# کثیر رکنی کے جذر معلوم کریں

> 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.

انٹرایکٹو صورت: https://www.calcopenly.com/ur/math/polynomial-root-finder
موضوع: ریاضی کے کیلکولیٹر

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.

## اندراجات

- **Coefficients, highest power first**: 1, -6, 11, -6 means x³ − 6x² + 11x − 6. Include zeros for missing powers.

## نتائج

- Roots — بنیادی نتیجہ
- Real roots (counted with multiplicity)
- Largest real root
- Smallest real root
- Degree

## فارمولا

$$
z_k \leftarrow z_k - \frac{p(z_k)}{\prod_{j \ne k} (z_k - z_j)}
$$

## حل شدہ مثالیں

### 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**
- جانچ کا ماخذ: ⁨(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**
- جانچ کا ماخذ: ⁨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**
- جانچ کا ماخذ: ⁨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**
- جانچ کا ماخذ: ⁨(x − 1)³(x + 2) = x⁴ − x³ − 3x² + 5x − 2 (expanded with Python fractions)⁩

### Quintic x⁵ − x − 1

- Coefficients, highest power first: 1, 0, 0, 0, -1, -1
- **Real roots (counted with multiplicity): 1**
- **Largest real root: 1.1673039783**
- جانچ کا ماخذ: ⁨Python decimal Newton iteration on x⁵ − x − 1 (60 digits)⁩

### Degree 6: x⁶ − 1

- Coefficients, highest power first: 1, 0, 0, 0, 0, 0, -1
- **Real roots (counted with multiplicity): 2**
- **Largest real root: 1**
- **Smallest real root: -1**
- جانچ کا ماخذ: ⁨Sixth roots of unity: ±1, ±1/2 ± i√3/2⁩

## سوالات

### 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.

### “⁨کثیر رکنی کے جذر معلوم کریں⁩” کتنا درست ہے؟

درستی آپ کی درج کردہ قدروں اور طریقے کے مفروضوں پر منحصر ہے۔ اعشاری حساب 50 بامعنی ہندسے استعمال کرتا ہے، مگر تخمینے، عددی طریقے اور ماخذ کا ڈیٹا کم درست ہو سکتے ہیں؛ دکھائی گئی قدروں کو راؤنڈ کرنے سے یہ حدود ختم نہیں ہوتیں۔ آزاد ذرائع کی حل شدہ مثالوں سے جانچ: 7۔ مثلاً، “⁨x³ − 6x² + 11x − 6⁩” کو ⁨(x − 1)(x − 2)(x − 3) expanded by hand⁩ سے جانچا جاتا ہے۔

### اس طریقے کا ماخذ کیا ہے؟

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

## ماخذ

- [Wolfram MathWorld — Durand-Kerner Method (Weierstrass iteration)](https://mathworld.wolfram.com/Durand-KernerMethod.html)
- [D. Y. Y. Yun, On square-free decomposition algorithms, SYMSAC 1976](https://doi.org/10.1145/800205.806320)