Equations
Cubic and Higher-Degree Equations
Finding one root, dividing it out, and finishing with the quadratic that remains — plus what to do when the cubic has no rational root.
What this page covers
- Solve a cubic by finding a root and reducing the degree
- Recognise special forms such as a difference of cubes
- Know how many roots to expect and in what kinds of pairs
- Say when the method fails and what remains
The strategy
There is no elementary formula worth using for a cubic. The practical method reduces the degree instead: find one root, divide out the corresponding factor, and solve the quadratic that remains.
Occasionally step 1 can be skipped because the cubic is already in a recognisable special form, which is where the source starts.
Special forms first
Worked example — a difference of cubes
Solve x3 - 8 = 0.
The zero-product property splits it. The linear factor gives x = 2. The quadratic has Δ = 4 - 16 = -12 < 0, so its roots are complex:
Three roots: x = 2 and the conjugate pair -1 ± i√3. The source notes 'no real solution' for the quadratic before giving the complex pair, which is the right order — the real question is answered first.
Sum should be 0 (no x2 term): 2 + (-1 + i√3) + (-1 - i√3) = 0. The imaginary parts cancel, as a conjugate pair must.
The general method
Worked example — x3 + 4x2 + 2x - 4 = 0
The constant is -4, so test ±1, ±2, ±4.
So (x + 2) is a factor. Dividing gives x2 + 2x - 2, which does not factor over the integers. The source completes the square:
Sum should be -4: -2 - 1 + √3 - 1 - √3 = -4. Product should be -(-4) = 4: -2 × ((-1)2 - 3) = -2 × (-2) = 4. Both hold.
Worked example — x3 - 5x2 + 7x - 12 = 0
Testing divisors of -12 finds f(4) = 64 - 80 + 28 - 12 = 0. Dividing out gives x2 - x + 3, whose discriminant is 1 - 12 = -11 < 0.
One real root and a complex pair — the commonest shape for a cubic that does not factor completely over the reals.
How many roots, and of what kind
A polynomial of degree n has exactly n roots over the complex numbers, counted with multiplicity. For a cubic with real coefficients that leaves only two possible shapes.
| Case | Real roots | Complex roots | Graph |
|---|---|---|---|
| Three real | 3 (possibly repeated) | 0 | Crosses or touches the axis three times, counted with multiplicity |
| One real | 1 | 2, a conjugate pair | Crosses the axis exactly once |
There is no third case. Complex roots of a real polynomial come in conjugate pairs, so their number is even; three minus an even number is odd, and so at least one real root always survives. That is why every cubic graph crosses the axis.
The same argument shows that every polynomial of odd degree with real coefficients has a real root. It is also why irrational roots arrive in conjugate surd pairs: in -1 ± √3 the radical enters with both signs.
When the method fails
Trial only finds rational roots. If a cubic has three irrational real roots, every candidate fails and the method produces nothing, despite the equation being perfectly solvable.
| Situation | Response |
|---|---|
| No rational root found | Use a numerical method; the graph shows where the roots lie |
| Coefficients are fractions | Multiply through to clear them, then the divisor argument applies |
| Leading coefficient not 1 | Also test ±pq with p | a0 and q | an |
| Degree 4 or higher | Same method; each division drops the degree by one |
A failed search is not proof that there are no real roots. It proves only that there are no rational ones. Sketching the graph, or evaluating at a few points to find a sign change, locates the real roots even when trial finds none.
Common mistakes
| Mistake | Correct | Why |
|---|---|---|
| Stopping after one root | Divide out and solve the quotient | A cubic has three roots |
| Testing f(2) for the factor x + 2 | Test f(-2) | x + 2 = x - (-2) |
| Concluding no real roots after a failed trial | Every real cubic has one | The root may be irrational |
| Missing the fractional candidates when an ≠ 1 | Test ±pq | 2x3 + 5x2 - 14x + 3 has root 32 |
| Reporting only real roots when complex were asked for | State both | Read the question's number system |
Frequently asked questions
How many roots does a cubic have?
Exactly three, counted with multiplicity, once complex roots are admitted. At least one is always real, because complex roots of a real polynomial arrive in conjugate pairs and three cannot be paired up.
What if I cannot find any root by trial?
Then there is no rational root, which the trial method is limited to. The equation still has three roots; finding them needs a numerical method or the cubic formula, neither of which is elementary.
Is there a formula like the quadratic one?
Yes, but it is long, and for a cubic with three real roots it routes through complex numbers even so. In practice degree reduction is the method used.
Does the method extend to quartics?
Yes, identically: find a root, divide out, and you are left with a cubic. Repeat. Each successful division drops the degree by one.
Source. Handwritten teaching notes, Week 3, page 7, and Week 7, page 1.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates, reorganises and verifies the mathematics of the supplied teaching notes; it is not a reproduction of them. Numerical values taken from the notes are identified as source examples and are not presented as engineering standards.