IGCSE · Practice questions · Solving cubic equations

Factor, divide, solve.

Six original Cambridge-style questions on solving cubic equations: finding a first root by the factor theorem, dividing to a quadratic factor, exact surd answers, one real root, hence equations and a volume problem.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Rearrange, find a root, divide, solve.

Rearrange to \(p(x) = 0\); never divide through by \(x\).
Find a root: try factors of the constant term until \(p(a) = 0\), so \((x - a)\) is a factor.
Divide to \((x - a)(bx^{2} + cx + d)\) and check by expanding one term.
Solve the quadratic: factorise, or use the formula and simplify the surd; a negative discriminant means no more real roots.
Hence or context: reuse the roots, undo any substitution, and reject only impossible values.

Recall

[3 marks]

Solve the equation \(x^{3} - 6x^{2} + 11x - 6 = 0\).

\(p(1) = 1 - 6 + 11 - 6 = 0\), so \((x - 1)\) is a factor. ✓
\(p(x) = (x - 1)(x^{2} - 5x + 6) = (x - 1)(x - 2)(x - 3)\). ✓
\(x = 1,\ 2\) or \(3\). ✓

Calculation

[4 marks]

Solve the equation \(2x^{3} - x^{2} - 13x - 6 = 0\).

\(p(-2) = -16 - 4 + 26 - 6 = 0\), so \((x + 2)\) is a factor. ✓
\(p(x) = (x + 2)(2x^{2} - 5x - 3)\). ✓
\(2x^{2} - 5x - 3 = (2x + 1)(x - 3)\). ✓
\(x = -2,\ -\dfrac{1}{2}\) or \(3\). ✓

Calculation

[4 marks]

Solve the equation \(x^{3} + 8 = 4x^{2} + 2x\), giving your answers in exact form.

Rearrange: \(p(x) = x^{3} - 4x^{2} - 2x + 8 = 0\). ✓
\(p(4) = 64 - 64 - 8 + 8 = 0\), so \((x - 4)\) is a factor. ✓
\(p(x) = (x - 4)(x^{2} - 2)\). ✓
\(x^{2} = 2\), so \(x = 4,\ \sqrt{2}\) or \(-\sqrt{2}\). ✓

Analysis

[4 marks]

Show that \(x = -3\) is the only real root of the equation \(x^{3} + 2x^{2} + x + 12 = 0\).

\(p(-3) = -27 + 18 - 3 + 12 = 0\), so \(x = -3\) is a root and \((x + 3)\) is a factor. ✓
\(p(x) = (x + 3)(x^{2} - x + 4)\). ✓
For \(x^{2} - x + 4 = 0\): \(b^{2} - 4ac = 1 - 16 = -15 < 0\), so it has no real roots. ✓
Hence \(x = -3\) is the only real root. ✓

Analysis

[7 marks]

\(p(x) = 2x^{3} + ax^{2} - 13x + 6\), where \(a\) is a constant. \((x - 2)\) is a factor of \(p(x)\).

(a) Find the value of \(a\).

(b) Factorise \(p(x)\) completely.

(d) Hence solve \(16y^{3} + 4y^{2} - 26y + 6 = 0\).

(a) \(p(2) = 16 + 4a - 26 + 6 = 0\), so \(4a = 4\), \(a = 1\). ✓
(b) \(p(x) = (x - 2)(2x^{2} + 5x - 3) = (x - 2)(2x - 1)(x + 3)\). ✓
(c) \(x = 2,\ \dfrac{1}{2}\) or \(-3\). ✓
(d) \(16y^{3} + 4y^{2} - 26y + 6 = 2(2y)^{3} + (2y)^{2} - 13(2y) + 6 = p(2y)\). ✓
So \(2y = 2,\ \dfrac{1}{2}\) or \(-3\), giving \(y = 1,\ \dfrac{1}{4}\) or \(-\dfrac{3}{2}\). ✓

Proof

[6 marks]

An open box is made from a square card of side 12 cm by cutting a square of side \(x\) cm from each corner and folding up the sides. The volume of the box is 128 cm³.

(a) Show that \(x^{3} - 12x^{2} + 36x - 32 = 0\).

(b) Solve this equation and find the value of \(x\), explaining why any other root is rejected.

(a) The base is \((12 - 2x)\) by \((12 - 2x)\) and the height is \(x\): \(x(12 - 2x)^{2} = 128\). ✓
\(x(144 - 48x + 4x^{2}) = 128\), so \(4x^{3} - 48x^{2} + 144x - 128 = 0\); divide by 4: \(x^{3} - 12x^{2} + 36x - 32 = 0\). ✓
(b) \(p(2) = 8 - 48 + 72 - 32 = 0\), so \(p(x) = (x - 2)(x^{2} - 10x + 16) = (x - 2)(x - 2)(x - 8)\). ✓
The roots are \(x = 2\) (repeated) and \(x = 8\). ✓
The card is 12 cm wide, so \(0 < x < 6\); \(x = 8\) would leave a negative base, so reject it. ✓
\(x = 2\) (check: \(2 \times 8 \times 8 = 128\)). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.