IGCSE · Practice questions · Cubic inequalities

Above the line or below it?

Six original Cambridge-style questions on cubic inequalities: products of three linear factors compared with zero, repeated factors, negative leading coefficients, and horizontal lines y = d found with the factor theorem.

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

Sketch, find the meeting points, read off.

Sketch \(y = f(x)\) through its roots; check the sign of the \(x^{3}\) coefficient.
If \(d \neq 0\), multiply out, rearrange \(f(x) - d = 0\) and factorise it with the factor theorem.
Mark the meeting points on the \(x\)-axis in order; a repeated root means the curve touches the line.
Read off where the curve is above (\(>\)) or below (\(<\)) the line.
Write the answer with 'or' between intervals; include end values only for \(\geqslant\) and \(\leqslant\), and watch any touching point.

Recall

[2 marks]

Solve the inequality \(x(x + 2)(x - 5) < 0\).

The roots are \(-2,\ 0,\ 5\), and the \(x^{3}\) coefficient is positive, so the curve starts bottom left. ✓
It is below the axis for \(x < -2\) and for \(0 < x < 5\). ✓
\(x < -2\) or \(0 < x < 5\). ✓

Calculation

[3 marks]

Solve the inequality \((x + 1)^{2}(x - 4) \geqslant 0\).

\((x + 1)^{2} \geqslant 0\) for all \(x\), and it equals \(0\) only at \(x = -1\) (the curve touches the axis there). ✓
For \(x \neq -1\) the sign is the sign of \(x - 4\), so the product is \(\geqslant 0\) when \(x \geqslant 4\). ✓
At \(x = -1\) the product is \(0\), which satisfies \(\geqslant 0\). ✓
\(x = -1\) or \(x \geqslant 4\). ✓

Calculation

[3 marks]

Solve the inequality \((3 - x)(x + 2)(2x + 1) > 0\).

The roots are \(-2,\ -\dfrac{1}{2},\ 3\). ✓
The \(x^{3}\) coefficient is \((-1)(1)(2) = -2\), negative, so the curve starts top left: above, below, above, below. ✓
Check \(x = 0\): \((3)(2)(1) = 6 > 0\), so the curve is above the axis between \(-\dfrac{1}{2}\) and \(3\). ✓
\(x < -2\) or \(-\dfrac{1}{2} < x < 3\). ✓

Analysis

[5 marks]

\(f(x) = x(x + 4)(x - 1)\).

(a) Show that \(f(x) - 12 = x^{3} + 3x^{2} - 4x - 12\), and factorise this expression completely.

(b) Hence solve \(f(x) < 12\).

(a) \(x(x + 4)(x - 1) = x(x^{2} + 3x - 4) = x^{3} + 3x^{2} - 4x\), so \(f(x) - 12 = x^{3} + 3x^{2} - 4x - 12\). ✓
\(x = 2\): \(8 + 12 - 8 - 12 = 0\), so \((x - 2)\) is a factor: \(x^{3} + 3x^{2} - 4x - 12 = (x - 2)(x^{2} + 5x + 6) = (x - 2)(x + 2)(x + 3)\). ✓
(b) The curve \(y = f(x)\) meets \(y = 12\) at \(x = -3,\ -2,\ 2\). It rises to the right, so it is below the line for \(x < -3\) and for \(-2 < x < 2\). ✓
Check \(x = 0\): \(f(0) = 0 < 12\). Correct. ✓
\(x < -3\) or \(-2 < x < 2\). ✓

Analysis

[6 marks]

\(f(x) = (x + 4)(x + 1)(x - 4)\).

(a) Show that \(f(x) + 36 = (x - 2)^{2}(x + 5)\).

(b) Explain what this tells you about the curve \(y = f(x)\) and the line \(y = -36\).

(a) \(f(x) = (x^{2} - 16)(x + 1) = x^{3} + x^{2} - 16x - 16\), so \(f(x) + 36 = x^{3} + x^{2} - 16x + 20\). ✓
\((x - 2)^{2}(x + 5) = (x^{2} - 4x + 4)(x + 5) = x^{3} + x^{2} - 16x + 20\). The two agree. ✓
(b) They meet where \(f(x) = -36\), at \(x = -5\) and \(x = 2\). The repeated factor \((x - 2)^{2}\) means the line touches the curve at \(x = 2\) (it is a tangent there) and crosses it at \(x = -5\). ✓
(c) \(f(x) \leqslant -36\) means \((x - 2)^{2}(x + 5) \leqslant 0\). ✓
\((x - 2)^{2} \geqslant 0\), so this holds when \(x + 5 \leqslant 0\), or when \((x - 2)^{2} = 0\). ✓
\(x \leqslant -5\) or \(x = 2\). ✓

Proof

[6 marks]

\(f(x) = (x + 1)(x - 2)(x - 3)\).

(a) Find, in exact form, the \(x\)-coordinates of the points where the curve \(y = f(x)\) meets the line \(y = 6\).

(b) Hence solve \(f(x) > 6\).

(a) \(f(x) = (x + 1)(x^{2} - 5x + 6) = x^{3} - 4x^{2} + x + 6\). ✓
\(f(x) = 6\) gives \(x^{3} - 4x^{2} + x = 0\), so \(x(x^{2} - 4x + 1) = 0\). ✓
\(x = 0\) or \(x = \dfrac{4 \pm \sqrt{16 - 4}}{2} = \dfrac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}\). ✓
(b) In order: \(0 < 2 - \sqrt{3} < 2 + \sqrt{3}\) (since \(\sqrt{3}\) is between 1 and 2). The curve rises to the right, so it is above \(y = 6\) between the first two meeting points and after the third. ✓
\(0 < x < 2 - \sqrt{3}\) or \(x > 2 + \sqrt{3}\). ✓

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.