How-to · Updated October 2, 2026
How do you find the zeros of a polynomial?
Direct answer
Set p(x) = 0. Factor, test rational candidates, then use the quadratic formula on what is left. A degree-n polynomial has n zeros, counting multiplicity and complex zeros. x³ − 15x² + 71x − 105 = (x − 3)(x − 5)(x − 7), so the zeros are 3, 5, 7.
Four polynomials
| Polynomial | Factored form | Zeros |
|---|---|---|
| x³ − 15x² + 71x − 105Cubic | (x − 3)(x − 5)(x − 7) | 3, 5, 7 |
| 3x² + x − 10Quadratic | (x + 2)(3x − 5) | −2, 5/3 |
| 2x² − 10x − 3Does not factor over the integers | 2x² − 10x − 3 | (5 − √31)/2, (5 + √31)/2 |
| x² − 3x + 3Complex pair | x² − 3x + 3 | (3 − i√3)/2, (3 + i√3)/2 |
2x² − 10x − 3 does not factor over the integers. Its zeros are (5 − √31)/2 ≈ −0.283882 and (5 + √31)/2 ≈ 5.28388. x² − 3x + 3 never crosses the x-axis; the zeros are (3 − i√3)/2, (3 + i√3)/2.
Rational candidates for the cubic
For x³ − 15x² + 71x − 105, the constant term is 105 and the leading coefficient is 1, so every rational zero is an integer factor of 105. Descartes' rule allows 3 or 1 positive zeros and 0 negative zeros. The three that work are 3, 5, 7. Their sum is 15, matching −(coefficient of x²) / (leading coefficient).
Check against the coefficients
For 3x² + x − 10, the zeros are −2, 5/3. Their product is −10/3, and c/a is (−10)/3 = −10/3. Real zeros are x-intercepts. Complex zeros are still zeros, and they are not x-intercepts.
Limits
- There is no general formula in radicals for degree 5 or higher. Divide out the rational zeros, then solve the rest numerically.
- A repeated factor counts more than once. Odd multiplicity crosses the x-axis; even multiplicity touches it.
- These examples use the same solver as the zeros calculator. A different written form can be the same polynomial.
Use the calculator
The reverse problem: how to write a polynomial with given zeros. Type any polynomial in the zeros calculator.