Topic Review — Algebra: Expanding and Factorising — Solutions

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1. Expanding expressions

   1. 4(3x − 7): 12x − 28
   2. −2x(x + 5): −2x² − 10x
   3. (x + 6)(x − 2): x² + 4x − 12
   4. (3a + 4)(2a − 3): 6a² − 9a + 8a − 12 = 6a² − a − 12
   5. (y + 7)²: y² + 14y + 49
   6. (5m − 3)(5m + 3): 25m² − 9

2. Factorising expressions

   1. 10x + 15: 5(2x + 3)
   2. 6a²b − 4ab²: 2ab(3a − 2b)
   3. x² + 6x + 9: (x + 3)²
   4. x² − 5x − 14: (x − 7)(x + 2)   [(−7)×2=−14, −7+2=−5]
   5. 4p² − 25: (2p + 5)(2p − 5)
   6. 3x² − 12x + 12: 3(x² − 4x + 4) = 3(x − 2)²

3. Mixed expanding and factorising

   1. Expand then verify (2x − 5)²: 4x² − 20x + 25. Factorises back: 4x²−20x+25 = (2x−5)² ✓
   2. 2x² + 14x + 24: 2(x² + 7x + 12) = 2(x + 3)(x + 4)
   3. 3xy + 6x − 4y − 8 grouping: 3x(y + 2) − 4(y + 2) = (3x − 4)(y + 2)
   4. x² − 16 vs (x−4)²: No. (x−4)² = x²−8x+16, not x²−16. The correct factorisation is x²−16 = (x+4)(x−4).
   5. Compare when x = 10: (x+3)(x−3) = x²−9 = 91; (x−3)² = x²−6x+9 = 49. So (x+3)(x−3) is larger. In general, x²−9 > x²−6x+9 when 6x > 18, i.e. x > 3.

4. Solving problems with algebra

   1. x² − 7x + 10 = 0: (x − 5)(x − 2) = 0  →  x = 5 or x = 2
   2. Rectangle length: 6x²+7x−3 = (2x+3)(3x−1). Since width = (2x+3), length = (3x−1) cm.
   3. 203 × 197: (200+3)(200−3) = 200²−3² = 40000−9 = 39991
   4. Right triangle: x² + (x+7)² = (x+8)² → x²+x²+14x+49 = x²+16x+64 → x²−2x−15 = 0 → (x−5)(x+3) = 0 → x=5. Legs: 5 cm and 12 cm, hypotenuse: 13 cm. Check: 25+144=169=13² ✓

5. Algebraic reasoning and proof

   1. Shaded area = (x+5)² − (x−1)² = (x+5+x−1)(x+5−x+1) = (2x+4)(6) = 12(x+2) m² ✓
   2. (2n+1)² − 1 = 4n²+4n+1−1 = 4n²+4n = 4n(n+1). Since n and (n+1) are consecutive integers, one is even, so n(n+1) is even. Therefore 4n(n+1) is always divisible by 8. ✓
   3. Alex’s 6(x²−1) is not fully factorised — (x²−1) is a difference of two squares. Bailey correctly applies DOTS. Fully factorised: 6(x+1)(x−1).