Solving and Modelling with Cubics — Solutions

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1. Null factor law. Fluency

   1. (a) (x−3)(x+1)(x+4)=0 x=3, x=−1, x=−4
   2. (b) x(2x−1)(x+5)=0 x=0, x=1/2, x=−5
   3. (c) (x+2)²(x−3)=0 x=−2 (double), x=3
   4. (d) −2(x−1)(x−2)(x+6)=0 x=1, x=2, x=−6

2. Factor theorem and factorising. Fluency

   1. (a) x³−6x+4=0, root x=2. Verify: 8−12+4=0 ✓. Divide: (x−2)(x²+2x−2)=0.
      x²+2x−2=0 → x=(−2±√12)/2 = −1±√3.
      Solutions: x=2, x=−1+√3, x=−1−√3.
   2. (b) x³+x²−4x−4=0, root x=2. Verify: 8+4−8−4=0 ✓. Divide: (x−2)(x²+3x+2) = (x−2)(x+1)(x+2)=0.
      x=2, x=−1, x=−2.
   3. (c) x³−7x+6=0, root x=1. Verify: 1−7+6=0 ✓. Divide: (x−1)(x²+x−6) = (x−1)(x+3)(x−2)=0.
      x=1, x=−3, x=2.

3. Finding a rational root first. Fluency

   1. (a) x³−2x²−x+2=0 Try x=1: 1−2−1+2=0 ✓. (x−1)(x²−x−2) = (x−1)(x−2)(x+1)=0.
      x=1, x=2, x=−1.
   2. (b) x³+3x²−4=0 Try x=1: 1+3−4=0 ✓. (x−1)(x²+4x+4) = (x−1)(x+2)²=0.
      x=1, x=−2 (double).
   3. (c) 2x³−3x²−11x+6=0 Try x=3: 54−27−33+6=0 ✓. (x−3)(2x²+3x−2) = (x−3)(2x−1)(x+2)=0.
      x=3, x=1/2, x=−2.

4. Graphical roots. Fluency

   1. (a) x-ints: −3, 1, 2. Solutions: x=−3, x=1, x=2. Read directly from factored form.
   2. (b) y=x³−9x=x(x²−9)=x(x−3)(x+3). Roots: x=0, x=3, x=−3.
   3. (c) y=−x³+4x²−4x=−x(x²−4x+4)=−x(x−2)². Roots: x=0, x=2 (double).

5. Mixed solving. Fluency

   1. (a) x³=8 x=2
   2. (b) 2x³=16 → x³=8 x=2
   3. (c) (x−2)³=27 x−2=3 → x=5
   4. (d) −(x+1)³+5=−3 −(x+1)³=−8 → (x+1)³=8 → x+1=2 → x=1

6. Open box problem. Understanding

   1. (a) V = x(30−2x)² = x(900−120x+4x²) = 4x³−120x²+900x
   2. (b) Practical domain. 0 < x < 15 (sides must be positive).
   3. (c) V=1000: 4x³−120x²+900x=1000 → 4x³−120x²+900x−1000=0 → x³−30x²+225x−250=0. Try x=5: 125−750+1125−250=250≠0. Try x=1: 1−30+225−250=−54≠0. Try x=2: 8−120+450−250=88≠0. Use technology: approximate roots in (0,15).
   4. (d) Maximum volume. Using technology (or calculus later): V is maximised at x = 5 cm, giving V = 5(20)² = 2000 cm³.

7. Revenue and profit. Understanding

   1. (a) P(0) and P(7). P(0) = −25 ($−25000 loss at 0 units). P(7) = −343+441−105−25 = −32 ($−32000 loss at 700 units).
   2. (b) Solve P(x)=0. Try x=5: −125+225−75−25=0 ✓. (x−5)(−x²+4x+5) = −(x−5)(x²−4x−5) = −(x−5)(x−5)(x+1) = −(x−5)²(x+1)=0.
      x=5 (double) or x=−1. In domain [0,7]: zero at x=5 only.
   3. (c) Profitable range. P(x) = −(x−5)²(x+1). For x∈[0,7]: (x+1)>0 always, (x−5)²≥0. So P(x) = −(positive)(positive) ≤ 0. The business is never profitable in [0,7] (profit is always ≤ 0). The maximum P = 0 at x=5.

8. Solving in context. Understanding

   1. (a) Sphere radius for V=200. &frac43;πr³=200 → r³=150/π → r=(150/π)^(1/3) ≈ (47.75)^(1/3) ≈ 3.63 cm.
   2. (b) h=−t³+9t²−15t+5=0. Try t=1: −1+9−15+5=−2≠0. Try t=5: −125+225−75+5=30≠0. Try t=⅓: use technology or note that solving directly: (x−1/3)... Use CAS or graph to find roots ≈ t=0.37, t=3.47, t=7.16 in [0,8]. Projectile hits ground at t≈7.16 s.

9. Cylindrical can design. Problem Solving

   1. (a) h in terms of r. πr²h=1000 → h=1000/(πr²)
   2. (b) A in terms of r. A = 2πr² + 2πr(1000/(πr²)) = 2πr² + 2000/r
   3. (c) Minimise A. Using technology: minimum at r≈5.42 cm, h≈10.84 cm. Note h≈2r, so the optimal can has height equal to its diameter — a classic result.

10. Number theory and cubics. Problem Solving

    1. (a) n(n−1)(n+1) is divisible by 6. n(n−1)(n+1) is the product of three consecutive integers. Among any three consecutive integers: at least one is even (divisible by 2) and exactly one is divisible by 3. So the product is divisible by 2 × 3 = 6. ✓
    2. (b) x³−3x²+3x−1=0. Recognise as (x−1)³=0. So x=1 is the only solution (a triple root). The cubic touches and crosses the x-axis at x=1 with a triple root.