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Cambridge IGCSE · Topic 3.3
IGCSE Add Maths 0606 / Topic 3 / IGCSE

One root unlocks the other two.

A cubic equation has no neat formula you need to learn. Find one root by the factor theorem, divide it out, and what is left is a quadratic you already know how to solve: by factorising, or by the formula when the roots are surds.

Topic 3All factors of polynomials lessons →
The key idea

To solve a cubic: rearrange to \(p(x) = 0\); try factors of the constant term until \(p(a) = 0\), so \((x - a)\) is a factor; divide (or compare coefficients) to write \(p(x) = (x - a)(bx^{2} + cx + d)\); then solve the quadratic. Its discriminant decides the rest: positive gives two more roots (exact surds if it will not factorise), zero gives a repeated root, negative gives no more real roots, so \(x = a\) is the only one.

x y O −1 2 −5 5 6 −2 1 3 y = x³ − 2x² − 5x + 6 p(1) = 0, so (x − 1) is a factor (x − 1)(x² − x − 6) = (x − 1)(x + 2)(x − 3) roots: x = −2, 1, 3 (where the curve crosses the x-axis) y-intercept p(0) = 6
Fig. 1: the curve y = x³ − 2x² − 5x + 6. Since p(1) = 0, (x − 1) is a factor, and p(x) = (x − 1)(x + 2)(x − 3). The three roots of p(x) = 0 are the x-coordinates where the curve crosses the x-axis
Section 01

Find one root, then divide.

Every cubic equation in this course is solved the same way. First collect every term on one side so that it reads \(p(x) = 0\). Then use the factor theorem from the first lesson of this topic: if \(p(a) = 0\), then \((x - a)\) is a factor. Integer roots must divide the constant term, so try \(\pm 1, \pm 2, \pm 3, \dots\) from its factors. Once one factor is found, divide it out (as in the previous lesson) to leave a quadratic factor, and solve that.

Worked example Three whole-number roots
Solve the equation \(x^{3} + 6 = 2x^{2} + 5x\).
  1. Rearrange: \(p(x) = x^{3} - 2x^{2} - 5x + 6 = 0\).
  2. Try \(x = 1\): \(p(1) = 1 - 2 - 5 + 6 = 0\), so \((x - 1)\) is a factor.
  3. Divide: \(x^{3} - 2x^{2} - 5x + 6 = (x - 1)(x^{2} - x - 6)\). Check: the constant term is \((-1)(-6) = 6\).
  4. Factorise the quadratic: \(x^{2} - x - 6 = (x - 3)(x + 2)\).
  5. So \((x - 1)(x - 3)(x + 2) = 0\), giving \(x = 1\), \(x = 3\) or \(x = -2\). These are where the curve in Fig. 1 crosses the \(x\)-axis.
Answer: \(x = -2,\ 1\) or \(3\)
Worked example A leading coefficient that is not 1
Solve \(2x^{3} + 3x^{2} - 8x + 3 = 0\).
  1. Try \(x = 1\): \(2 + 3 - 8 + 3 = 0\), so \((x - 1)\) is a factor.
  2. Compare coefficients: \(2x^{3} + 3x^{2} - 8x + 3 = (x - 1)(2x^{2} + bx - 3)\). The \(x^{2}\) terms give \(b - 2 = 3\), so \(b = 5\).
  3. Check the \(x\) terms: \(-3 - 5 = -8\). Correct.
  4. Factorise: \(2x^{2} + 5x - 3 = (2x - 1)(x + 3)\).
  5. So \(x = 1\), \(x = \dfrac{1}{2}\) or \(x = -3\).
Answer: \(x = -3,\ \frac{1}{2}\) or \(1\)
Your turn
Solve \(x^{3} + 2x^{2} - 5x - 6 = 0\).
Examiner trap

Never divide both sides by \(x\). In \(x^{3} = 4x\), dividing by \(x\) gives \(x^{2} = 4\) and loses the root \(x = 0\). Rearrange instead: \(x^{3} - 4x = x(x - 2)(x + 2) = 0\), so \(x = 0,\ 2\) or \(-2\). And a factor \((x + 2)\) gives the root \(x = -2\), not \(x = 2\).

Section 02

When the quadratic factor will not factorise.

The quadratic factor need not split into nice brackets. Check its discriminant \(b^{2} - 4ac\) before you hunt for factors: it tells you how many more roots there are. Paper 1 is non-calculator, so surd roots are left in exact form.

Worked example Exact surd roots
Solve \(x^{3} - 5x^{2} + 5x + 3 = 0\), giving your answers in exact form.
  1. Factors of \(3\): try \(x = 3\): \(27 - 45 + 15 + 3 = 0\), so \((x - 3)\) is a factor.
  2. Divide: \(x^{3} - 5x^{2} + 5x + 3 = (x - 3)(x^{2} - 2x - 1)\).
  3. \(x^{2} - 2x - 1\): \(b^{2} - 4ac = 4 + 4 = 8\), not a perfect square, so use the formula.
  4. \(x = \dfrac{2 \pm \sqrt{8}}{2} = \dfrac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}\).
Answer: \(x = 3\), \(x = 1 + \sqrt{2}\) or \(x = 1 - \sqrt{2}\)
Worked example Only one real root
Show that the equation \(x^{3} + x^{2} - 4x + 6 = 0\) has exactly one real root, and state it.
  1. Factors of \(6\): \(p(1) = 4\), \(p(-1) = 10\), \(p(2) = 10\), \(p(-2) = 10\), \(p(-3) = -27 + 9 + 12 + 6 = 0\).
  2. So \((x + 3)\) is a factor: \(x^{3} + x^{2} - 4x + 6 = (x + 3)(x^{2} - 2x + 2)\).
  3. \(x^{2} - 2x + 2 = 0\): \(b^{2} - 4ac = 4 - 8 = -4 < 0\), so it has no real roots.
  4. The only real root comes from \(x + 3 = 0\).
Answer: Exactly one real root: \(x = -3\)
Your turn
Solve \(x^{3} - 3x + 2 = 0\) and say which root is repeated.
Examiner trap

A negative discriminant does not mean 'no solution': the root you found first is still a real answer, so the cubic has exactly one real root. With the formula, divide the WHOLE top by \(2a\): \(\dfrac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}\), never \(2 \pm \sqrt{2}\) or \(2 \pm 2\sqrt{2}\).

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Quick check+10 XP

Which of these is a factor of \(x^{3} - 4x^{2} + x + 6\)?

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\(x^{3} - 2x^{2} - 3x + 10 = (x + 2)(x^{2} - 4x + 5)\). How many real roots does \(x^{3} - 2x^{2} - 3x + 10 = 0\) have?

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Solve \(x^{3} = 9x\).

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What are the roots of \((x - 2)(x^{2} - 4x + 1) = 0\)?

Section 03

Show, factorise, hence solve.

Exam questions build in steps. 'Show that \((x - a)\) is a factor' means work out \(p(a)\) and show it is zero. 'Factorise completely' means a product of linear factors, or one linear factor times a quadratic that will not factorise. 'Hence' means you must use the work you have just done: an equation that is the same cubic in disguise (with \(x\) replaced by \(2x\), say, or by \(e^{x}\)) has the same roots, now as values of \(2x\) or \(e^{x}\) instead of \(x\). In a real-world context, keep only the roots that make sense.

Worked example A full exam-style question
\(p(x) = 2x^{3} - 3x^{2} - 11x + 6\). (a) Show that \((x - 3)\) is a factor of \(p(x)\). (b) Factorise \(p(x)\) completely. (c) Hence solve \(p(x) = 0\). (d) Hence solve \(16y^{3} - 12y^{2} - 22y + 6 = 0\).
  1. (a) \(p(3) = 54 - 27 - 33 + 6 = 0\), so by the factor theorem \((x - 3)\) is a factor.
  2. (b) Divide: \(p(x) = (x - 3)(2x^{2} + 3x - 2) = (x - 3)(2x - 1)(x + 2)\).
  3. (c) \(x = 3\), \(x = \dfrac{1}{2}\) or \(x = -2\).
  4. (d) Notice that \(16y^{3} - 12y^{2} - 22y + 6 = 2(2y)^{3} - 3(2y)^{2} - 11(2y) + 6 = p(2y)\).
  5. So \(2y = 3,\ \dfrac{1}{2}\) or \(-2\), giving \(y = \dfrac{3}{2},\ \dfrac{1}{4}\) or \(-1\).
  6. In the same way, \(2e^{3t} - 3e^{2t} - 11e^{t} + 6 = 0\) is \(p(e^{t}) = 0\), so \(e^{t} = 3\) or \(\dfrac{1}{2}\) (reject \(e^{t} = -2\), since \(e^{t} > 0\)), giving \(t = \ln 3\) or \(t = -\ln 2\).
Answer: (b) \((x - 3)(2x - 1)(x + 2)\)   (c) \(x = -2,\ \frac{1}{2},\ 3\)   (d) \(y = -1,\ \frac{1}{4},\ \frac{3}{2}\)
Worked example Volume of a box
An open box is made from a rectangular card 10 cm by 16 cm by cutting a square of side \(x\) cm from each corner and folding up the sides. The volume of the box is 112 cm3. Show that \(x^{3} - 13x^{2} + 40x - 28 = 0\) and find the possible values of \(x\), in exact form.
  1. The base is \((16 - 2x)\) by \((10 - 2x)\) and the height is \(x\), so \(V = x(10 - 2x)(16 - 2x) = 4x^{3} - 52x^{2} + 160x\).
  2. \(4x^{3} - 52x^{2} + 160x = 112\); divide by 4 and rearrange: \(x^{3} - 13x^{2} + 40x - 28 = 0\).
  3. \(p(1) = 1 - 13 + 40 - 28 = 0\), so \(p(x) = (x - 1)(x^{2} - 12x + 28)\).
  4. \(x = \dfrac{12 \pm \sqrt{144 - 112}}{2} = \dfrac{12 \pm 4\sqrt{2}}{2} = 6 \pm 2\sqrt{2}\).
  5. The card is only 10 cm wide, so \(0 < x < 5\). \(6 + 2\sqrt{2}\) is more than 5, so reject it. \(6 - 2\sqrt{2}\) is between 3 and 4 (since \(2\sqrt{2} = \sqrt{8}\) is between 2 and 3), so keep it.
Answer: \(x = 1\) or \(x = 6 - 2\sqrt{2}\)
Your turn
Show that \((x + 1)\) is a factor of \(p(x) = x^{3} - 4x^{2} + x + 6\), and hence solve \(p(x) = 0\).
Examiner trap

'Hence' is an instruction, not a hint: start from the roots you already have, or you lose the marks even with correct answers. Undo the substitution at the end (\(2y = 3\) gives \(y = \frac{3}{2}\), not \(y = 3\)), and reject only what is impossible: \(e^{t}\) is never negative, and a length must fit the card. Do not reject a negative root of an ordinary equation just because it is negative.

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\((x - 2)\) is a factor of \(p(x) = 2x^{3} + ax^{2} - 7x - 6\). Find \(a\) and solve \(p(x) = 0\).

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Solve \(x^{3} - x^{2} - 7x + 3 = 0\), giving your answers in exact form.

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\(x^{3} - 7x + 6 = (x - 1)(x - 2)(x + 3)\). Hence solve \(e^{3t} - 7e^{t} + 6 = 0\).

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A cuboid has edges \(x\) cm, \((x + 2)\) cm and \((x - 1)\) cm and volume 72 cm3. Find \(x\).

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One root found, the cubic falls apart.

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Six original Cambridge-style questions
Three-root cubics, a leading coefficient of 2, exact surd roots, a cubic with only one real root, a full show-factorise-hence question and a box whose volume gives a repeated root.