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Algebraic identities · 4 marks

Expand the following using the identity
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca:

  1. (i) (p + 3q + 7r)2
  2. (ii) (3x − 2y + 4z)2
Answer: (i) p2 + 9q2 + 49r2 + 6pq + 42qr + 14pr (ii) 9x2 + 4y2 + 16z2 − 12xy − 16yz + 24xz

Step-by-step solution

Idea: Name the three terms a, b, c, keeping each term’s sign with it. In (ii), b = −2y, not 2y. Then write the three squares (always positive) and the three cross products 2ab, 2bc, 2ca (their signs come out by themselves).

(i) (p + 3q + 7r)2

  1. Here a = p, b = 3q, c = 7r.
  2. = p2 + (3q)2 + (7r)2 + 2(p)(3q) + 2(3q)(7r) + 2(7r)(p)1 mark
  3. = p2 + 9q2 + 49r2 + 6pq + 42qr + 14pr1 mark
p2 + 9q2 + 49r2 + 6pq + 42qr + 14pr

(ii) (3x − 2y + 4z)2

  1. Write it as (3x + (−2y) + 4z)2. So a = 3x, b = −2y, c = 4z.
  2. = (3x)2 + (−2y)2 + (4z)2 + 2(3x)(−2y) + 2(−2y)(4z) + 2(4z)(3x)1 mark
  3. (−2y)2 = +4y2 (a negative times a negative is positive). The two products that contain −2y come out negative.
  4. = 9x2 + 4y2 + 16z2 − 12xy − 16yz + 24xz1 mark
9x2 + 4y2 + 16z2 − 12xy − 16yz + 24xz
(i) p² + 9q² + 49r² + 6pq + 42qr + 14pr (ii) 9x² + 4y² + 16z² − 12xy − 16yz + 24xz

Check: Put every letter equal to 1. (i) (1 + 3 + 7)2 = 121 and 1 + 9 + 49 + 6 + 42 + 14 = 121 ✓. (ii) (3 − 2 + 4)2 = 25 and 9 + 4 + 16 − 12 − 16 + 24 = 25 ✓.

Answer to write in the exam

(i)

(p + 3q + 7r)2 = p2 + (3q)2 + (7r)2 + 2(p)(3q) + 2(3q)(7r) + 2(7r)(p) [(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]

∴ (p + 3q + 7r)2 = p2 + 9q2 + 49r2 + 6pq + 42qr + 14pr

(ii)

(3x − 2y + 4z)2 = (3x)2 + (−2y)2 + (4z)2 + 2(3x)(−2y) + 2(−2y)(4z) + 2(4z)(3x) [(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]

∴ (3x − 2y + 4z)2 = 9x2 + 4y2 + 16z2 − 12xy − 16yz + 24xz

Common mistakes that cost marks

  • Writing −4y2 for (−2y)2. A square is never negative: (−2y)2 = +4y2.
  • Making all cross terms negative in (ii) just because there is a minus sign inside. Only the products that include −2y are negative; 2(4z)(3x) = +24xz.
  • Forgetting the factor 2 in the cross terms, e.g. writing 3pq instead of 6pq.

How this can come in the exam

MCQ (1 mark)

The coefficient of yz in the expansion of (x − 3y + 2z)2 is

  1. 12
  2. −12
  3. −6
  4. 6
Show answer

(B) −12
2bc = 2(−3y)(2z) = −12yz.

Assertion–Reason (1 mark)

Assertion (A): (a − b − c)2 = a2 + b2 + c2 − 2ab + 2bc − 2ca.
Reason (R): (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca holds for all values, including negative ones.

  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 −b for b and −c for c in R: 2ab becomes −2ab, 2bc becomes 2(−b)(−c) = +2bc, 2ca becomes −2ca. This gives A.

Try one yourself

Expand (2a + b − 3c)2.

Show answer

4a2 + b2 + 9c2 + 4ab − 6bc − 12ca (with a → 2a, b → b, c → −3c).

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