CalcOpenly
Tiếng Việt

Công cụ tìm nghiệm đa thức

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.

Cập nhật Ví dụ đã kiểm tra: 7

1, -6, 11, -6 means x³ − 6x² + 11x − 6. Include zeros for missing powers.
Thử
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

-2-1012123xp(x)x = 1x = 2x = 3

Roots in the complex plane (dashed: unit circle)

ReIm123
Roots Số hàng: 3
#Real partImaginary partMultiplicity|p(root)|
11010
22010
33010
Cách tính S

  1. Polynomial

    p(x)=x3−6x2+11x−6p(x) = x^{3} - 6x^{2} + 11x - 6

  2. Durand–Kerner iteration

    zk←zk−p(zk)∏j≠k(zk−zj)z_k \leftarrow z_k - \frac{p(z_k)}{\prod_{j \ne k}(z_k - z_j)}

    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.

  3. Exact rational roots

    x=1,x=2,x=3x = 1,\quad x = 2,\quad x = 3

    Each was confirmed by substituting the fraction into p(x) and getting exactly 0.

  4. Check with Vieta's formulas

    ∑zk=6=−an−1an=6,∏zk=6=(−1)na0an=6\sum z_k = 6 = -\tfrac{a_{n-1}}{a_n} = 6,\qquad \prod z_k = 6 = (-1)^{n}\tfrac{a_0}{a_n} = 6

Giới thiệu Công cụ tìm nghiệm đa thức

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.

Ví dụ có lời giải

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

Nguồn đối chiếu: (x − 1)(x − 2)(x − 3) expanded by hand

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

Nguồn đối chiếu: 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

Nguồn đối chiếu: 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

Nguồn đối chiếu: (x − 1)³(x + 2) = x⁴ − x³ − 3x² + 5x − 2 (expanded with Python fractions)

Câu hỏi

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.

“Công cụ tìm nghiệm đa thức” chính xác đến mức nào?

Độ chính xác phụ thuộc vào dữ liệu nhập và giả định của phương pháp. Phép tính thập phân dùng 50 chữ số có nghĩa, nhưng ước lượng, phương pháp số và dữ liệu nguồn có thể kém chính xác hơn; làm tròn khi hiển thị không loại bỏ các giới hạn đó. Ví dụ có lời giải đã đối chiếu với nguồn độc lập: 7. Ví dụ, “x³ − 6x² + 11x − 6” được kiểm tra bằng (x − 1)(x − 2)(x − 3) expanded by hand.

Phương pháp này lấy từ đâu?

Wolfram MathWorld — Durand-Kerner Method (Weierstrass iteration); D. Y. Y. Yun, On square-free decomposition algorithms, SYMSAC 1976.

Về công cụ tính này

zk←zk−p(zk)∏j≠k(zk−zj)z_k \leftarrow z_k - \frac{p(z_k)}{\prod_{j \ne k} (z_k - z_j)}

Nguồn

  1. Wolfram MathWorld — Durand-Kerner Method (Weierstrass iteration)
  2. D. Y. Y. Yun, On square-free decomposition algorithms, SYMSAC 1976

Đã đối chiếu nguồn tham khảo

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