Practice Maths

Solving and Modelling with Cubics

Key Terms

Solving a cubic algebraically
is only straightforward when the cubic is already factorised (or can be easily factorised).
Factor theorem
(x − a) is a factor of p(x) if and only if p(a) = 0. Use this to find a root by trial, then factorise.
Once one factor (x − a) is found, divide out to get a quadratic, then factorise or use the quadratic formula for the rest.
Technology
for cubic equations that don’t factorise nicely, solve numerically using CAS/graphing technology.
When modelling with cubics, always consider the practical domain and interpret solutions in context.
Volume, container design, and profit/loss problems commonly lead to cubic equations.
StrategyWhen to use
Null factor lawCubic is already in factored form
Factor theorem + divideCan guess one rational root by trial
Technology (CAS/graph)No rational root; approximate solutions needed
Hot Tip To find rational roots to try, test factors of the constant term divided by factors of the leading coefficient. For x³ − 6x + 4: try ±1, ±2, ±4. Here p(2) = 8 − 12 + 4 = 0, so (x − 2) is a factor.

Worked Example 1 — Solving a Cubic by Factorising

Question: Solve x³ − 2x² − 5x + 6 = 0.

Step 1 (find a root): Test x = 1: 1 − 2 − 5 + 6 = 0. ✓ So (x − 1) is a factor.

Step 2 (divide): x³ − 2x² − 5x + 6 = (x − 1)(x² − x − 6).

Step 3 (factorise quadratic): x² − x − 6 = (x − 3)(x + 2).

Answer: (x − 1)(x − 3)(x + 2) = 0 → x = 1, x = 3, or x = −2.

Worked Example 2 — Cubic Modelling Problem

Question: A square piece of cardboard has side length 20 cm. Squares of side x cm are cut from each corner and the sides folded up to form an open box. Find x so that the volume is 500 cm³.

Step 1 (model): V = x(20 − 2x)².

Step 2 (equation): x(20 − 2x)² = 500. Let u = 20 − 2x, or expand directly.

Step 3 (expand): x(400 − 80x + 4x²) = 500 → 4x³ − 80x² + 400x − 500 = 0 → 4x³ − 80x² + 400x − 500 = 0.

Step 4 (technology or trial): Divide by 4: x³ − 20x² + 100x − 125 = 0. Try x = 5: 125 − 500 + 500 − 125 = 0. ✓

Step 5: (x − 5)(x² − 15x + 25) = 0 → x = 5 or x = (15 ± √125)/2 ≈ 13.1 or 1.9.

Practical domain: 0 < x < 10. So valid solutions: x = 5 cm or x ≈ 1.9 cm (reject x ≈ 13.1 as it violates 0 < x < 10).

The Factor Theorem: Why It Works

The factor theorem states: (x − a) is a factor of the polynomial p(x) if and only if p(a) = 0. This follows directly from the polynomial division algorithm: we can always write p(x) = (x − a) × q(x) + R, where R is the remainder. If we substitute x = a: p(a) = (a − a) × q(a) + R = 0 + R = R. So the remainder R equals p(a). If p(a) = 0, then R = 0, meaning (x − a) divides p(x) exactly with no remainder — confirming (x − a) is a factor.

The factor theorem is also known as a special case of the remainder theorem: p(a) equals the remainder when p(x) is divided by (x − a). This gives a quick way to find remainders without full division.

Rational Root Theorem: Systematic Testing

For a polynomial p(x) = axn + … + d with integer coefficients, any rational root p/q (in lowest terms) must satisfy: p divides the constant term d, and q divides the leading coefficient a.

For x³ − 2x² − 5x + 6 (leading coefficient 1, constant term 6): rational roots must be factors of 6, which are ±1, ±2, ±3, ±6. Test these in order. p(1) = 1 − 2 − 5 + 6 = 0 ✓. Found one: (x − 1) is a factor.

Practical order of testing: try ±1 first (easiest to evaluate), then ±2, then ±3, etc. For cubics with leading coefficient 1, only the factors of the constant term need testing.

Polynomial Long Division: The Full Process

Once you find a root x = a, divide p(x) by (x − a) to get the remaining quadratic. Long division proceeds like numerical long division:

  1. Divide the leading term of p(x) by the leading term of the divisor.
  2. Multiply the result by the full divisor.
  3. Subtract from p(x), bringing down the next term.
  4. Repeat until the degree of the remainder is less than the divisor.

For example, dividing x³ − 2x² − 5x + 6 by (x − 1): x³ ÷ x = x². Multiply: x²(x−1) = x³−x². Subtract: −2x²−(−x²) = −x². Bring down −5x: −x²−5x. Divide: −x² ÷ x = −x. Multiply: −x(x−1) = −x²+x. Subtract: −5x−x = −6x. Bring down 6: −6x+6. Divide: −6x ÷ x = −6. Multiply: −6(x−1) = −6x+6. Subtract: 0. Result: x²−x−6.

Setting Up Cubic Models

Real-world problems involving cubics most commonly arise from volume. If a box is made by cutting squares of side x from a rectangular sheet and folding up the sides, the volume is a product of three linear expressions in x, giving a cubic. The practical domain restricts x to positive values that keep all dimensions positive.

Always define variables clearly, write the model, state the domain (usually x > 0 and bounded by the dimensions of the problem), and interpret all solutions in context. A root that falls outside the practical domain is mathematically valid but physically impossible — reject it with an explanation.

Solving Cubic Equations: The Complete Approach

The full process: (1) Look for a rational root using the rational root theorem and factor theorem. (2) Divide by (x − a) to get a quadratic. (3) Use factorisation or the quadratic formula on the quadratic. (4) Check the discriminant of the quadratic — if Δ < 0, the cubic has only one real root. (5) Verify solutions by substitution or inspection.

Exam Tip: Always try x = 1 and x = −1 first — they are the easiest to test mentally. p(1) = (sum of all coefficients); p(−1) = (sum of coefficients with alternating signs). For x³ − 2x² − 5x + 6: p(1) = 1−2−5+6 = 0 immediately. If neither ±1 works, try ±2 next.
Exam Tip: After dividing out one linear factor to get a quadratic, always calculate the discriminant Δ = b²−4ac of the quadratic before attempting to factorise. If Δ < 0, there are no more real roots and the cubic has exactly one real root. Stating this explicitly shows clear mathematical understanding and avoids wasted algebraic effort.

Mastery Practice

See Answers ➔

  1. Solve each cubic equation by the null factor law (equations given in factored form). Fluency

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

  2. Use the factor theorem to verify the given root, then fully factorise and solve. Fluency

    1. (a) x³ − 6x + 4 = 0, given that x = 2 is a root.
    2. (b) x³ + x² − 4x − 4 = 0, given that x = 2 is a root.
    3. (c) x³ − 7x + 6 = 0, given that x = 1 is a root.

  3. Find all solutions by first finding a rational root using the factor theorem. Fluency

    1. (a) x³ − 2x² − x + 2 = 0
    2. (b) x³ + 3x² − 4 = 0
    3. (c) 2x³ − 3x² − 11x + 6 = 0

  4. Solve graphically using the features of the cubic. Fluency

    1. (a) From the factored form y = (x + 3)(x − 1)(x − 2), state all x-intercepts. Hence write the solutions to (x + 3)(x − 1)(x − 2) = 0.
    2. (b) y = x³ − 9x. Factorise and state all roots.
    3. (c) y = −x³ + 4x² − 4x. Factorise and solve y = 0.

  5. Mixed solving. Fluency

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

  6. Open box problem. Understanding

    Modelling volume. An open box is made from a 30 cm × 30 cm square sheet of cardboard by cutting equal squares of side x cm from each corner and folding up the sides.
    1. (a) Write an expression for the volume V of the box in terms of x.
    2. (b) State the practical domain for x.
    3. (c) Find the value(s) of x for which V = 1000 cm³.
    4. (d) Find the value of x that makes the volume a maximum (use technology or graph).

  7. Revenue and profit model. Understanding

    Business context. A manufacturer’s profit in thousands of dollars is modelled by P(x) = −x³ + 9x² − 15x − 25, where x is the number of units (hundreds) produced, 0 ≤ x ≤ 7.
    1. (a) Find P(0) and P(7) and interpret each value in context.
    2. (b) Solve P(x) = 0 to find when the profit is zero. (Try x = 5 as a factor theorem candidate.)
    3. (c) Determine the production levels for which the business is profitable in this range.

  8. Solving in context. Understanding

    Interpreting solutions.
    1. (a) A sphere has volume V = &frac43;πr³. Find the radius (correct to 2 decimal places) of a sphere with volume 200 cm³.
    2. (b) The height of a projectile at time t seconds is h = −t³ + 9t² − 15t + 5 metres. Solve h = 0 using the factor theorem to find when it hits the ground, 0 ≤ t ≤ 8.

  9. Constructing and solving a cubic. Problem Solving

    Design problem. A cylindrical can (with top and bottom) must hold 1 litre (1000 cm³) of liquid. The radius is r cm and the height is h cm.
    1. (a) Write h in terms of r using the volume formula V = πr²h.
    2. (b) The total surface area is A = 2πr² + 2πrh. Substitute your expression from (a) to write A in terms of r only.
    3. (c) To minimise material, A must be minimised. Use technology (or calculus if known) to find the radius and height that give minimum surface area. Comment on the shape of the optimal can.

  10. Number theory and cubics. Problem Solving

    Algebraic investigation.
    1. (a) Show that n³ − n = n(n − 1)(n + 1) for all integers n, and hence prove that the product of any three consecutive integers is always divisible by 6.
    2. (b) Solve x³ − 3x² + 3x − 1 = 0 and explain why there is only one solution. (Hint: recognise the left side as a perfect cube.)