Quartic equations, polynomials of degree four, can always be solved with a formula using basic arithmetic and radicals (square and cube roots). The core strategy is to reduce the quartic into simpler equations you already know how to solve: a cubic and two quadratics. This makes the quartic the highest-degree polynomial with a general closed-form solution. For degree five and above, no such formula exists.
The process involves three main stages: simplify the equation by removing the cubic term, solve an intermediate cubic equation (called the “resolvent cubic”), then use that result to split the quartic into two quadratic equations you can finish with the quadratic formula.
Step 1: Remove the Cubic Term
Start with the general quartic equation:
Ax⁴ + Bx³ + Cx² + Dx + E = 0
First, divide everything by A so the leading coefficient is 1. Then substitute x = u − B/(4A). This substitution shifts the variable just enough to eliminate the x³ term entirely, producing what’s called a “depressed quartic”:
u⁴ + pu² + qu + r = 0
The new coefficients p, q, and r are expressions built from the original coefficients. This is the same trick used when solving cubics (where you substitute to kill the x² term). With the cubic term gone, the algebra in every subsequent step becomes far more manageable.
If q happens to equal zero, you’re lucky: the depressed quartic is actually a quadratic in disguise. Just let w = u² and solve w² + pw + r = 0, then take square roots. The interesting case is when q ≠ 0, which requires the full method.
Step 2: Set Up a Perfect Square
This is where the real ingenuity of the solution appears. The idea, first discovered by Lodovico Ferrari in 1540, is to rearrange the depressed quartic so that both sides of the equation become perfect squares. Once you have (something)² = (something else)², you can take square roots and reduce to quadratics.
Start by rewriting u⁴ + pu² + qu + r = 0. Notice that u⁴ + pu² is close to a perfect square: (u² + p/2)² = u⁴ + pu² + p²/4. So you can write:
(u² + p/2)² = −qu − r + p²/4
The left side is a perfect square, but the right side is linear in u, not a perfect square. To fix this, introduce a new variable α and add 2αu² + α² + pα to both sides:
(u² + p/2 + α)² = 2αu² − qu + (α² + pα + p²/4 − r)
The left side is still a perfect square regardless of what α is. The right side is a quadratic expression in u with coefficients that depend on α. A quadratic Au² + Bu + C is a perfect square exactly when its discriminant B² − 4AC equals zero. Setting that discriminant to zero gives you a cubic equation in α.
Step 3: Solve the Resolvent Cubic
The right side of the equation from Step 2 has A = 2α, B = −q, and C = α² + pα + p²/4 − r. Setting the discriminant to zero:
q² − 8α(α² + pα + p²/4 − r) = 0
Expanding and rearranging produces a cubic in α, known as the resolvent cubic. You only need one root of this cubic, not all three. Solve it using the cubic formula (Cardano’s method) or, for specific numerical problems, by inspection or numerical methods.
Call that root α₁. Plugging it back in makes the right side a perfect square, so the equation now looks like:
(u² + p/2 + α₁)² = 2α₁(u − q/(4α₁))²
Step 4: Factor Into Two Quadratics
With both sides as perfect squares, you have a difference of squares: X² − Y² = 0, which factors as (X + Y)(X − Y) = 0. This gives two separate quadratic equations:
u² + p/2 + α₁ = +√(2α₁) · (u − q/(4α₁))
u² + p/2 + α₁ = −√(2α₁) · (u − q/(4α₁))
Each of these is a standard quadratic equation in u. Apply the quadratic formula to each one, and you get four solutions total (two from each quadratic). These are the four roots of the depressed quartic. To get the roots of the original equation, reverse the substitution from Step 1: x = u − B/(4A).
Descartes’ Alternative Approach
Ferrari’s method isn’t the only route. René Descartes proposed a different strategy: assume the depressed quartic factors directly into two quadratics with unknown coefficients.
Write u⁴ + pu² + qu + r = (u² + ku + l)(u² − ku + m), where k, l, and m are unknowns. The minus sign on k in the second factor is forced because the cubic term must vanish when you expand the product. Expanding and matching coefficients gives three equations:
- u² coefficient: m + l − k² = p
- u coefficient: k(m − l) = q
- constant: lm = r
From these, you can express l and m in terms of k: 2m = p + k² + q/k, and 2l = p + k² − q/k. Substituting into the third equation (lm = r) produces a cubic equation in j = k², called the resolvent cubic for Descartes’ method:
j³ + 2pj² + (p² − 4r)j − q² = 0
Solve this cubic for j, take k = √j, compute l and m, and you have your two quadratic factors. Finish with the quadratic formula. The final answers are the same four roots you’d get from Ferrari’s method; the path is just organized differently. Some people find Descartes’ approach more intuitive because the goal (split the quartic into two quadratics) is stated upfront rather than emerging through algebraic manipulation.
A Worked Example
Consider x⁴ − 4x³ + 3x² + 4x − 4 = 0. First, depress it by substituting x = u + 1 (since B/(4A) = −4/4 = −1, so x = u − (−1) = u + 1). After expanding and simplifying, you get a depressed quartic in u.
For simpler problems, though, you can often skip the full machinery. Try the rational root theorem first: if the quartic has integer coefficients, any rational root must divide the constant term. Testing small values like ±1, ±2, ±4 often reveals a root. If you find one root, factor it out to get a cubic, then find another root and reduce to a quadratic. The general method described above is essential when no rational roots exist, which is the typical case.
How Many Real Roots to Expect
A quartic always has exactly four roots when you count complex roots and multiplicities. But the number of real roots varies: you might get four, two, or zero real roots.
The discriminant of the quartic, a long expression in the coefficients, determines which case you’re in. For the depressed quartic u⁴ + pu² + qu + r = 0, the discriminant is:
δ = 256r³ − 128p²r² + 144pq²r + 16p⁴r − 27q⁴ − 4p³q²
When δ is positive, the quartic has either four real roots or none. When δ is negative, it has exactly two real roots and two complex ones. When δ is zero, the equation has a repeated root. A second quantity, L = 8pr − 9q² − 2p³, helps distinguish further within these cases. For practical purposes, if you just need the roots, solving the resolvent cubic and completing the method will naturally produce whatever real and complex roots exist.
Why This Only Works Up to Degree Four
The quartic holds a special place in algebra: it’s the highest-degree polynomial equation with a general solution using addition, subtraction, multiplication, division, and radicals. The Abel-Ruffini theorem, proved in the early 1800s, establishes that no such formula can exist for polynomials of degree five or higher. The proof relies on group theory: the mathematical structure (called the Galois group) underlying fifth-degree and higher polynomials doesn’t have the right properties to allow a radical solution. For those equations, you need numerical approximation methods or special functions instead.
This means the quartic formula is the end of the line for a tradition that started with the quadratic formula most people learn in school, extended through Cardano’s cubic formula in the 1500s, and was completed by Ferrari’s quartic solution that same decade. Every step up in degree made the algebra dramatically more complex, and the quartic already pushes the limits of what’s practical to compute by hand.