Polynomial Calculator (Division and Synthetic Division)

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roots of a cubic polynomial

How to find the roots of a cubic polynomial

Type the polynomial on the Roots and factoring tab. The tool lists the rational candidates p/q, tests them one by one, removes each root found with synthetic division and solves what is left with the quadratic formula.

From rational candidate to complex pair

For a polynomial with integer coefficients, any rational root p/q (in lowest terms) has p dividing the constant term and q dividing the leading coefficient. In 2x³ − 3x² − 11x + 6, p divides 6 and q divides 2, leaving 12 candidates: ±1, ±2, ±3, ±6, ±1/2 and ±3/2. Testing each one means computing P(c), and the tool table shows the value of P at all of them, with the ones that vanish highlighted. The roots are −2, 1/2 and 3.
Once the first root is found, the degree drops. You divide P by x − r (synthetic division) and the quotient becomes the new polynomial, of degree 2 in the cubic case, solved with the quadratic formula. Multiplicity comes out for free: in x³ − 3x + 2, after removing the root 1 the quotient x² + x − 2 still vanishes at 1, so 1 is a double root and the other root is −2.
Not every cubic has a rational root. x³ − 2 has none, and its real root is the cube root of 2, about 1.2599, accompanied by a pair of complex conjugate roots (approximately −0.6300 ± 1.0911i). In those cases the tool computes the roots numerically. By contrast, x³ − 1 has the root 1 and, from the quotient x² + x + 1, the complex roots −1/2 ± (√3/2)i in exact form. Every cubic with real coefficients has at least one real root, and the complex ones come in pairs.

Solved cubics

Three integer roots

Input: x^3 - 6x^2 + 11x - 6
Expected output: x = 1, 2, 3

Sum 6 and product 6, as Vieta’s formulas require.

Rational roots that are not integers

Input: 2x^3 - 3x^2 - 11x + 6
Expected output: x = −2, 1/2, 3

Twelve candidates tested; three of them zero the polynomial.

Double root

Input: x^3 - 3x + 2
Expected output: x = −2; x = 1 (dupla)

It factors as (x + 2)(x − 1)².

Exact complex roots

Input: x^3 - 1
Expected output: x = 1; x = −1/2 ± (√3/2)i

The factor x² + x + 1 has discriminant −3.

No rational root

Input: x^3 - 2
Expected output: x = ∛2 (real) e duas raízes complexas conjugadas

The real root is about 1.2599; the complex ones are −0.6300 ± 1.0911i.

Three fractional roots

Input: 6x^3 + x^2 - 5x - 2
Expected output: x = −2/3, −1/2, 1

It factors as (3x + 2)(2x + 1)(x − 1).

Full tool FAQ

How do you divide polynomials with long division?

Divide the leading term of the dividend by the leading term of the divisor, multiply the result by the whole divisor, subtract it from the dividend and repeat with the running remainder until the remainder has a lower degree than the divisor. Example: (2x³ − 3x + 1) ÷ (x² + x − 2). First 2x³ ÷ x² = 2x; 2x·(x² + x − 2) = 2x³ + 2x² − 4x; what is left is −2x² + x + 1. Then −2x² ÷ x² = −2; −2·(x² + x − 2) = −2x² − 2x + 4; what is left is 3x − 3. Quotient 2x − 2, remainder 3x − 3. Missing terms, like the x² in the dividend, go in with a zero coefficient.

How does synthetic division work?

What is the remainder theorem and how does it help?

How do you find the roots of a polynomial of degree 3 or higher?

How do you divide when the divisor is ax + b with a different from 1?

What are Vieta's formulas?

What does the tool do with complex roots?

Can I use fractional or decimal coefficients?

How do you find the GCD of two polynomials?

What are the limits and what is saved?

Frequently asked questions

Does every cubic equation have a real root?

Yes, when the coefficients are real. The graph of a cubic goes from −∞ to +∞ (or the reverse) and crosses the x-axis at least once. The other two roots are either real or a complex conjugate pair.

What if the constant term is zero?

Then x = 0 is a root and x is a common factor. The tool removes the factor x before testing candidates. In x³ − x, what is left is x² − 1, and the roots are −1, 0 and 1.

What do you do when the coefficients are fractions?

Multiply the polynomial by the least common multiple of the denominators to get integer coefficients; the roots do not change. The tool does this step on its own. In x²/2 − 3x/2 + 1, multiplying by 2 gives x² − 3x + 2, with roots 1 and 2.

Next steps

From rational candidate to complex pair

For a polynomial with integer coefficients, any rational root p/q (in lowest terms) has p dividing the constant term and q dividing the leading coefficient. In 2x³ − 3x² − 11x + 6, p divides 6 and q divides 2, leaving 12 candidates: ±1, ±2, ±3, ±6, ±1/2 and ±3/2. Testing each one means computing P(c), and the tool table shows the value of P at all of them, with the ones that vanish highlighted. The roots are −2, 1/2 and 3.

Once the first root is found, the degree drops. You divide P by x − r (synthetic division) and the quotient becomes the new polynomial, of degree 2 in the cubic case, solved with the quadratic formula. Multiplicity comes out for free: in x³ − 3x + 2, after removing the root 1 the quotient x² + x − 2 still vanishes at 1, so 1 is a double root and the other root is −2.

Not every cubic has a rational root. x³ − 2 has none, and its real root is the cube root of 2, about 1.2599, accompanied by a pair of complex conjugate roots (approximately −0.6300 ± 1.0911i). In those cases the tool computes the roots numerically. By contrast, x³ − 1 has the root 1 and, from the quotient x² + x + 1, the complex roots −1/2 ± (√3/2)i in exact form. Every cubic with real coefficients has at least one real root, and the complex ones come in pairs.

Solved cubics

Three integer roots

Input: x^3 - 6x^2 + 11x - 6
Expected output: x = 1, 2, 3

Sum 6 and product 6, as Vieta’s formulas require.

Rational roots that are not integers

Input: 2x^3 - 3x^2 - 11x + 6
Expected output: x = −2, 1/2, 3

Twelve candidates tested; three of them zero the polynomial.

Double root

Input: x^3 - 3x + 2
Expected output: x = −2; x = 1 (dupla)

It factors as (x + 2)(x − 1)².

Exact complex roots

Input: x^3 - 1
Expected output: x = 1; x = −1/2 ± (√3/2)i

The factor x² + x + 1 has discriminant −3.

No rational root

Input: x^3 - 2
Expected output: x = ∛2 (real) e duas raízes complexas conjugadas

The real root is about 1.2599; the complex ones are −0.6300 ± 1.0911i.

Three fractional roots

Input: 6x^3 + x^2 - 5x - 2
Expected output: x = −2/3, −1/2, 1

It factors as (3x + 2)(2x + 1)(x − 1).

Full tool FAQ

How do you divide polynomials with long division?

How does synthetic division work?

What is the remainder theorem and how does it help?

How do you find the roots of a polynomial of degree 3 or higher?

How do you divide when the divisor is ax + b with a different from 1?

What are Vieta's formulas?

What does the tool do with complex roots?

Can I use fractional or decimal coefficients?

How do you find the GCD of two polynomials?

What are the limits and what is saved?

Frequently asked questions

Does every cubic equation have a real root?

What if the constant term is zero?

Then x = 0 is a root and x is a common factor. The tool removes the factor x before testing candidates. In x³ − x, what is left is x² − 1, and the roots are −1, 0 and 1.