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Cubic Equation Solver

Cubic Equation Solver

Depressed Cubic solver

Find real and complex roots, discriminant, and cubic graphs in seconds with our precision solver.

The Equation

Input your cubic equation's coefficients to eliminate the x² term and obtain the depressed form t³ + pt + q = 0.

f(x) = 0 = 0

Enter coefficients and solve to view the calculation, graph, and step-by-step explanation.

The Solution

Calculation: Depressed Cubic Calculator

Depressed Cubic Calculator

Enter your polynomial coefficients above and click "Depress Cubic" to see results.
Graph will appear here after you solve.

The Treatise

What isDepressed Cubic Calculator?

  • For a school student: A depressed cubic is a cubic equation with its "bulge" removed. The ordinary cubic ax³ + bx² + cx + d = 0 has an awkward x² term pushing its graph sideways. Shifting the graph left or right by exactly the right amount centers it, leaving the cleaner form t³ + pt + q = 0. Same answers, shifted; much easier to work with.
  • For a college student: The substitution x = t − b/3a eliminates the quadratic term of any cubic with a ≠ 0. The new coefficients are p = (3ac − b²)/3a² and q = (2b³ − 9abc + 27a²d)/27a³. This is the mandatory first step of Cardano's method, and it also simplifies the discriminant to Δ = −(4p³ + 27q²).
  • For the advanced reader: This is the cubic case of a Tschirnhaus transformation (1683): a polynomial change of variables that kills a chosen coefficient. The shift −b/3a is exactly the average of the three roots, so depressing centers the equation at the mean of its roots, which is also the inflection point of the graph. The general theory extends to quartics and, in the Bring-Jerrard form, to quintics, though Abel-Ruffini limits what radicals can then achieve.
  • Why it matters in practice: Never attack a general cubic directly when a two-line shift removes a quarter of its complexity. Depress first, solve the simpler equation with Cardano's method or our cubic roots calculator, then shift the answers back.

The Formula

Formula / Method

  • Substitution: x = t − b/3a converts ax³ + bx² + cx + d = 0 into t³ + pt + q = 0.
  • New coefficients: p = (3ac − b²)/3a² and q = (2b³ − 9abc + 27a²d)/27a³.
  • Simplified discriminant: In depressed form, Δ = −(4p³ + 27q²), two terms instead of five. Our discriminant calculator uses exactly this shortcut internally.
  • A common trap: The roots shift too. If t₀ solves the depressed cubic, then x₀ = t₀ − b/3a solves the original. Solving the easy equation and forgetting to shift back is the classic error.

The Method

How To Use

  1. Step 1. Write your cubic as ax³ + bx² + cx + d = 0 and read off a, b, c, d.
  2. Step 2. Enter the four coefficients into the tool.
  3. Step 3. Press "Depress Cubic" to get p, q, and the shifted equation t³ + pt + q = 0.
  4. Step 4. Solve the depressed form with Cardano's method, then shift each root back via x = t − b/3a.

Key Features

  • Computes the exact shift −b/3a and the reduced coefficients p and q.
  • Shows the full substitution algebra so you can follow every step.
  • Displays the depressed equation ready for Cardano's method.
  • Verifies the shift by expanding back to your original cubic.
  • Handles fractional and negative coefficients without rounding the shift.

The Illustration

Visual Diagram

Before (General Form) ax³ + bx² + cx + d = 0 4 coefficients: a, b, c, d Inflection point off-center Harder to solve directly → x = t − b/(3a) After (Depressed Form) t³ + pt + q = 0 2 parameters: p, q Inflection point at origin Ready for Cardano / trig
Example Concept
Depress x³ − 6x² + 11x − 6 = 0. Here a = 1, b = −6, c = 11, d = −6. Shift: x = t − (−6)/3 = t + 2. p = (3·1·11 − 36)/3 = (33 − 36)/3 = −1 q = (2(−216) − 9(1)(−6)(11) + 27(1)(−6))/27 = (−432 + 594 − 162)/27 = 0 The depressed cubic is t³ − t = 0, which factors instantly as t(t − 1)(t + 1) = 0. Shifting back (x = t + 2) gives roots x = 2, 3, 1. Confirm them with our cubic roots calculator.

The Deep Dive

Interactive Deep Dive

The transformation is named for Ehrenfried Walther von Tschirnhaus, who published it in 1683 as part of a grand program to solve all polynomial equations by clever substitutions. His program ultimately failed for the quintic, but the cubic case survived as an indispensable tool. The idea is disarmingly simple: because the x² coefficient is, up to a factor, the sum of the roots, shifting the variable by the average root centers the whole configuration at the origin. What looks like an algebraic trick is really a geometric recentering.

Why does x = t − b/3a kill exactly the quadratic term? Expand a(t − b/3a)³ + b(t − b/3a)² + c(t − b/3a) + d and collect the t² coefficient: from the cubic term comes 3a(−b/3a) = −b, from the quadratic term comes +b. They cancel exactly, and no other choice of shift does this. The t and constant terms then assemble themselves into p and q. It is worth doing this expansion once by hand: the cancellation feels like watching a lock click open.

The formulas p = (3ac − b²)/3a² and q = (2b³ − 9abc + 27a²d)/27a³ look intimidating, but each has meaning. The numerator of p is, up to sign, the discriminant of the quadratic part, measuring how "spread" the equation is. The expression for q mixes all four coefficients because the constant term of the shifted equation genuinely depends on everything. Notice the denominators are powers of a: for a monic cubic they vanish into p = 3c − b² over 3 style simplicity, which is why textbooks usually present the monic case first.

Depressing also centers the graph. The inflection point of ax³ + bx² + cx + d sits at x = −b/3a, precisely the shift amount, so the depressed cubic's graph is symmetric about its inflection point at the origin. Our inflection point calculator locates this center, and the graphing tool shows the before-and-after: the same S-curve, slid horizontally until its midpoint sits on the y-axis. This symmetry is why the depressed form is easier to analyze qualitatively.

The same spirit removes terms from higher-degree equations: a quartic can be depressed to t⁴ + pt² + qt + r = 0, which is Ferrari's entry point. For the quintic, the Bring-Jerrard transformation reduces it to t⁵ + pt + q = 0, yet no radical formula finishes the job, as Abel and Ruffini proved. The depressed cubic is thus the sweet spot of the whole theory: the hardest equation that a shift plus radicals can completely conquer.

The Applications

Real-World Applications

📚

Cardano Preparation

Every application of Cardano's method begins here: no depressed form, no formula. The reduction is so standard that computer algebra systems perform it silently before displaying Cardano's steps. Mastering this one transformation opens up the entire classical solution theory of the cubic.

🎨

Graph Analysis

Centering a cubic at its inflection point makes sketching dramatically easier: the turning points become symmetric about the origin, and the number of real roots can be read from the values at those symmetric points. Graphing tools use this internally when choosing viewing windows.

⚙

Numerical Stability

When coefficients vary wildly in magnitude, the shift often produces smaller, better-balanced p and q. Numerical solvers converge faster and roundoff error shrinks, because the computation no longer mixes huge and tiny terms. Preconditioning by depression is a quiet workhorse of numerical libraries.

🏫

Teaching Algebra

Instructors use depression to split a frightening problem into two manageable ones: a routine substitution, then a cleaner equation. Students who can depress a cubic on sight have internalized the relationship between coefficients and roots that Vieta's formulas describe.

⚡

Physics Coordinates

Physicists routinely shift coordinates to remove linear or quadratic terms: completing the square is the quadratic version of this exact move. In perturbation theory and normal-mode analysis, centering the coordinate system at the natural midpoint simplifies every subsequent expansion.

The Cautions

Common Mistakes to Avoid

MISTAKE Nº 1

Forgetting to shift back: The depressed cubic has different roots than the original: each is shifted by −b/3a. Solving t³ − t = 0 gives t = 0, ±1, but the original equation's roots are x = 2, 3, 1. Always apply x = t − b/3a at the end.

MISTAKE Nº 2

Sign error in the shift: The substitution is x = t − b/3a, with b carrying its own sign. For x³ − 6x² + …, b = −6, so the shift is t + 2, not t − 2. Write b with its sign before substituting.

MISTAKE Nº 3

Arithmetic slips in p and q: The q formula has three signed terms and powers up to b³. It is the most error-prone hand computation in elementary cubic theory. Let the tool compute them, or check by expanding back.

MISTAKE Nº 4

Dividing by a zero leading coefficient: If a = 0 the equation is not cubic and the formulas divide by zero. Confirm the degree is genuinely three before depressing.

MISTAKE Nº 5

Mixing discriminant formulas: The five-term discriminant belongs to the general cubic; the two-term version Δ = −(4p³ + 27q²) belongs to the depressed one. Applying the short formula to unshifted coefficients gives a wrong sign and wrong verdict.

The Reference

Quick Reference Table

Quick Reference TableREF
Substitutionx = t − b/(3a)
Result Formt³ + pt + q = 0
p formula(3ac − b²) / (3a²)
q formula(2b³ − 9abc + 27a²d) / (27a³)
Geometric EffectCenters inflection point on y-axis

The Questions

Frequently Asked Questions

What is a depressed cubic? +

A cubic with no x² term: t³ + pt + q = 0. Every cubic ax³ + bx² + cx + d = 0 can be converted to this form by the shift x = t − b/3a.

Why depress a cubic at all? +

Because every classical solution method, especially Cardano's method, requires the depressed form. It also simplifies the discriminant and centers the graph at its inflection point.

Can every cubic be depressed? +

Yes, provided the leading coefficient a is nonzero. The shift −b/3a always exists and always kills the quadratic term exactly.

What are p and q? +

Fixed combinations of the original coefficients: p = (3ac − b²)/3a² and q = (2b³ − 9abc + 27a²d)/27a³. They carry all the equation's information in compressed form.

Is the shift the only one that works? +

Yes. Expanding a(t + h)³ + b(t + h)² + … shows the t² coefficient vanishes only when h = −b/3a. Any other shift leaves a quadratic term behind.

Does depression change the roots? +

It shifts each root by −b/3a but preserves their relationships: differences, multiplicities, and reality are unchanged. See how roots relate with our roots relationship calculator.

Can I solve the depressed cubic directly? +

Yes, with Cardano's formula, which was designed for exactly this form. Many depressed cubics also factor nicely by inspection.

What happens to the discriminant? +

It collapses from five terms to two: Δ = −(4p³ + 27q²). Verify the classification with our discriminant calculator.