Solving Cubic Equations

Solve cubic equations by finding one root with the factor theorem, dividing to a quadratic, and solving it, including surd roots, one-real-root proofs and hence questions.

In the exam for Edexcel Further Pure Maths 4PM1, IGCSE Additional Maths 0606 · 22 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

The plan for a cubic

A cubic equation has an $x^{3}$ term as its highest power, like $x^{3} - 2 x^{2} - 5 x + 6 = 0$ . It can have up to three real roots.

It always has at least one. The graph of a cubic starts on one side of the x-axis and ends on the other, so it must cross the axis at least once.

Rule

The plan never changes: find one root, divide it out, then solve the quadratic that is left.

Steps

1
1. Find a value a with $P (a) = 0$ . By the factor theorem, $x - a$ is a factor.
2. Divide by $x - a$ to get a quadratic factor.
3. Factorise the quadratic, or use the quadratic formula.
4. Set each factor equal to zero and list every root.

Key idea

In 1545 the Italian scholar Gerolamo Cardano published a formula that solves any cubic. It is long and messy, so nobody uses it in an exam. The factor theorem gets you there in a few lines.

Finding the first root

Where do you look for the first root? If the cubic starts with $x^{3}$ and has whole-number coefficients, any whole-number root must divide the number term exactly.

Rule

Try the divisors of the number term, positive and negative: for a number term of 6, try $\pm 1, \pm 2, \pm 3, \pm 6$ . Start with the small ones.

Worked example

1
Solve $x^{3} - 2 x^{2} - 5 x + 6 = 0$ .
2
$P (1) = 1 - 2 - 5 + 6 = 0$ — Try x = 1 first. It gives 0, so x = 1 is a root.
3
$(x - 1) is a factor$ — Factor theorem.
4
$(x - 1) (x^{2} - x - 6) = 0$ — Divide by x − 1, by long division or by comparing coefficients.
5
$(x - 1) (x - 3) (x + 2) = 0$ — Factorise the quadratic: −3 and 2 multiply to −6 and add to −1.
6
$x = 1, x = 3, x = - 2$ — Set each bracket equal to zero.

Common mistake

Stopping at the brackets. Solve means give the values of x, so finish with $x = 1, 3, - 2$ .

Key idea

Small divisors of the number term first. One success is all you need.

When x³ has a number in front

When the cubic starts with a number other than 1, such as $2 x^{3}$ , the same plan works. Whole-number roots still divide the number term. A root may also be a fraction, so if no whole number works, try fractions like $\pm \frac{1}{2}$ or $\pm \frac{3}{2}$ .

Worked example

1
Solve $2 x^{3} + x^{2} - 13 x + 6 = 0$ .
2
$P (1) = 2 + 1 - 13 + 6 = - 4$ — Not zero, so x = 1 is not a root. Try again.
3
$P (2) = 16 + 4 - 26 + 6 = 0$ — x = 2 works, so (x − 2) is a factor.
4
$(x - 2) (2 x^{2} + 5 x - 3) = 0$ — Divide by x − 2. The quadratic keeps the 2 in front.
5
$(x - 2) (2 x - 1) (x + 3) = 0$ — Factorise the quadratic: (2x − 1)(x + 3) = 2x² + 5x − 3.
6
$x = 2, x = \frac{1}{2}, x = - 3$ — Each bracket equal to zero. 2x − 1 = 0 gives one half.

Mark schemes give marks for finding a factor, for the quadratic factor, for solving, and for the roots. When a question says “use algebra”, correct roots with no working can score nothing, even on the 4PM1 calculator papers.

Common mistake

Losing the first root. The answer has three values: the one you found by trial, and the two from the quadratic.

Rule

0606 Paper 1 has no calculator, so test values by hand. Write each test line out: it also earns method marks.

Key idea

A cubic with a whole-number root is a quadratic you have not met yet. One division introduces you.

Surd roots

Sometimes the quadratic factor will not factorise. Then use the quadratic formula on it, and give the roots as exact surds.

Formula

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

$a$ — the number in front of x² in the quadratic factor.
$b$ — the number in front of x.
$c$ — the number term.

Rule

0606 and 4037 print this formula on the paper. 4PM1 does not: learn it.

Worked example

1
Solve $x^{3} - 3 x^{2} - 2 x + 4 = 0$ , giving exact answers.
2
$P (1) = 1 - 3 - 2 + 4 = 0$ — x = 1 is a root, so (x − 1) is a factor.
3
$(x - 1) (x^{2} - 2 x - 4) = 0$ — Divide by x − 1.
4
$x = \frac{2 \pm 4 + 16}{2}$ — x² − 2x − 4 does not factorise, so use the formula: a = 1, b = −2, c = −4.
5
$= \frac{2 \pm 2 5}{2} = 1 \pm 5$ — √20 = √4 × √5 = 2√5. Then divide both terms by 2.
6
$x = 1, x = 1 + 5, x = 1 - 5$ — All three exact roots.

Common mistake

Dividing only the 2 by 2. Both parts of the top are divided: $\frac{2 \pm 2 5}{2} = 1 \pm 5$ .

Key idea

Exact means surds, not decimals. On a non-calculator paper this is the only way to finish.

Only one real root

Another favourite: show that a cubic equation has only one real root. Find the root you can, divide, then show the quadratic factor has no real roots.

The part under the square root, $b^{2} - 4 a c$ , decides this. If it is negative, there is no real square root, so the quadratic has no real roots.

Worked example

1
Show that $x^{3} + x^{2} + x - 3 = 0$ has only one real root.
2
$P (1) = 1 + 1 + 1 - 3 = 0$ — x = 1 is a root.
3
$(x - 1) (x^{2} + 2 x + 3) = 0$ — Divide by x − 1.
4
$b^{2} - 4 a c = 2^{2} - 4 (1) (3) = - 8$ — Work out b² − 4ac for the quadratic factor: a = 1, b = 2, c = 3.
5
$- 8 < 0$ — Negative, so x² + 2x + 3 = 0 has no real roots.
6
$Only real root: x = 1$ — Write the conclusion in words.

Common mistake

Working out $b^{2} - 4 a c$ and stopping. The mark needs the value, the comparison with 0, and the conclusion.

Key idea

One root from the bracket, none from the quadratic: one real root in total.

Hence: cubics in disguise

A question may solve a cubic, then say “hence solve” a longer equation. Look closely: the longer one is often the same cubic with $u^{2}$ written in place of x.

For example, $2 u^{6} + u^{4} - 13 u^{2} + 6 = 0$ is $2 x^{3} + x^{2} - 13 x + 6 = 0$ with $x = u^{2}$ , because $u^{6} = (u^{2})^{3}$ and $u^{4} = (u^{2})^{2}$ .

Worked example

1
You know $2 x^{3} + x^{2} - 13 x + 6 = (x - 2) (2 x - 1) (x + 3)$ . Hence solve $2 u^{6} + u^{4} - 13 u^{2} + 6 = 0$ .
2
$(u^{2} - 2) (2 u^{2} - 1) (u^{2} + 3) = 0$ — Put u² in place of x in each factor.
3
$u^{2} = 2 \Rightarrow u = \pm 2$ — A square has two square roots: positive and negative.
4
$u^{2} = \frac{1}{2} \Rightarrow u = \pm \frac{1}{2} = \pm \frac{2}{2}$ — Rationalise: multiply top and bottom by √2.
5
$u^{2} = - 3$ — No real solution: a square is never negative.
6
$u = \pm 2, \pm \frac{2}{2}$ — Four real solutions.

Common mistake

Forgetting the minus sign. $u^{2} = 2$ has two solutions, $2$ and $- 2$ .

Rule

Reject any value the new letter cannot take. A square cannot be negative, and if the cubic uses $e^{2 y}$ in place of x, a negative value is rejected too, because $e^{2 y}$ is always positive.

Key idea

Hence means: use what you just found. Swap the letter back, and check each value can really happen.

Which method? Read the question

Read the question first. The wording tells you how far to go.

You see	You do
“Solve” a cubic, no factor given	Try divisors of the number term until $P (a) = 0$
A factor is given	Divide by it straight away
Number in front of $x^{3}$ is not 1	Same plan; try fractions if no whole number works
Quadratic factor will not factorise	Quadratic formula; give exact surds
“Only one real root”	Show $b^{2} - 4 a c < 0$ for the quadratic factor, then conclude
“Hence solve” with $u^{6}, u^{4}, u^{2}$	Put $u^{2}$ in place of x; take ± square roots; reject negatives
“Use algebra”	Show every stage; solver answers alone score nothing

Common mistake

Giving the factorised form when the question asks for the roots, or the roots when it asks for the factorised form. Answer exactly what is asked.

Key idea

Find one, divide once, solve the quadratic. Every cubic in this course yields to those three moves.

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Real past-paper questions on solving cubic equations

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