★ Topic Review — Binomial Expansion and Cubic Functions — Solutions

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This review covers all four lessons in this topic: Pascal’s Triangle & Combinations, The Binomial Theorem, Cubic Functions — Graphs and Features, and Solving & Modelling with Cubics. Questions are mixed across difficulty levels.

1. Fluency

   1. (a) Evaluate ⁷C₃.
   2. (b) Use the symmetry property to evaluate ¹¹C₉.
   3. (c) Write out row 5 of Pascal’s triangle and state the sum of all entries.
   4. (d) Use Pascal’s rule to find ⁸C₄, given ⁷C₃ = 35 and ⁷C₄ = 35.
   (a) 35   (b) ¹¹C₂=55   (c) 1,5,10,10,5,1; sum=32=2&sup5;   (d) 35+35=70

2. Fluency

   1. (a) Expand (x + 3)³ fully.
   2. (b) Expand (2x − 1)⁴ fully.
   3. (c) Find the coefficient of x³ in the expansion of (1 + 2x)⁵.
   4. (d) Find the constant term in the expansion of (x + 1/x)⁴.
   (a) x³+9x²+27x+27   (b) 16x&sup4;−32x³+24x²−8x+1   (c) T₄=⁵C₃(2)³x³=10×8=80   (d) r=2: ⁴C₂(x²)(1/x²)=6

3. Fluency

   1. (a) For y = (x + 1)(x − 2)(x − 4), state the x-intercepts, y-intercept, and end behaviour.
   2. (b) For y = −2(x − 3)³ + 1, state the inflection point and end behaviour.
   3. (c) Expand y = (x − 1)(x + 3)(x − 2) into standard form.
   (a) x-ints: −1,2,4; y-int: 8; a=1>0 so ↗.   (b) inflection (3,1); a=−2<0 so ↘.   (c) x³−7x+6

4. Fluency

   1. (a) Solve (x + 2)(x − 1)(x + 5) = 0.
   2. (b) Solve x³ − 4x = 0.
   3. (c) Solve (x − 3)³ = −8.
   (a) x=−2, x=1, x=−5   (b) x(x−2)(x+2)=0: x=0, x=±2   (c) x−3=−2: x=1

5. Understanding

   1. (a) Find the coefficient of x² in (1 + x + x²)³. (Hint: use (1 + (x + x²))³ and expand carefully.)
   2. (b) A cubic passes through (−2, 0), (1, 0), and (4, 0), and has y-intercept −16. Find its equation and verify the y-intercept.
   3. (c) Use the factor theorem to show that (x − 2) is a factor of x³ − 2x² − 4x + 8, then fully factorise.
   (a) (1+u)³ where u=x+x²: coefficient of x² from 3u+3u² terms: 3(1 from x² term of u) + 3(1 from u²=x² term) = 3+3=6. Actually: (1+x+x²)³ expanded: coefficient of x² = ³C₂(1)(1)+³C₁(1)=3+3=6.   (b) y=a(x+2)(x−1)(x−4); x=0: a(2)(−1)(−4)=8a=−16, a=−2. y=−2(x+2)(x−1)(x−4).   (c) p(2)=8−8−8+8=0 ✓. (x−2)(x²−4)=(x−2)(x−2)(x+2)=(x−2)²(x+2).

6. Understanding

   1. (a) Solve P(x) = 0 where P(x) = x³ + 3x² − 10x − 24. (Try x = 3 or x = −4.)
   2. (b) Sketch the graph of P(x) from part (a), labelling all intercepts and describing end behaviour.
   3. (c) For what values of x is P(x) > 0?
   (a) Try x=3: 27+27−30−24=0 ✓. (x−3)(x²+6x+8)=(x−3)(x+2)(x+4)=0. Roots: x=3, −2, −4.   (b) x-ints −4, −2, 3; y-int −24; a=1>0 ↗.   (c) P>0 when −4<x<−2 or x>3.

7. Connecting binomial theorem and cubics. Understanding

   1. (a) Use the binomial theorem to expand (x + 1)³ and verify that it matches the factored-to-expanded form directly.
   2. (b) Use (a − b)³ = a³ − 3a²b + 3ab² − b³ to show that (x − 1)³ + 1 can be factorised as (x)(x² − 2x + 3). Verify by expansion.
   (a) (x+1)³=x³+3x²+3x+1 ✓.   (b) (x−1)³+1 = x³−3x²+3x−1+1 = x³−3x²+3x = x(x²−3x+3). Note: question states x²−2x+3 but correct factor is x²−3x+3.

8. Problem Solving

   Multi-step problem.
   1. (a) An open rectangular box is constructed from a 24 cm × 18 cm sheet of metal by removing squares of side x from each corner. Find the value of x that gives a volume of 640 cm³. (Use technology or trial and error.)
   2. (b) The middle term of (2x + 3)⁴ contains a specific coefficient. Find it, and determine which term number it is.
   3. (c) A cubic equation has roots α, β, γ where α + β + γ = 4 and αβγ = −6. Given that two of the roots are 3 and −1, find the third root and write the cubic as a factored expression with integer coefficients.
   (a) V=x(24−2x)(18−2x)=640. Using technology: x≈2 (exact: try x=2: 2×20×14=560≠640. x=2.5: 2.5×19×13=617.5. x=3: 3×18×12=648. Approximately x≈2.9 cm.)   (b) (2x+3)&sup4; has 5 terms; middle is T₃: ⁴C₂(2x)²(3)²=6×4x²×9=216x². Coefficient: 216.   (c) Third root = 4−3−(−1)=2. αβγ=3×−1×2=−6 ✓. Cubic: (x−3)(x+1)(x−2).

9. Problem Solving

   Pascal’s triangle investigation.
   1. (a) The sum of entries in row n of Pascal’s triangle is 2ⁿ. Using the binomial theorem with appropriate values, prove this result.
   2. (b) By choosing x = 10 in the expansion of (1 + x)ⁿ, explain how Pascal’s triangle is related to powers of 11. Verify for n = 4.
   (a) Set x=1, y=1 in (x+y)&sup n;: 2&sup n;=∑ⁿC_r ✓.   (b) (1+10)&sup4;=11&sup4;=14641. Expansion: 1+4(10)+6(100)+4(1000)+10000=1+40+600+4000+10000=14641 ✓. Each coefficient is a digit (works when no carry-over occurs).

10. Understanding — Sketching and solving cubics

    1. (a) For P(x) = 2x³ − 3x² − 11x + 6, verify that x = 3 is a root and fully factorise.
    2. (b) Sketch the graph of P(x), labelling all intercepts and describing end behaviour.
    3. (c) Solve 2x³ − 3x² − 11x + 6 ≤ 0.
    (a) P(3)=54−27−33+6=0 ✓. Divide: P(x)=(x−3)(2x²+3x−2)=(x−3)(2x−1)(x+2).   (b) x-ints: x=−2, x=1/2, x=3. y-int: P(0)=6. Leading coeff 2>0 so as x→−∞, y→−∞; as x→+∞, y→+∞.   (c) P(x)≤0 when x≤−2 or 1/2≤x≤3.

11. Problem Solving — Finding unknowns in binomial expansions

    Multi-step.
    1. (a) In the expansion of (1 + ax)^6, the coefficient of x² is 60. Find the value of a.
    2. (b) In the expansion of (2 + bx)^5, the coefficient of x³ equals the coefficient of x². Find b.
    3. (c) The sum of the first two terms in the expansion of (1 + x)^n is 1 + 12x. Find n.
    (a) T₃=&sup6;C₂a²x²=15a²=60 → a²=4 → a=±2.   (b) Coeff x³=&sup5;C₃(2)²b³=10×4b³=40b³. Coeff x²=&sup5;C₂(2)³b²=10×8b²=80b². 40b³=80b² → b=2 (b≠0). b=2.   (c) (1+x)^n=1+nx+... First two terms: 1+nx. So nx=12x, n=12.