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
Doğrulama kaynağı: (x − 1)(x − 2)(x − 3) expanded by hand
Polinom kökü hesaplayıcı
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.
Güncellendi Doğrulanan örnekler: 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 (counting multiplicity).
p(x) = x³ − 6x² + 11x − 6
Roots in the complex plane (dashed: 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 |
Nasıl hesaplanır S
Polynomial
Durand–Kerner iteration
Started from powers of 0.4 + 0.9i scaled by the Cauchy bound; 12 sweeps until every correction was below 10⁻³⁰ of the root, then Newton polishing.
Exact rational roots
Each was confirmed by substituting the fraction into p(x) and getting exactly 0.
Check with Vieta's formulas
Polinom kökü hesaplayıcı hakkında
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.
Çözümlü örnekler
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
Doğrulama kaynağı: Python decimal: 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
Doğrulama kaynağı: 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
Doğrulama kaynağı: (x − 1)³(x + 2) = x⁴ − x³ − 3x² + 5x − 2 (expanded with Python fractions)
Sorular
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.
“Polinom kökü hesaplayıcı” ne kadar doğru sonuç verir?
Doğruluk, girdilerinize ve yöntemin varsayımlarına bağlıdır. Ondalık aritmetik 50 anlamlı basamak kullanır; ancak tahminler, sayısal yöntemler ve kaynak veriler daha az hassas olabilir. Gösterilen değerin yuvarlanması bu sınırları ortadan kaldırmaz. Bağımsız kaynaklarla doğrulanan çözümlü örnek sayısı: 7. Örneğin “x³ − 6x² + 11x − 6”, (x − 1)(x − 2)(x − 3) expanded by hand ile karşılaştırılarak doğrulanır.
Yöntemin kaynağı nedir?
Wolfram MathWorld — Durand-Kerner Method (Weierstrass iteration); D. Y. Y. Yun, On square-free decomposition algorithms, SYMSAC 1976.
Bu hesaplayıcı hakkında
Kaynaklar
Kaynaklarla doğrulandı
Bu hesaplayıcı, yanıtları bağımsız kaynaklardan alınan 7 çözümlü örnek içerir. Bunlar test paketinde çalıştırılır; burada da çalıştırabilirsiniz.
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