 = (a − b)2]

∴ 16s2 + 25t2 − 40st = (4s − 5t)2

(iii)

r2 − r − 42 = r2 + (−7 + 6)r + (−7)(6) [(x + a)(x + b) = x2 + (a + b)x + ab]

∴ r2 − r − 42 = (r − 7)(r + 6)

(iv)

49g2 + 14gh + h2 = (7g)2 + 2(7g)(h) + h2 [a2 + 2ab + b2 = (a + b)2]

∴ 49g2 + 14gh + h2 = (7g + h)2

(v)

64u2 + 121v2 + 4w2 − 176uv − 32uw + 44vw = (8u)2 + (−11v)2 + (−2w)2 + 2(8u)(−11v) + 2(−11v)(−2w) + 2(−2w)(8u) [(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]

∴ 64u2 + 121v2 + 4w2 − 176uv − 32uw + 44vw = (8u − 11v − 2w)2

Common mistakes that cost marks

  • In (iii), writing (r + 7)(r − 6). That expands to r2 + r − 42: the middle sign is wrong. The bigger number must carry the sign of the middle term (−).
  • In (v), putting the minus sign only on 11v: (8u − 11v + 2w)2 would give +32uw and −44vw, the opposite of what is printed.
  • In (ii), not noticing the terms are out of the usual order and trying (4s + 5t)2. The −40st is the middle term, so it must be (4s − 5t)2.

How this can come in the exam

MCQ (1 mark)

The factors of r2 + 2r − 48 are

  1. (r − 8)(r + 6)
  2. (r + 12)(r − 4)
  3. (r + 8)(r − 6)
  4. (r + 16)(r − 3)
Show answer

(C) (r + 8)(r − 6)
We need product −48 and sum +2: 8 and −6. So (r + 8)(r − 6).

Assertion–Reason (1 mark)

Assertion (A): 4x2 + 9y2 + z2 − 12xy − 6yz + 4xz = (2x − 3y + z)2.
Reason (R): a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2.

  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.
Show answer

(A) Both A and R are true, and R is the correct explanation of A.
Put a = 2x, b = −3y, c = z in R: 2ab = −12xy, 2bc = −6yz, 2ca = 4xz. This gives exactly A.

Try one yourself

Factor: x2 + 4y2 + 9z2 − 4xy + 12yz − 6xz.

Show answer

Square roots x, 2y, 3z. The xy and xz terms are negative and yz is positive, so x has the opposite sign to the others: (x − 2y − 3z)2.

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