A cubic inequality is a question about a graph: where is the curve above (or below) a line? Sketch the cubic, find where it meets the line, and the answer is a set of intervals on the \(x\)-axis, often in two parts.
To solve \(f(x) > d\) (or \(\geqslant, <, \leqslant\)) where \(f(x)\) is a product of three linear factors: sketch \(y = f(x)\) and the line \(y = d\); find the \(x\)-coordinates where they meet (for \(d = 0\) these are the roots; otherwise solve \(f(x) = d\), which is \(f(x) - d = 0\), a cubic you factorise with the factor theorem); then read off the \(x\)-values where the curve is above the line (for \(>\)) or below it (for \(<\)). Strict signs give open ends, \(\geqslant\) and \(\leqslant\) include the end values, and a repeated factor makes the curve touch, not cross.
Fig. 1: y = (x + 2)(x + 1)(x - 3) and the line y = -12. They meet where x³ - 7x + 6 = 0, that is x = -3, 1 and 2. The curve is above the line (blue) for -3 < x < 1 and for x > 2, so these intervals, marked on the x-axis, solve (x + 2)(x + 1)(x - 3) > -12
Section 01
When d = 0, the roots split the axis.
For \(f(x) > 0\) or \(f(x) < 0\) with \(f(x)\) already in three linear factors, you need no algebra beyond the roots. Sketch the cubic through its three roots (from the previous lesson): with a positive \(x^{3}\) coefficient it starts low and ends high, with a negative one it starts high and ends low. The roots cut the \(x\)-axis into four pieces, and the curve changes side at each root, so the signs alternate. \(f(x) > 0\) is where the curve is above the axis, \(f(x) < 0\) where it is below.
Find the roots from the brackets and put them in order, smallest first.
Sketch: positive \(x^{3}\) coefficient means bottom left to top right; negative means top left to bottom right.
Read off the \(x\)-values where the curve is on the side you want. Usually that is one bounded interval and one unbounded one.
Write it properly: \(-2 < x < 1\) or \(x > 3\). Use \(\leqslant\) or \(\geqslant\) at the roots only if the question does.
Worked example Below the axis
Solve \((x + 2)(x - 1)(x - 3) < 0\).
The roots are \(x = -2,\ 1\) and \(3\). The \(x^{3}\) coefficient is \(+1\), so the curve comes up from the bottom left.
So the curve is below the axis to the left of \(-2\), above it between \(-2\) and \(1\), below between \(1\) and \(3\), and above to the right of \(3\).
The sign is \(\geqslant\), so the roots are included.
Answer: \(x \leqslant -4\) or \(-1 \leqslant x \leqslant 2\)
Your turn
Solve \(x(x - 2)(x + 3) > 0\).
Roots \(-3,\ 0,\ 2\); positive \(x^{3}\) coefficient, so the curve is above the axis for \(-3 < x < 0\) and for \(x > 2\). Answer: \(-3 < x < 0\) or \(x > 2\).
Examiner trap
Two separate intervals must be written as two separate inequalities joined by 'or'. Never write \(1 < x > 3\) or \(-2 > x > 3\): no number satisfies the second, and the first means nothing. And look at the sign of the \(x^{3}\) term before you sketch: \((2 - x)\) flips the whole curve, which swaps every answer.
Section 02
A repeated factor touches, it does not cross.
If a factor is squared, as in \((x - 1)^{2}(x + 2)\), the curve touches the \(x\)-axis at \(x = 1\) and turns back. A square is never negative, so the curve stays on the same side on both sides of that root. Now the strict and non-strict versions give different answers at that one point, and you must think about it: the touching point is a single value where \(f(x) = 0\).
\((x - 1)^{2}(x + 2) \geqslant 0\): \(x \geqslant -2\). The point \(x = 1\) is already inside.
\((x - 1)^{2}(x + 2) > 0\): \(-2 < x < 1\) or \(x > 1\). At \(x = 1\) the value is \(0\), which is not \(> 0\), so it is cut out.
\((x - 1)^{2}(x + 2) \leqslant 0\): \(x \leqslant -2\) or \(x = 1\). The touching point is an isolated solution.
\((x - 1)^{2}(x + 2) < 0\): \(x < -2\). The touching point is not a solution.
Worked example An isolated point
Solve \((x - 1)^{2}(x + 2) \leqslant 0\).
Roots: \(x = -2\) (single, the curve crosses) and \(x = 1\) (repeated, the curve touches).
The \(x^{3}\) coefficient is positive, so the curve is below the axis for \(x < -2\), then above it for all \(x > -2\) except at \(x = 1\), where it just touches.
\(\leqslant 0\) means on or below the axis: every \(x \leqslant -2\), and also the single point \(x = 1\), where \(f(1) = 0\).
The \(x^{3}\) coefficient is \((1)(-1) = -1\), so the curve starts top left: above the axis for \(x < -1\), touches at \(-1\), stays above until it crosses at \(3\), then below.
The inequality is strict and \(f(-1) = 0\), so \(x = -1\) must be left out.
Answer: \(x < -1\) or \(-1 < x < 3\)
Your turn
Solve \((x - 2)^{2}(x + 1) < 0\).
\((x - 2)^{2} \geqslant 0\), so the sign comes from \((x + 1)\), except at \(x = 2\) where the product is \(0\). \(< 0\) needs \(x + 1 < 0\) and \(x \neq 2\): the answer is \(x < -1\).
Examiner trap
Do not treat a squared factor like two different roots: the sign does NOT change at a repeated root. And do not forget the touching point itself: with \(\leqslant\) or \(\geqslant\) it may be an extra isolated answer (\(x = 1\)); with \(<\) or \(>\) it may have to be cut out of an interval (\(x \neq -1\)).
Stage 1 · Learn
Check your understanding
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Solve \((x + 1)(x - 2)(x - 4) > 0\).
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Solve \((x - 3)^{2}(x + 1) < 0\).
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Solve \((1 - x)(x + 2)(x - 4) \geqslant 0\).
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A cubic \(y = f(x)\) with positive \(x^{3}\) coefficient meets the line \(y = 5\) at \(x = -2,\ 1\) and \(4\) only. Solve \(f(x) < 5\).
Section 03
A line y = d: find where they meet, then read off.
For \(f(x) > d\) with \(d \neq 0\), compare the curve with the horizontal line \(y = d\), not with the \(x\)-axis. The roots of \(f(x)\) no longer give the answer. Instead, multiply out \(f(x)\), rearrange \(f(x) = d\) to \(f(x) - d = 0\), and factorise that cubic with the factor theorem (topic 3): try factors of the constant term. Its roots are the \(x\)-coordinates where the curve meets the line. Then read the answer from the sketch, exactly as before, using 'above the line' or 'below the line'.
Where the curve meets \(y = -12\): \(x^{3} - 7x - 6 = -12\), so \(x^{3} - 7x + 6 = 0\).
Try \(x = 1\): \(1 - 7 + 6 = 0\), so \((x - 1)\) is a factor: \(x^{3} - 7x + 6 = (x - 1)(x^{2} + x - 6) = (x - 1)(x - 2)(x + 3)\).
They meet at \(x = -3,\ 1\) and \(2\). The cubic rises to the right, so it is above the line between \(-3\) and \(1\), dips below between \(1\) and \(2\), and is above again for \(x > 2\) (see Fig. 1).
Strict inequality, so the meeting points are not included.
Try \(x = -1\): \(-1 + 13 - 12 = 0\), so \((x + 1)\) is a factor: \(x^{3} - 13x - 12 = (x + 1)(x^{2} - x - 12) = (x + 1)(x + 3)(x - 4)\).
So the curve meets \(y = 24\) at \(x = -3,\ -1\) and \(4\). Check: at \(x = -2\) (between \(-3\) and \(-1\)), \(f(-2) = (2)(-3)(-5) = 30 \geqslant 24\), so the curve is above the line there.
Above or on the line: between \(-3\) and \(-1\), and to the right of \(4\). Include the ends.
Answer: \(-3 \leqslant x \leqslant -1\) or \(x \geqslant 4\)
\(x^{3} - 7x + 6 = 12\) gives \(x^{3} - 7x - 6 = 0\); \(x = -1\) is a root, and \(x^{3} - 7x - 6 = (x + 1)(x^{2} - x - 6) = (x + 1)(x + 2)(x - 3)\). Below the line: \(x < -2\) or \(-1 < x < 3\).
Examiner trap
Never set each bracket equal to \(d\): \((x + 2)(x + 1)(x - 3) = -12\) does NOT mean \(x + 2 = -12\) or \(x + 1 = -12\). That trick only works for \(= 0\). Multiply out, take \(d\) to the left, and factorise again. And the roots of \(f(x)\) (here \(-2,\ -1,\ 3\)) are where the curve meets the \(x\)-axis, not the line, so they do not appear in the answer.
Stage 2 · Exam
Exam-style questions
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Solve \((x + 2)(x - 2)(x - 3) > 12\).
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Given that \(x^{3} - x - 6 = (x - 2)(x^{2} + 2x + 3)\), solve \(x(x - 1)(x + 1) > 6\).