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Algebraic identities · 3 marks

Simplify the following:
Note: Assume that the denominators are not equal to 0.

  1. (i) 4x2 + 4x + 14x2 − 1
  2. (ii) 9(3a3 − 24b3)9a2 − 36b2
  3. (iii) s3 + 125t3s2 − 2st − 35t2
Answer: (i) 2x + 12x − 1 (ii) 3(a2 + 2ab + 4b2)a + 2b (iii) s2 − 5st + 25t2s − 7t

Step-by-step solution

Idea: To simplify a fraction of algebraic expressions, factorise the top and the bottom completely using identities, then cancel any factor that appears in both. Cancelling is allowed because the note says the denominator is never 0, so the common factor is not 0.

(i) 4x2 + 4x + 14x2 − 1

  1. Top: 4x2 + 4x + 1 = (2x)2 + 2(2x)(1) + 12 = (2x + 1)2, using a2 + 2ab + b2 = (a + b)2, because the first and last terms are squares and the middle term is 2 × 2x × 1.½ mark
  2. Bottom: 4x2 − 1 = (2x)2 − 12 = (2x + 1)(2x − 1), using a2 − b2 = (a + b)(a − b), a difference of two squares.
  3. (2x + 1)(2x + 1)(2x + 1)(2x − 1). Cancel one (2x + 1) from top and bottom:½ mark
  4. = 2x + 12x − 1
2x + 12x − 1

(ii) 9(3a3 − 24b3)9a2 − 36b2

  1. First take out common number factors. Top: 9(3a3 − 24b3) = 9 × 3(a3 − 8b3) = 27(a3 − 8b3). Bottom: 9a2 − 36b2 = 9(a2 − 4b2).
  2. Top: a3 − 8b3 = a3 − (2b)3 = (a − 2b)(a2 + 2ab + 4b2), using the difference of cubes A3 − B3 = (A − B)(A2 + AB + B2) with A = a, B = 2b.½ mark
  3. Bottom: a2 − 4b2 = (a − 2b)(a + 2b), difference of two squares.
  4. 27(a − 2b)(a2 + 2ab + 4b2)9(a − 2b)(a + 2b). Cancel (a − 2b), and 27 ÷ 9 = 3:½ mark
  5. = 3(a2 + 2ab + 4b2)a + 2b
3(a2 + 2ab + 4b2)a + 2b

(iii) s3 + 125t3s2 − 2st − 35t2

  1. Top: s3 + 125t3 = s3 + (5t)3 = (s + 5t)(s2 − 5st + 25t2), using the sum of cubes a3 + b3 = (a + b)(a2 − ab + b2) with a = s, b = 5t.½ mark
  2. Bottom: s2 − 2st − 35t2 is of the form s2 + (p + q)s + pq, so use (s + p)(s + q). We need two terms that add to −2t and multiply to −35t2: these are −7t and +5t. So s2 − 2st − 35t2 = (s − 7t)(s + 5t).
  3. (s + 5t)(s2 − 5st + 25t2)(s − 7t)(s + 5t). Cancel (s + 5t):½ mark
  4. = s2 − 5st + 25t2s − 7t
s2 − 5st + 25t2s − 7t
(i) (2x + 1)/(2x − 1) (ii) 3(a² + 2ab + 4b²)/(a + 2b) (iii) (s² − 5st + 25t²)/(s − 7t)

Check: (i) x = 1: 4 + 4 + 14 − 1 = 93 = 3 and 31 = 3 ✓. (ii) a = 3, b = 1: 9(81 − 24)81 − 36 = 51345 = 11.4 and 3(9 + 6 + 4)5 = 575 = 11.4 ✓. (iii) s = 8, t = 1: 512 + 12564 − 16 − 35 = 63713 = 49 and 64 − 40 + 251 = 49 ✓.

Answer to write in the exam

(i)

4x2 + 4x + 1 = (2x)2 + 2(2x)(1) + 12 = (2x + 1)2 [a2 + 2ab + b2 = (a + b)2]

4x2 − 1 = (2x)2 − 12 = (2x + 1)(2x − 1) [a2 − b2 = (a + b)(a − b)]

4x2 + 4x + 14x2 − 1 = (2x + 1)(2x + 1)(2x + 1)(2x − 1)

∴ 4x2 + 4x + 14x2 − 1 = 2x + 12x − 1

(ii)

9(3a3 − 24b3) = 27(a3 − 8b3) = 27[a3 − (2b)3] = 27(a − 2b)(a2 + 2ab + 4b2) [A3 − B3 = (A − B)(A2 + AB + B2)]

9a2 − 36b2 = 9(a2 − 4b2) = 9(a − 2b)(a + 2b) [a2 − b2 = (a + b)(a − b)]

9(3a3 − 24b3)9a2 − 36b2 = 27(a − 2b)(a2 + 2ab + 4b2)9(a − 2b)(a + 2b)

∴ 9(3a3 − 24b3)9a2 − 36b2 = 3(a2 + 2ab + 4b2)a + 2b

(iii)

s3 + 125t3 = s3 + (5t)3 = (s + 5t)(s2 − 5st + 25t2) [a3 + b3 = (a + b)(a2 − ab + b2)]

s2 − 2st − 35t2 = s2 + (−7t + 5t)s + (−7t)(5t) = (s − 7t)(s + 5t) [(x + a)(x + b) = x2 + (a + b)x + ab]

s3 + 125t3s2 − 2st − 35t2 = (s + 5t)(s2 − 5st + 25t2)(s − 7t)(s + 5t)

∴ s3 + 125t3s2 − 2st − 35t2 = s2 − 5st + 25t2s − 7t

Common mistakes that cost marks

  • Cancelling terms instead of factors, e.g. crossing out 4x2 from the top and bottom of (i). You may only cancel something that multiplies the whole top and the whole bottom.
  • Using the wrong middle sign in the cube identities: a3 − 8b3 has +2ab in its second bracket, while s3 + 125t3 has −5st.
  • In (ii), forgetting the 9 outside the bracket on top, which gives the wrong number in front (1 instead of 3).

How this can come in the exam

MCQ (1 mark)

x2 − 16x2 − 8x + 16, where the denominator is not 0, simplifies to

  1. x − 4x + 4
  2. x + 4x − 4
  3. −1
  4. 2x
Show answer

(B) x + 4x − 4
Top = (x − 4)(x + 4); bottom = (x − 4)2. Cancel one (x − 4) to get x + 4x − 4. (Cancelling terms such as x2 or 16 instead of factors is the mistake behind the other options.)

Short answer (2 marks)

Simplify a3 − b3a2 − b2, assuming the denominator is not 0.

Show answerTop: (a − b)(a2 + ab + b2) (½ mark). Bottom: (a − b)(a + b) (½ mark). Cancel (a − b): a2 + ab + b2a + b (1 mark).

Try one yourself

Simplify 8x3 + 274x2 − 9 (assume the denominator is not 0).

Show answer

Top = (2x + 3)(4x2 − 6x + 9), bottom = (2x + 3)(2x − 3). Answer: 4x2 − 6x + 92x − 3.

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