Everything You Need to Know About Cubic Equations

A cubic equation is a third-degree polynomial written in standard cubic form, where the leading coefficient cannot be zero.

Yes. If the equation has one real root and a complex-conjugate pair, the results section shows them clearly and labels them as complex.

If a = 0, the equation is no longer cubic. The UI validates this immediately and explains why the solver cannot proceed.

It summarizes the normalized equation, depressed cubic transformation, Cardano's discriminant D, and final interpretation so the solver feels more transparent.

No. It tells you the type of roots to expect. For the values themselves, use our cubic roots calculator.

Yes, and it is informative when it happens. A zero discriminant means at least two roots coincide. Try x³ − 3x² + 3x − 1 = (x − 1)³ to see a triple root in action.

Because it must be symmetric in all four coefficients while encoding the product of squared root differences. The depressed cubic form Δ = −(4p³ + 27q²) is much shorter; our depressed cubic calculator handles the conversion.

No, that is quadratic thinking. Every cubic has at least one real root. A negative discriminant means exactly one real root plus a complex conjugate pair, which the complex roots calculator finds.

The discriminant is computed from the same p and q that appear in Cardano's formula, so it predicts which branch of the solution you will land on. See the full derivation in our Cardano's method calculator.

Then it is not a cubic at all, just a quadratic or lower in disguise, and this formula does not apply. Check the degree first.

Indirectly. If Δ = 0 there is a repeated root, and the rational root theorem calculator will usually find it quickly. A positive discriminant means three distinct real roots; rational candidates are worth testing, but an irreducible cubic with Δ > 0 can have no rational roots at all (try x³ − 3x + 1).

Almost. Modern textbooks agree on the convention used here, but some older sources flip the sign. When comparing values across sources, verify the convention first.

A 1545 formula that expresses the roots of any cubic in radicals. It works on the depressed form t³ + pt + q = 0, giving t = ∛(−q/2 + √D) + ∛(−q/2 − √D) with D = (q/2)² + (p/3)³.

Yes. The formula only handles t³ + pt + q = 0. Use our depressed cubic calculator to convert any cubic first; it takes seconds.

This is the casus irreducibilis: with three real roots and no rational root, real radicals alone cannot express them. The formula detours through complex cube roots whose imaginary parts cancel. Our complex roots calculator shows the full picture.

Scipione del Ferro found it first around 1515, Tartaglia re-derived it independently, and Cardano published it in Ars Magna (1545). Credit is shared, though the priority dispute was bitter.

Yes. Ferrari, Cardano's student, extended the method to quartics by reducing them to a resolvent cubic. For degree five and above, the Abel-Ruffini theorem proves no general radical formula exists.

Rarely. Try the rational root theorem first; it cracks most textbook cubics in seconds. Reserve Cardano for irreducible cases or when exact radicals are required.

With Δ = −(4p³ + 27q²) < 0 you get one real root and a complex conjugate pair from real radicals, the simplest case. Check the sign anytime with our discriminant calculator.

Only if misapplied: the leading coefficient must be nonzero, the cubic must be depressed first, and cube-root branches must satisfy uv = −p/3. Applied correctly, it cannot fail.

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.

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.

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

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.

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.

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.

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

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

Three, counting multiplicity and including complex roots. That is the fundamental theorem of algebra applied to degree three.

No. A cubic has either three real roots or one real root plus a complex conjugate pair. The discriminant tells you which case you have before you compute anything.

The rational root theorem: list candidates, test them, then factor. It solves most textbook cubics in under a minute.

Use Cardano's method for exact radical forms, or a numerical method like Newton's for decimal approximations. Both always work.

Substitute each root back into the equation, or use Vieta's formulas: the three roots must satisfy the sum, pairwise-sum, and product relations.

A root that appears more than once in the factorization, like the double root 2 in (x − 2)²(x + 1). A zero discriminant is the exact detector of this situation.

Counting multiplicity, no: the possibilities are three real roots (one possibly repeated) or one real root with a complex pair. Complex roots of real polynomials always come in pairs.

No. Understand the rational-root-then-factor ladder for hand work, and let tools handle Cardano's formula. Memorizing the formula without understanding the ladder is the worst of both worlds.

That cubic has one real root and two complex ones, which do not appear on a real x-y plot. The discriminant will be negative for such equations.

Yes. The plot renders as a clean SVG that you can download and reuse in reports or homework.

Yes, both turning points are marked with their coordinates, computed algebraically rather than guessed from pixels.

All three crossings are drawn and labeled. Try x³ − 6x² + 11x − 6 to see crossings at 1, 2, and 3.

The graph page focuses on the picture; for step-by-step algebra use the Cardano method calculator.

It is the sign of the leading coefficient a. Positive a gives the S rising to the right; negative a mirrors it into an N shape falling to the right.

A larger |a| makes outputs grow faster, so the curve climbs more steeply. The roots and turning points spread or squeeze accordingly.

No. Complex roots live off the real plane, so they never appear as crossings. Use the complex roots calculator to find them.

Yes, exactly one. The second derivative of a cubic is a straight line, which always crosses zero once and changes sign there.

No. The formula x = −b/(3a) needs only the coefficients a and b. The calculus runs inside the tool.

It is the cubic cousin of the quadratic vertex formula −b/(2a). Both locate the center of symmetry of their curve.

The concavity flips: the curve changes from bending one way to bending the other, like the middle of the letter S.

Yes, exactly halfway between them whenever two turning points exist. The turning points calculator shows this symmetry.

Shifting x by the inflection x-coordinate removes the quadratic term. That shift is precisely what the depressed cubic calculator does.

Yes, by Vieta's formulas. See the Vieta's formulas calculator and the roots relationship calculator for the details.

Yes, when b = 0 and d = 0, as in y = x³ − 3x, the classic centered S-curve.

No. The derivative is quadratic, so it has zero or two distinct real zeros. The in-between case is a single saddle point, which is not a max or min.

With three real roots, one turning point sits between each neighboring pair of roots. Check the pattern with the cubic roots calculator.

No, but they help. Evaluating f at the turning points reveals the sign pattern that brackets each root, which is how many numerical solvers start.

The sign of b² − 3ac. Positive gives two turning points, zero gives a saddle point, negative gives none. It mirrors the discriminant classification.

Yes. Take y = x³ − 3x + 10: the turning values are 12 and 8, both positive, so the curve crosses the axis only once. The graph generator shows this clearly.

Only local. A cubic is unbounded in both directions, so it has no global maximum or minimum. The turning points are the highest and lowest points nearby, not overall.

Use the ordering rule: for a > 0 the left turning point is the max and the right one is the min; for a < 0 it reverses.

Like y = x³ at the origin: the curve goes flat for an instant, then continues the same way. It is a horizontal inflection point, not an extremum.

The roots, immediately, each with its multiplicity. Expanded form hides them; factored form displays them. For the numerical values, see our cubic roots calculator.

Then its roots are irrational or complex, and no integer-coefficient method will find them. The discriminant tells you which case you are in.

Nearly. Each linear factor (x - r) hands you the root r directly, so a fully factored cubic is a solved cubic.

Yes. For example (x - 1)²(x + 2) = x³ - 3x + 2 has the double root 1. A zero discriminant always signals a repeated factor.

Because a(x - r₁)(x - r₂)(x - r₃) with a ≠ 1 is a different polynomial than the product of the factors alone. Dropping it changes every coefficient.

List rational candidates, test them with synthetic division, deflate to a quadratic, and finish with the quadratic formula.

As an irreducible quadratic factor, which splits into conjugate linear factors over the complex numbers. The complex roots calculator finds them.

Multiply the factors back out. Every coefficient of the product must match the original polynomial exactly.

Dividing a polynomial by (x - c) quickly, and as a bonus, evaluating P(c). It replaces long division whenever the divisor is linear and monic, which covers the majority of hand computations.

When the divisor is not of the form (x - c): quadratics, higher-degree divisors, or non-monic linears need adjustment. Use our long division calculator for those.

It is the remainder, and by the remainder theorem it equals P(c). Zero means (x - c) is a factor.

Use c = -5. Always rewrite the divisor as (x - c) first and read c off carefully.

Factor out the 3 to get 3(x - 2/3), run the tableau with c = 2/3, then divide the quotient coefficients by 3.

Every power of x owns a column. Skipping a missing term shifts all later columns left and corrupts the entire tableau.

Yes, for linear divisors it performs identical arithmetic with the x's stripped out, roughly half the writing and far fewer sign slips.

It tests candidates but does not invent them. Generate candidates with the rational root theorem, then let synthetic division judge each one.

Whenever the divisor is not of the form (x - c): quadratic divisors, higher-degree divisors, or non-monic linears that you prefer not to adjust.

The divisor is a genuine factor. Feed it to the factorization calculator to continue breaking the polynomial down.

The divisor is not a factor. When the divisor is (x - c), that remainder still equals P(c) by the remainder theorem, so the work was not wasted: you evaluated the polynomial at c.

Yes, with zero placeholders for missing powers. Order and alignment are what make the column arithmetic valid.

Then the quotient is 0 and the remainder is the dividend itself. No computation needed.

Compute (Divisor × Quotient) + Remainder and confirm it reproduces the dividend term for term. Because the quotient-remainder pair is unique, a successful reconstruction is a complete proof of correctness.

For linear divisors, yes: synthetic division is the same computation with the powers of x stripped out. The quotient and remainder are identical; only the bookkeeping differs.

Yes. Clearing denominators first keeps the arithmetic tidier, but the algorithm works over any field.

No. The cubic x³ - 2 has none at all; its real root ∛2 is irrational. The theorem lists candidates, it does not guarantee winners.

The denominator of a root in lowest terms must divide the leading coefficient, just as the numerator must divide the constant term. Both constraints come from the same proof.

At most (divisors of the constant term) × (divisors of the leading coefficient), counting ± once, minus duplicates from unreduced fractions.

No. Stop as soon as you have found as many roots as the degree. Each find also deflates the polynomial via synthetic division, shrinking later tests, and the quadratic formula finishes whatever remains.

The roots are irrational or complex. Use the cubic roots calculator for values and the discriminant for their types. An empty candidate result is itself information: it rules out the easy cases.

Multiply through to clear decimals into integers first, then apply the theorem to the scaled polynomial.

Because then every rational root is an integer dividing the constant term, which keeps candidate lists short and hand-testing pleasant.

Yes. Any p/q in lowest terms counts as rational, including negatives and fractions. The theorem was built exactly for cases like this.

Dividing P(x) by (x - c) leaves remainder P(c): just plug c in. No division tableau required unless you also want the quotient.

The factor theorem is the special case where the remainder is zero. Zero remainder means (x - c) is a factor; see the factorization calculator. Any other remainder is just a number, still useful for verification.

Yes. P(1/2) is the remainder on division by (x - 1/2), no special handling needed. Fractions, negatives, and irrationals all work the same way.

Any degree. The proof never uses the degree except to guarantee the division terminates, so degree-50 polynomials obey it exactly like cubics do.

Plugging in is O(n) writing versus a full tableau. Same number, far less work, which is why the theorem exists. Reserve the tableau for when you also need the quotient coefficients.

The remainder is P(2). Always solve divisor = 0 for x first, then evaluate there. For (2x - 4), x = 2, so one substitution gives the remainder with no division at all.

Only when the divisor has degree 2 or higher. That case belongs to our long division calculator, which handles the full quotient-plus-remainder computation step by step.

Generate candidates with the rational root theorem, then test each c for P(c) = 0. Every hit is a root, and the synthetic division tableau for that c doubles as the deflation step.

Three relations linking a cubic's coefficients to its roots: the roots sum to −b/a, their pairwise products sum to c/a, and their product is −d/a.

Because expanding a(x − r₁)(x − r₂)(x − r₃) introduces a minus with each factor. Odd-position relations inherit the minus; the even one does not.

Yes, after dividing by the leading coefficient a. The relations are stated with /a precisely so they hold for every cubic.

The three relations are three equations in three unknowns, so in principle yes, but solving that system is itself hard. In practice use them to check roots found by the cubic roots calculator or guessed via the rational root theorem.

François Viète (1540–1603), French mathematician who published the relations in 1591 and pioneered symbolic algebra, using letters for both unknowns and parameters.

Yes. For degree n there are n relations with alternating signs, each an elementary symmetric sum of the roots. The cubic case is the most used.

Absolutely. The relations are pure algebra and hold for complex roots unchanged. Check examples with our complex roots calculator.

Add your three candidate roots: the sum must equal −b/a. This one addition catches the majority of arithmetic slips before you ever substitute back.

A root involving i = √(−1), written a + bi. It is a perfectly valid solution of the equation; "complex" describes the number system, not the difficulty.

For polynomials with real coefficients, conjugating the equation f(r) = 0 gives f(r̄) = 0, so the conjugate of any root is also a root. This is the conjugate root theorem.

Not with real coefficients: complex roots pair up, and three is odd, so at least one root is always real. With complex coefficients, anything goes.

Find one real root with the rational root theorem, divide it out with synthetic division, and apply the quadratic formula to what remains.

The imaginary unit, defined by i² = −1. It extends the real numbers to the complex plane, where every polynomial equation has its full set of roots.

Deeply useful. Electrical engineers, control theorists, and signal processors compute with them daily: oscillation frequencies and decay rates live in the imaginary parts.

Exactly this case: one real root plus a complex conjugate pair. Our discriminant calculator detects it before you compute anything.

Not honestly. Even with three real roots, Cardano's formula may route through complex intermediates. The complex plane is part of the cubic's native landscape.

The window is probably far too wide, flattening the S into invisibility. Let the tool auto-frame around the turning points and the curve's shape will appear.

Yes. Overlay mode draws multiple cubics together, which is ideal for seeing how one coefficient change reshapes the graph.

Yes. Both x-intercepts and the y-intercept (0, d) are computed exactly and labeled on the plot.

The sign of a sets the end behavior, the discriminant sets the crossing pattern, and b, c, d position the S. See the discriminant calculator for the classification.

Because ax³ dominates at the extremes, flipping the sign of a mirrors the whole curve top to bottom. Try a = 1 versus a = −1 to watch it happen.

No, complex roots do not lie on the real plane. Find them with the complex roots calculator.

That is likely a multiple root, where the curve grazes the axis. A zero discriminant confirms it.

Yes, the SVG downloads cleanly for reports, slides, and homework, staying sharp at any size.

A repeated root: two roots coincide exactly. The discriminant will be zero in that case.

Yes. Distances are measured in the complex plane, so a conjugate pair 2|v| apart is perfectly meaningful. See the complex roots calculator.

Yes. Changing d slides the graph vertically, which moves the roots and reshapes every gap. Only the centroid's formula stays fixed in form.

Tiny gaps mean the roots are numerically fragile: small coefficient changes swing them wildly. Wide gaps mean stable, trustworthy roots.

The discriminant is essentially the squared product of the three gaps times a⁴. Wide gaps give a large positive discriminant; shrinking gaps drive it toward zero.

Not generally. The centroid is the average of all three, which equals −b/(3a). For roots 1, 2, 4 the centroid is 7/3 ≈ 2.333, while the middle root is 2.

No. Gaps are absolute distances, always non-negative. The ordering r₁ ≤ r₂ ≤ r₃ keeps every reported gap positive or zero.

Three pairs: (r₁,r₂), (r₂,r₃), and (r₁,r₃). The last one is the total spread, always the largest of the three.