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Algebraic identities · 5 marks

Use suitable identities to find the following products:

  1. (i) (−3x + 4)2
  2. (ii) (2s + 7) (2s − 7)
  3. (iii) (p2 + 12)(p2 − 12)
  4. (iv) (2n + 7) (2n − 7)
  5. (v) (s − 2t) (s2 + 2st + 4t2)
  6. (vi) (12r − 4r)2
  7. (vii) (−3m + 4k − l)2
  8. (viii) (x − 13y)3
  9. (ix) (72k − 23m)3
Answer: (i) 9x2 − 24x + 16 (ii) 4s2 − 49 (iii) p4 − 14 (iv) 4n2 − 49 (v) s3 − 8t3 (vi) 14r2 − 4 + 16r2 (vii) 9m2 + 16k2 + l2 − 24mk − 8kl + 6ml (viii) x3 − x2y + 13xy2 − 127y3 (ix) 3438k3 − 492k2m + 143km2 − 827m3

Step-by-step solution

Idea: First look at the shape of each product and pick the identity that matches it: a square of two terms → (a ± b)2; (something + something)(same − same) → (a + b)(a − b) = a2 − b2; (a − b)(a2 + ab + b2) → a3 − b3; a square of three terms → (a + b + c)2; a cube of two terms → (a − b)3. Then put the actual terms (with their signs) in place of a, b, c.

(i) (−3x + 4)2

  1. Identity: (a + b)2 = a2 + 2ab + b2, because this is the square of a sum of two terms. Take a = −3x and b = 4 (keep the minus sign with 3x).
  2. (−3x + 4)2 = (−3x)2 + 2(−3x)(4) + 42½ mark
  3. (−3x)2 = 9x2 (minus × minus = plus), and 2 × (−3x) × 4 = −24x.
  4. = 9x2 − 24x + 16
9x2 − 24x + 16

(ii) (2s + 7) (2s − 7)

  1. Identity: (a + b)(a − b) = a2 − b2, because the two brackets have the same terms, one with + and one with −. Here a = 2s, b = 7.
  2. (2s + 7)(2s − 7) = (2s)2 − 72½ mark
  3. = 4s2 − 49
4s2 − 49

(iii) (p2 + 12)(p2 − 12)

  1. Identity: (a + b)(a − b) = a2 − b2, again a sum times a difference of the same two terms. Here a = p2, b = 12.
  2. = (p2)2 − (12)2½ mark
  3. (p2)2 = p4 (multiply the powers: 2 × 2 = 4), and (12)2 = 14.
  4. = p4 − 14
p4 − 14

(iv) (2n + 7) (2n − 7)

  1. Identity: (a + b)(a − b) = a2 − b2 with a = 2n, b = 7.
  2. = (2n)2 − 72½ mark
  3. = 4n2 − 49
4n2 − 49

(v) (s − 2t) (s2 + 2st + 4t2)

  1. Identity: (a − b)(a2 + ab + b2) = a3 − b3. Check that it fits: with a = s and b = 2t, we get a2 = s2, ab = 2st, b2 = 4t2. That is exactly the second bracket.
  2. So the product = s3 − (2t)3½ mark
  3. (2t)3 = 2 × 2 × 2 × t3 = 8t3, so the answer is s3 − 8t3.
s3 − 8t3

(vi) (12r − 4r)2

  1. Identity: (a − b)2 = a2 − 2ab + b2, because it is the square of a difference. Here a = 12r, b = 4r.
  2. = (12r)2 − 2(12r)(4r) + (4r)2½ mark
  3. (12r)2 = 14r2. Middle term: 2 × 12r × 4r = 8r2r = 4 (the r cancels). (4r)2 = 16r2.
  4. = 14r2 − 4 + 16r2
14r2 − 4 + 16r2

(vii) (−3m + 4k − l)2

  1. Identity: (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca, because there are three terms inside the square. Take each term with its own sign: a = −3m, b = 4k, c = −l.
  2. = (−3m)2 + (4k)2 + (−l)2 + 2(−3m)(4k) + 2(4k)(−l) + 2(−l)(−3m)½ mark
  3. Squares are always positive: 9m2, 16k2, l2. Products: 2 × (−3) × 4 = −24, so −24mk; 2 × 4 × (−1) = −8, so −8kl; 2 × (−1) × (−3) = +6, so +6ml.
  4. = 9m2 + 16k2 + l2 − 24mk − 8kl + 6ml
9m2 + 16k2 + l2 − 24mk − 8kl + 6ml

(viii) (x − 13y)3

  1. Identity: (a − b)3 = a3 − 3a2b + 3ab2 − b3, because it is the cube of a difference of two terms. Here a = x, b = 13y.
  2. = x3 − 3(x2)(13y) + 3(x)(13y)2 − (13y)3½ mark
  3. 3 × 13 = 1, so the second term is x2y. (13y)2 = 19y2 and 3 × 19 = 13. (13y)3 = 127y3.
  4. = x3 − x2y + 13xy2 − 127y3
x3 − x2y + 13xy2 − 127y3

(ix) (72k − 23m)3

  1. Identity: (a − b)3 = a3 − 3a2b + 3ab2 − b3, the cube of a difference. Here a = 72k, b = 23m.
  2. = (72k)3 − 3(72k)2(23m) + 3(72k)(23m)2 − (23m)31 mark
  3. Work out each number separately:
    (72)3 = 3438
    3 × 494 × 23 = 29412 = 492
    3 × 72 × 49 = 8418 = 143
    (23)3 = 827
  4. = 3438k3 − 492k2m + 143km2 − 827m3
3438k3 − 492k2m + 143km2 − 827m3
(i) 9x² − 24x + 16 (ii) 4s² − 49 (iii) p⁴ − 1/4 (iv) 4n² − 49 (v) s³ − 8t³ (vi) 1/(4r²) − 4 + 16r² (vii) 9m² + 16k² + l² − 24mk − 8kl + 6ml (viii) x³ − x²y + (1/3)xy² − (1/27)y³ (ix) (343/8)k³ − (49/2)k²m + (14/3)km² − (8/27)m³

Check: Put x = 1 in (i): (−3 + 4)2 = 1 and 9 − 24 + 16 = 1 ✓. Put s = 3, t = 1 in (v): (3 − 2)(9 + 6 + 4) = 19 and 27 − 8 = 19 ✓. Put m = k = l = 1 in (vii): (−3 + 4 − 1)2 = 0 and 9 + 16 + 1 − 24 − 8 + 6 = 0 ✓. Put x = 1, y = 3 in (viii): (1 − 1)3 = 0 and 1 − 3 + 3 − 1 = 0 ✓.

Answer to write in the exam

(i)

(−3x + 4)2 = (−3x)2 + 2(−3x)(4) + 42 [(a + b)2 = a2 + 2ab + b2]

∴ (−3x + 4)2 = 9x2 − 24x + 16

(ii)

(2s + 7)(2s − 7) = (2s)2 − 72 [(a + b)(a − b) = a2 − b2]

∴ (2s + 7)(2s − 7) = 4s2 − 49

(iii)

(p2 + 12)(p2 − 12) = (p2)2 − (12)2 [(a + b)(a − b) = a2 − b2]

∴ (p2 + 12)(p2 − 12) = p4 − 14

(iv)

(2n + 7)(2n − 7) = (2n)2 − 72 [(a + b)(a − b) = a2 − b2]

∴ (2n + 7)(2n − 7) = 4n2 − 49

(v)

(s − 2t)(s2 + 2st + 4t2) = (s − 2t)[s2 + s(2t) + (2t)2]

= s3 − (2t)3 [(a − b)(a2 + ab + b2) = a3 − b3]

∴ (s − 2t)(s2 + 2st + 4t2) = s3 − 8t3

(vi)

(12r − 4r)2 = (12r)2 − 2(12r)(4r) + (4r)2 [(a − b)2 = a2 − 2ab + b2]

∴ (12r − 4r)2 = 14r2 − 4 + 16r2

(vii)

(−3m + 4k − l)2 = (−3m)2 + (4k)2 + (−l)2 + 2(−3m)(4k) + 2(4k)(−l) + 2(−l)(−3m) [(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]

∴ (−3m + 4k − l)2 = 9m2 + 16k2 + l2 − 24mk − 8kl + 6ml

(viii)

(x − 13y)3 = x3 − 3(x2)(13y) + 3(x)(13y)2 − (13y)3 [(a − b)3 = a3 − 3a2b + 3ab2 − b3]

∴ (x − 13y)3 = x3 − x2y + 13xy2 − 127y3

(ix)

(72k − 23m)3 = (72k)3 − 3(72k)2(23m) + 3(72k)(23m)2 − (23m)3 [(a − b)3 = a3 − 3a2b + 3ab2 − b3]

= 3438k3 − 3 × 494 × 23k2m + 3 × 72 × 49km2 − 827m3

∴ (72k − 23m)3 = 3438k3 − 492k2m + 143km2 − 827m3

Common mistakes that cost marks

  • Dropping the minus sign that belongs to a term. In (vii), c = −l, so 2bc = 2(4k)(−l) = −8kl, not +8kl; and 2ca = 2(−l)(−3m) = +6ml.
  • Cubing only the letter: writing (72k)3 = 72k3 or 216k3. The fraction is cubed too: 73 = 343 and 23 = 8.
  • In (v), multiplying out all six terms the long way and making a sign slip, instead of spotting that the second bracket is a2 + ab + b2 with a = s, b = 2t, which gives s3 − 8t3 in one line.

How this can come in the exam

MCQ (1 mark)

The coefficient of xy2 in the expansion of (x − 2y)3 is

  1. 6
  2. 12
  3. −12
  4. 8
Show answer

(B) 12
(a − b)3 has the term +3ab2. With a = x, b = 2y: 3 × x × 4y2 = 12xy2.

Assertion–Reason (1 mark)

Assertion (A): (2s + 7)(2s − 7) = 4s2 − 49.
Reason (R): (a + b)(a − b) = a2 − b2 for all values of a and b.

  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.
R is a standard identity. Putting a = 2s, b = 7 in R gives exactly A.

Try one yourself

Use a suitable identity to find (2a − 3b)3.

Show answer

(a − b)3 with 2a and 3b: (2a)3 − 3(2a)2(3b) + 3(2a)(3b)2 − (3b)3 = 8a3 − 36a2b + 54ab2 − 27b3.

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