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Simplifying rational expressions · 5 marks

Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero:

  1. (i) 3p2 − 3pq − 18q2p2 + 3pq − 10q2
  2. (ii) n3 − 3n2m + 3nm2 − m35m2 − 10mn + 5n2
  3. (iii) w3 − v3 + x3 + 3wvxw2 + v2 + x2 − 2wv − 2vx + 2wx
  4. (iv) 4y2 − 20yz + 25z2(25z2 − 4y2)
  5. (v) (x2 + x − 6)(x2 − 7x + 12)(x2 − 6x + 8)(x2 − 9)
  6. (vi) p4 − 16p2 − 4p + 4
Answer: (i) 3(p − 3q)(p + 2q)(p + 5q)(p − 2q) (no common factor, so it does not reduce) (ii) n − m5 (iii) w2 + v2 + x2 + wv + vx − wxw − v + x (iv) 5z − 2y5z + 2y (v) 1 (vi) (p + 2)(p2 + 4)p − 2

Step-by-step solution

Idea: To simplify a fraction of expressions, factor the top and the bottom completely, then cancel any factor that appears in both (this is allowed because the denominator is not zero). Identities used: a2 − b2 = (a + b)(a − b), (a − b)2, (a − b)3 = a3 − 3a2b + 3ab2 − b3, (a + b + c)2, a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca), and splitting the middle term for quadratics.

(i) 3p2 − 3pq − 18q2p2 + 3pq − 10q2

  1. Numerator: take out 3, then find two numbers with product −6 and sum −1 (−3 and 2):
    3p2 − 3pq − 18q2 = 3(p2 − pq − 6q2) = 3(p − 3q)(p + 2q)½ mark
  2. Denominator: two numbers with product −10 and sum 3 (5 and −2):
    p2 + 3pq − 10q2 = (p + 5q)(p − 2q)
  3. So the expression = 3(p − 3q)(p + 2q)(p + 5q)(p − 2q).
  4. Note: the top has factors 3, (p − 3q), (p + 2q) and the bottom has (p + 5q), (p − 2q). No factor is common (note (p + 2q) and (p − 2q) are different), so nothing cancels. The factored form above is the simplest form. Because every other part cancels neatly, this part may contain a printing error, but as printed this is the answer.
3(p − 3q)(p + 2q)(p + 5q)(p − 2q) (cannot be reduced further)

(ii) n3 − 3n2m + 3nm2 − m35m2 − 10mn + 5n2

  1. Numerator matches (a − b)3 = a3 − 3a2b + 3ab2 − b3 with a = n, b = m: it is (n − m)3.½ mark
  2. Denominator: 5(m2 − 2mn + n2) = 5(m − n)2 = 5(n − m)2 (a square is the same whichever way round you subtract).½ mark
  3. (n − m)35(n − m)2 = n − m5
n − m5

(iii) w3 − v3 + x3 + 3wvxw2 + v2 + x2 − 2wv − 2vx + 2wx

  1. Denominator: by (a + b + c)2 with a = w, b = −v, c = x: 2ab = −2wv, 2bc = −2vx, 2ca = 2wx ✓. So the denominator is (w − v + x)2.½ mark
  2. Numerator: with a = w, b = −v, c = x, we have a3 + b3 + c3 − 3abc = w3 − v3 + x3 − 3(w)(−v)(x) = w3 − v3 + x3 + 3wvx ✓.
  3. So the numerator = (a + b + c)(a2 + b2 + c2 − ab − bc − ca) = (w − v + x)(w2 + v2 + x2 + wv + vx − wx).
    (−ab = −w(−v) = +wv; −bc = −(−v)x = +vx; −ca = −wx.)½ mark
  4. Cancel one factor (w − v + x):
    = w2 + v2 + x2 + wv + vx − wxw − v + x
w2 + v2 + x2 + wv + vx − wxw − v + x

(iv) 4y2 − 20yz + 25z2(25z2 − 4y2)

  1. Numerator: (2y)2 − 2(2y)(5z) + (5z)2 = (2y − 5z)2 = (5z − 2y)2.½ mark
  2. Denominator: (5z)2 − (2y)2 = (5z − 2y)(5z + 2y).
  3. (5z − 2y)2(5z − 2y)(5z + 2y) = 5z − 2y5z + 2y
5z − 2y5z + 2y

(v) (x2 + x − 6)(x2 − 7x + 12)(x2 − 6x + 8)(x2 − 9)

  1. Factor each bracket:
    x2 + x − 6 = (x + 3)(x − 2)
    x2 − 7x + 12 = (x − 3)(x − 4)
    x2 − 6x + 8 = (x − 2)(x − 4)
    x2 − 9 = (x − 3)(x + 3)1 mark
  2. (x + 3)(x − 2)(x − 3)(x − 4)(x − 2)(x − 4)(x − 3)(x + 3). Every factor on top also appears on the bottom, so all cancel: = 1
1

(vi) p4 − 16p2 − 4p + 4

  1. Numerator: use a2 − b2 twice. p4 − 16 = (p2)2 − 42 = (p2 − 4)(p2 + 4) = (p − 2)(p + 2)(p2 + 4).½ mark
  2. Denominator: p2 − 4p + 4 = (p − 2)2.½ mark
  3. Cancel one (p − 2): = (p + 2)(p2 + 4)p − 2
(p + 2)(p2 + 4)p − 2
(i) 3(p − 3q)(p + 2q) / [(p + 5q)(p − 2q)] — no common factor, so this factored form is the simplest form (ii) (n − m)/5 (iii) (w² + v² + x² + wv + vx − wx)/(w − v + x) (iv) (5z − 2y)/(5z + 2y) (v) 1 (vi) (p + 2)(p² + 4)/(p − 2)

Check: Substitute numbers. (ii) n = 3, m = 1: top = 27 − 27 + 9 − 1 = 8, bottom = 5 − 30 + 45 = 20, 8 ÷ 20 = 25 and (3 − 1) ÷ 5 = 25 ✓. (vi) p = 3: (81 − 16) ÷ (9 − 12 + 4) = 65 and (5 × 13) ÷ 1 = 65 ✓. (iii) w = 2, v = 1, x = 1: top = 8 − 1 + 1 + 6 = 14, bottom = 4 + 1 + 1 − 4 − 2 + 4 = 4, 14 ÷ 4 = 72; answer = (4 + 1 + 1 + 2 + 1 − 2) ÷ 2 = 72 ✓.

Answer to write in the exam

(i)

3p2 − 3pq − 18q2 = 3(p2 − 3pq + 2pq − 6q2) = 3(p − 3q)(p + 2q)

p2 + 3pq − 10q2 = p2 + 5pq − 2pq − 10q2 = (p + 5q)(p − 2q)

∴ 3p2 − 3pq − 18q2p2 + 3pq − 10q2 = 3(p − 3q)(p + 2q)(p + 5q)(p − 2q) (no common factor)

(ii)

n3 − 3n2m + 3nm2 − m3 = (n − m)3 [a3 − 3a2b + 3ab2 − b3 = (a − b)3]

5m2 − 10mn + 5n2 = 5(m − n)2 = 5(n − m)2

Given expression = (n − m)35(n − m)2

∴ n3 − 3n2m + 3nm2 − m35m2 − 10mn + 5n2 = n − m5

(iii)

Numerator = w3 + (−v)3 + x3 − 3(w)(−v)(x)

= (w − v + x)(w2 + v2 + x2 + wv + vx − wx) [a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca)]

Denominator = w2 + (−v)2 + x2 + 2(w)(−v) + 2(−v)(x) + 2(x)(w) = (w − v + x)2 [(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]

∴ w3 − v3 + x3 + 3wvxw2 + v2 + x2 − 2wv − 2vx + 2wx = w2 + v2 + x2 + wv + vx − wxw − v + x

(iv)

Numerator = (2y)2 − 2(2y)(5z) + (5z)2 = (2y − 5z)2 = (5z − 2y)2 [a2 − 2ab + b2 = (a − b)2]

Denominator = (5z)2 − (2y)2 = (5z − 2y)(5z + 2y) [a2 − b2 = (a + b)(a − b)]

∴ 4y2 − 20yz + 25z2(25z2 − 4y2) = 5z − 2y5z + 2y

(v)

x2 + x − 6 = (x + 3)(x − 2)

x2 − 7x + 12 = (x − 3)(x − 4)

x2 − 6x + 8 = (x − 2)(x − 4)

x2 − 9 = (x − 3)(x + 3)

Given expression = (x + 3)(x − 2)(x − 3)(x − 4)(x − 2)(x − 4)(x − 3)(x + 3)

∴ Given expression = 1

(vi)

p4 − 16 = (p2)2 − 42 = (p2 − 4)(p2 + 4) = (p − 2)(p + 2)(p2 + 4) [a2 − b2 = (a + b)(a − b)]

p2 − 4p + 4 = (p − 2)2 [a2 − 2ab + b2 = (a − b)2]

∴ p4 − 16p2 − 4p + 4 = (p + 2)(p2 + 4)p − 2

Common mistakes that cost marks

  • Cancelling terms instead of factors, e.g. crossing out p2 from top and bottom in (i). Only whole factors (things multiplied) can be cancelled, never terms that are added or subtracted.
  • In (iv), thinking (2y − 5z) and (5z − 2y) cannot cancel. Inside a square they are equal: (2y − 5z)2 = (5z − 2y)2.
  • In (i), treating (p + 2q) and (p − 2q) as the same factor and cancelling them. They differ by a sign inside, so they do not cancel.

How this can come in the exam

MCQ (1 mark)

x2 − 9x2 + 6x + 9, where x ≠ −3, simplifies to

  1. x + 3x − 3
  2. −1
  3. x − 3x + 3
  4. 16x
Show answer

(C) x − 3x + 3
(x − 3)(x + 3)(x + 3)2 = x − 3x + 3.

Short answer (2 marks)

Simplify a2 − 4a2 + a − 6, assuming the denominator is not zero.

Show answerTop: (a − 2)(a + 2) (½ mark). Bottom: (a + 3)(a − 2) (½ mark). Cancel (a − 2): a + 2a + 3 (1 mark).

Try one yourself

Simplify x3 − 8x2 − 4, assuming the denominator is not zero.

Show answer

x3 − 8 = (x − 2)(x2 + 2x + 4) and x2 − 4 = (x − 2)(x + 2). Answer: x2 + 2x + 4x + 2.

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