x³ − 6x² + 11x − 6
- Coefficients, highest power first
- 1, -6, 11, -6
- Roots
- 1, 2, 3
- Real roots (counted with multiplicity)
- 3
- Largest real root
- 3
- Smallest real root
- 1
Checked against: (x − 1)(x − 2)(x − 3) expanded by hand
Polynomial root finder
All real and complex roots of a polynomial up to degree 6, such as a cubic or quartic equation, with repeated roots found exactly and 30-digit accuracy.
By AdminDrafted with AI writing toolsUpdated Checked examples: 7
Roots
1, 2, 3
Roots: 1, 2, 3
- Real roots (counted with multiplicity)
- 3
- Largest real root
- 3
- Smallest real root
- 1
- Degree
- 3
This degree-3 polynomial has 3 real roots and 0 non-real complex roots (counting multiplicity).
p(x) = x³ − 6x² + 11x − 6; xₖ denotes root k in the table.
Complex plane: zₖ denotes root k in the table; dashed circle: unit circle
Roots
| # | Real part | Imaginary part | Multiplicity | |p(root)| |
|---|---|---|---|---|
| 1 | 1 | 0 | 1 | 0 |
| 2 | 2 | 0 | 1 | 0 |
| 3 | 3 | 0 | 1 | 0 |
How it's calculated S
Polynomial
Durand–Kerner iteration
Started from powers of 0.4 + 0.9i scaled by the Cauchy bound; ran 12 sweeps, then Newton polishing. Every correction met the 10⁻³⁰ relative tolerance.
Exact rational roots
Each was confirmed by substituting the fraction into p(x) and getting exactly 0.
Check with Vieta's formulas
About the polynomial root finder
A polynomial of degree n has exactly n roots among the complex numbers, counted with multiplicity; this is the fundamental theorem of algebra. The calculator first splits off repeated factors exactly with Yun's square-free algorithm, then finds all remaining roots at once with the Durand–Kerner (Weierstrass) iteration, which refines n guesses together until each correction is below 10⁻³⁰ of the root. Rational roots are confirmed by exact substitution.
Cubic and quartic equations from engineering, physics and algebra courses are typical inputs. The default, 1, −6, 11, −6, is x³ − 6x² + 11x − 6 = (x − 1)(x − 2)(x − 3), so the roots are 1, 2 and 3; they sum to 6 and multiply to 6, as Vieta's formulas require.
Enter coefficients from the highest power down, with zeros for missing powers: x³ − 2 is 1, 0, 0, −2. Degrees 1 to 6 are accepted.
How to find the roots of a polynomial
Set the polynomial equal to zero and list its coefficients from the highest power down, writing 0 for any missing power. x³ − 6x² + 11x − 6 becomes 1, -6, 11, -6, and its roots are 1, 2 and 3. A polynomial of degree n has n roots once complex and repeated roots are counted.
How to write the coefficient list
- Highest power first.
1, -6, 11, -6means x³ − 6x² + 11x − 6, and the sign belongs to its coefficient. - A zero for every missing power. x³ − 2 is
1, 0, 0, -2, and x⁴ − 1 is1, 0, 0, 0, -1. Leaving out a zero describes a different polynomial. - Everything on one side. x³ = 2 becomes x³ − 2 = 0 before you read off the coefficients.
- Products multiplied out. (x − 1)(x − 2) is x² − 3x + 2, entered as
1, -3, 2. - Degree 1 to 6. That means 2 to 7 coefficients. Fractions and expressions such as 1/2 or sqrt(2) work as coefficients.
Worked examples
| Polynomial | Coefficients, highest power first | Roots |
|---|---|---|
| 2x − 3 | 2, -3 | 3/2 |
| x² + 1 | 1, 0, 1 | ±i |
| x³ − 6x² + 11x − 6 | 1, -6, 11, -6 | 1, 2, 3 |
| (x − 1)³(x + 2) | 1, -1, -3, 5, -2 | −2, 1 (×3) |
| x⁴ − 1 | 1, 0, 0, 0, -1 | −1, 1, ±i |
Check a root by substituting it. For the cubic at x = 2, 8 − 24 + 22 − 6 = 0. Vieta's formulas give a second check: the roots 1, 2 and 3 add to 6 and multiply to 6, matching the coefficients.
x⁴ − 1 has four roots but only two real ones, 1 and −1. The graph of a polynomial crosses the x-axis only at real roots, so the pair ±i never shows on it.
How to express a polynomial as a product of linear factors
Each root r gives a factor (x − r). Keep the leading coefficient a and repeat each factor by its multiplicity m: p(x) = a(x − r₁)^m₁ … (x − rₖ)^mₖ. The multiplicities add up to the degree.
So x³ − 6x² + 11x − 6 = (x − 1)(x − 2)(x − 3). For 2x − 3 the root is 3/2 and the product is 2(x − 3/2), since dropping the 2 changes the polynomial. (x + 2)(x − 1)³ multiplies back to x⁴ − x³ − 3x² + 5x − 2.
Over the complex numbers, x² + 1 = (x − i)(x + i). If every factor must have real coefficients, keep x² + 1 as a quadratic factor: x⁴ − 1 = (x − 1)(x + 1)(x² + 1).
The answer lists roots, so you write the product from them. Exact rational roots give exact factors. A rounded root such as 1.25992105 gives an approximate factor; check a guessed fraction in the original polynomial before using it.
The OpenStax lesson on zeros of polynomial functions explains the factor theorem behind this.
Multiplicity and repeated roots
A root of multiplicity 3 appears three times in the factored form. (x − 1)³(x + 2) has two distinct real roots but four real roots counted with multiplicity, which is why its roots read −2, 1 (×3).
Multiplicity shows on a graph. At a root of odd multiplicity the curve crosses the x-axis; at even multiplicity it touches the axis and turns back. x³ − 3x + 2 = (x − 1)²(x + 2) touches at 1 and crosses at −2.
How the roots are computed
Repeated factors are split off first, exactly, by Yun's square-free algorithm in rational arithmetic, so repeated roots keep full accuracy.
The Durand–Kerner (Weierstrass) iteration then refines all remaining roots together until each correction is below 10⁻³⁰ of the root. If the iteration limit comes first, a warning says the last digits may be off.
Newton steps polish each root, and a rational root is confirmed by exact substitution, which is why 2x − 3 gives exactly 3/2.
Each root's residual |p(root)| is a consistency check, not a bound on its error. Scaling every coefficient changes the residual, and nearly repeated roots can shift a lot when a coefficient changes slightly.
Coefficients must lie between 10⁻⁴⁰ and 10⁴⁰ in size, and leading zeros are dropped. A constant has no roots, and the zero polynomial has every x as a root, so neither gives a root list.
Related calculators
For a single quadratic with the formula worked out, use the quadratic calculator. To do arithmetic with complex roots, use the complex number calculator. For several linear equations in several unknowns, use the linear-system solver.
Worked examples
x³ − 2 (one real, two complex)
- Coefficients, highest power first
- 1, 0, 0, -2
- Roots
- 1.25992105, −0.6299605249 ± 1.091123636i
- Real roots (counted with multiplicity)
- 1
- Largest real root
- 1.2599210499
Checked against: 2^(1/3) and 2^(1/3)·(−1/2 ± i√3/2)
x⁴ − 1
- Coefficients, highest power first
- 1, 0, 0, 0, -1
- Roots
- −1, 1, ±i
- Real roots (counted with multiplicity)
- 2
Checked against: Fourth roots of unity: ±1, ±i
Repeated root (x − 1)³(x + 2)
- Coefficients, highest power first
- 1, -1, -3, 5, -2
- Roots
- −2, 1 (×3)
- Real roots (counted with multiplicity)
- 4
- Smallest real root
- -2
Checked against: (x − 1)³(x + 2) = x⁴ − x³ − 3x² + 5x − 2
Questions
How many roots does a polynomial have?
Exactly as many as its degree, counted with multiplicity, once complex roots are included; this is the fundamental theorem of algebra. x⁴ − 1 has four roots: −1, 1, i and −i. The number of real roots can be smaller: x³ − 2 has one real root, ∛2 ≈ 1.259921, and two complex ones. A root of multiplicity 3, such as x = 1 in (x − 1)³(x + 2), counts three times.
How do you solve a cubic equation?
Look for a rational root first. By the rational root theorem, any rational root p/q of a polynomial with integer coefficients has p dividing the constant term and q dividing the leading coefficient. For x³ − 6x² + 11x − 6, trying divisors of 6 finds x = 1, and dividing by (x − 1) leaves x² − 5x + 6 = (x − 2)(x − 3). Without a rational root, Cardano's formula or a numerical method is needed.
Is there a formula for the roots of a quintic?
No general formula using radicals exists for degree 5 or higher. The Abel–Ruffini theorem, proved by Abel in 1824, shows this, and Galois theory explains which equations can be solved that way. x⁵ − x − 1 is a standard example whose roots cannot be written with radicals. Numerical methods still find them: its only real root is about 1.167304.
What are Vieta's formulas?
They link the roots to the coefficients. For aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₀, the roots add up to −aₙ₋₁/aₙ and multiply to (−1)ⁿa₀/aₙ. For x³ − 6x² + 11x − 6 the roots 1, 2 and 3 sum to 6 and multiply to 6, matching −(−6)/1 and (−1)³ × (−6)/1. The calculator uses both as a check on the roots it finds.
Why do complex roots come in conjugate pairs?
When every coefficient is real, conjugating the equation p(z) = 0 gives p(z̄) = 0, so the conjugate of a root is also a root. x³ − 2 therefore has the pair −0.629961 ± 1.091124i alongside its real root. It follows that a polynomial of odd degree with real coefficients always has at least one real root.
About this calculator
Sources
Checked against references
This calculator includes 7 worked examples with independently sourced answers. They run in the test suite, and you can run them here too.