A cubic equation is any equation where the highest power of the variable is 3, written in the standard form ax³ + bx² + cx + d = 0 (where a ≠ 0). Solving one is more involved than solving a quadratic, but there are several reliable methods depending on the equation’s complexity. The approach you choose depends on whether the equation has “nice” rational roots, whether you need exact answers, or whether a decimal approximation is good enough.
Check for Rational Roots First
Before reaching for any formula, try the simplest approach: test whether the equation has roots that are whole numbers or simple fractions. The Rational Root Theorem tells you that any rational root of the equation ax³ + bx² + cx + d = 0 must be a factor of d divided by a factor of a. For example, if your equation is x³ − 6x² + 11x − 6 = 0, the possible rational roots are ±1, ±2, ±3, and ±6. Plug each one in, and you’ll find that x = 1 works.
Once you find a single root, you can reduce the cubic to a quadratic. If x = 1 is a root, then (x − 1) is a factor. Use synthetic division or polynomial long division to divide out that factor, and you’re left with a quadratic that you can solve with the standard quadratic formula. This “find one root, then divide” strategy is by far the most common method taught in algebra courses, and it works cleanly whenever at least one root is rational.
Synthetic Division in Practice
Synthetic division is a shortcut for dividing a polynomial by a linear factor (x − c). You write down the coefficients of the cubic, bring down the first one, then repeatedly multiply and add. If the remainder comes out to zero, you’ve confirmed that (x − c) is a factor. The remaining coefficients form your quadratic quotient. For instance, dividing x³ − 6x² + 11x − 6 by (x − 1) gives you x² − 5x + 6, which factors neatly into (x − 2)(x − 3). So the three roots are 1, 2, and 3.
The Depressed Cubic and Cardano’s Formula
When no rational root exists, you need a general algebraic method. The classic approach, developed in the 16th century, works in two stages.
First, eliminate the x² term. Substitute y = x + b/(3a) into the original equation ax³ + bx² + cx + d = 0. This always produces a simpler equation called the “depressed cubic,” which has the form y³ + py + q = 0 with no squared term. This substitution is just a horizontal shift of the equation, and it makes the algebra that follows much cleaner.
Next, solve the depressed cubic using a clever trick. Let y = u + v, and substitute into y³ + py + q = 0. After expanding, you can set 3uv + p = 0 as an additional constraint, which collapses the equation into a system:
- u³ + v³ = −q
- u³ · v³ = −p³/27
These two equations tell you the sum and product of u³ and v³, which means u³ and v³ are themselves roots of a quadratic equation: t² + qt − p³/27 = 0. Solve that with the quadratic formula, take cube roots to get u and v, and then y = u + v gives you a root of the depressed cubic. Shift back by subtracting b/(3a) to get the root of your original equation.
This process, known as Cardano’s method, always produces at least one root. The remaining two roots can then be found by dividing out the known factor and solving the resulting quadratic, or by carefully choosing the cube roots of u³ and v³ (there are three cube roots of each, but only specific pairings satisfy the constraint 3uv = −p).
Use the Discriminant to Know What to Expect
Before solving, you can predict what kind of roots you’ll get. For the depressed cubic y³ + py + q = 0, the discriminant is:
D = −4p³ − 27q²
What D tells you:
- D > 0: Three distinct real roots
- D = 0: All roots are real, but at least two are equal
- D < 0: One real root and two complex conjugate roots
This matters because when D > 0, Cardano’s formula runs into an awkward situation called the “casus irreducibilis.” The formula technically works, but it expresses real numbers using intermediate complex numbers (you’ll see square roots of negative numbers inside cube roots). The final answer is real, but getting there through Cardano’s formula requires complex arithmetic. In that case, a trigonometric method is cleaner.
The Trigonometric Method for Three Real Roots
When all three roots are real (D > 0), a trigonometric approach avoids complex numbers entirely. Start with the depressed cubic y³ + py + q = 0, where p must be negative for this case to apply. Write p = −3k², then transform the equation so it looks like 4z³ − 3z = e, where z = y/(2k) and e = −q/(2k³).
The key insight is that the identity cos(3α) = 4cos³(α) − 3cos(α) has exactly the same structure as 4z³ − 3z = e. So if |e| ≤ 1, you set e = cos(3α), which means z = cos(α) is a solution. The other two solutions come from the fact that cosine repeats with period 360°:
- z₁ = cos(α)
- z₂ = cos(120° − α)
- z₃ = cos(120° + α)
Then convert back to y by multiplying each z by 2k, and shift back to x by subtracting b/(3a). This method gives you all three real roots directly using a calculator’s cosine and inverse cosine functions, no complex numbers required. The condition |e| ≤ 1 is guaranteed whenever 4p³ + 27q² ≤ 0, which is precisely the case when all three roots are real.
Numerical Approximation With Newton’s Method
In practice, especially when you just need a decimal answer, numerical methods are the fastest route. Newton’s method (also called Newton-Raphson) starts with an initial guess x₀ and improves it repeatedly using the formula:
x(n+1) = x(n) − f(x(n)) / f′(x(n))
For the cubic f(x) = ax³ + bx² + cx + d, the derivative is f′(x) = 3ax² + 2bx + c. Pick a starting value (graphing the function helps you choose one near a root), then iterate. Each step typically doubles the number of correct decimal places, so you’ll have an accurate answer in just a few iterations.
For example, to solve x³ − 2x − 5 = 0, you might start with x₀ = 2 (since f(2) = −1 and f(3) = 16, so a root lies between 2 and 3). One iteration gives x₁ = 2 − (−1)/10 = 2.1. Another gives roughly 2.0946. Within four or five steps, you’ll have the root to many decimal places. Once you’ve found one real root this way, divide it out and solve the remaining quadratic for the other two.
Relationships Between Roots and Coefficients
Even without solving, you can extract useful information from the coefficients. For ax³ + bx² + cx + d = 0 with roots r₁, r₂, and r₃, three relationships always hold:
- Sum of roots: r₁ + r₂ + r₃ = −b/a
- Sum of pairwise products: r₁r₂ + r₂r₃ + r₃r₁ = c/a
- Product of all three roots: r₁r₂r₃ = −d/a
These are called Vieta’s formulas. They’re particularly handy for checking your work: after finding all three roots, verify that their sum, pairwise products, and total product match the ratios above. They also let you reconstruct a cubic from its roots, or find a missing root if you already know two.
Complex Roots Always Come in Pairs
A cubic equation with real coefficients always has at least one real root. The other two roots are either both real or both complex. When they’re complex, they always appear as conjugate pairs: if a + bi is a root, then a − bi is also a root. For example, x³ − 27 = 0 has the real root x = 3, and its other two roots are 3(−1/2 + i√3/2) and 3(−1/2 − i√3/2), which are conjugates of each other.
This conjugate pairing isn’t a coincidence. It’s a consequence of having real coefficients, and it holds for polynomials of any degree. For cubics specifically, it means you’ll never encounter a situation with exactly two real roots and one complex root. It’s always three real roots or one real root plus a conjugate pair.
Choosing the Right Method
For homework and competition problems, start by testing rational roots. Most textbook cubics are designed to have at least one. If you find it, divide and solve the quadratic. For equations with no rational roots, Cardano’s method gives an exact algebraic answer, but the trigonometric method is cleaner when all three roots are real. For applied problems in science or engineering where you just need a number, Newton’s method on a calculator or computer gets you there in seconds. Graphing the cubic first, even roughly, helps you see how many real roots to expect and where they approximately lie, regardless of which solving method you choose.