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Quick answer
The zeros of a polynomial are the x-values where p(x) = 0. To find them, factor out common terms, test rational candidates ±(factors of the constant)/(factors of the leading coefficient) with synthetic division, then solve the leftover quadratic with the quadratic formula. A degree-n polynomial has exactly n zeros counting multiplicity and complex zeros. Example: x³ − 15x² + 71x − 105 = (x − 3)(x − 5)(x − 7), so the zeros are 3, 5, 7.
Type any polynomial — such as x^3 - 6x^2 + 11x - 6, 4(x+3)^2(x-1) or 2x² − 6.9x + 1 — or a rational function like (x^2-4)/(x^2-x-2). You get exact zeros, multiplicities, the factored form, a graph and every step. Switch tabs to build a polynomial from its zeros.
Last updated October 1, 2026. Checked against OpenStax College Algebra and NCERT Class 10. · how to find zeros · polynomial from zeros
Use ^ for powers (x^3), implied multiplication 4(x+3)^2(x−1), fractions 1/2x², √3, decimals, or a fraction like (2x−5)/(2x−1).
Zeros of f(x) = x³ − 15x² + 71x − 105
Factored: f(x) = (x − 3)(x − 5)(x − 7)
| Zero | Decimal | Mult. | Type / graph |
|---|---|---|---|
| 3 | 3 | 1 | rational · crosses the x-axis |
| 5 | 5 | 1 | rational · crosses the x-axis |
| 7 | 7 | 1 | rational · crosses the x-axis |
Degree · zeros
3 · 3 real, 0 complex
y-intercept
(0, −105)
End behavior
Falls left, rises right: f(x) → −∞ as x → −∞ and +∞ as x → +∞
Descartes’ rule of signs
Positive real zeros: 3 or 1 · negative: 0
| Sum of zeros | 15 | −(−15)/1 = 15 ✓ |
| Sum of products of pairs | 71 | 71/1 = 71 ✓ |
| Product of zeros | 105 | −(−105)/1 = 105 ✓ |
Step-by-step
Every row below is computed by the same engine as the calculator.
| Polynomial | Factored form | Zeros | Why it's useful |
|---|---|---|---|
| x³ − 15x² + 71x − 105 | (x − 3)(x − 5)(x − 7) | 3; 5; 7 | Arithmetic-progression cubic (k = 105) |
| x³ − 3x² − 13x + 15 | (x + 3)(x − 1)(x − 5) | −3; 1; 5 | 1 and −3 are zeros → a = 3, b = 15 |
| 3x² + x − 10 | (x + 2)(3x − 5) | −2; 5/3 | Verify zeros vs coefficients |
| x³ − 36x | x(x + 6)(x − 6) | −6; 0; 6 | Factor out x first |
| 4x³ + 20x² + 12x − 36 | 4(x + 3)²(x − 1) | −3 (multiplicity 2); 1 | Zero with multiplicity 2 |
| −x⁴ + 3x³ + 9x² − 23x + 12 | −(x + 3)(x − 1)²(x − 4) | −3; 1 (multiplicity 2); 4 | Repeated rational zero |
| x⁴ + x³ − 12x² | x²(x + 4)(x − 3) | −4; 0 (multiplicity 2); 3 | x² factor + quadratic |
| x⁵ + 10x⁴ + 25x³ | x³(x + 5)² | −5 (multiplicity 2); 0 (multiplicity 3) | x³(x + 5)² |
| 2x² − 10x − 3 | 2x² − 10x − 3 | (5 − √31)/2 ≈ −0.283882; (5 + √31)/2 ≈ 5.28388 | Irrational zeros (surds) |
| 2x² − 6.9x + 1 | 0.1(20x² − 69x + 10) | (69 − √3961)/40 ≈ 0.151588; (69 + √3961)/40 ≈ 3.29841 | Decimal coefficients |
| x² − 3x + 3 | x² − 3x + 3 | (3 − i√3)/2 ≈ 1.5 − 0.866025i; (3 + i√3)/2 ≈ 1.5 + 0.866025i | Complex conjugate zeros |
| 1.73205x² − 8x + 6.9282 | √3(x − 2√3/3)(x − 2√3) | 2√3/3 ≈ 1.1547; 2√3 ≈ 3.4641 | Irrational coefficients |
Finding zeros means solving p(x) = 0. Work from the easiest method to the hardest, and divide out every zero you find so the polynomial gets smaller.
1. Factor out the GCF
Take out common numbers and powers of x. x⁴ + x³ − 12x² = x²(x² + x − 12) = x²(x + 4)(x − 3), so 0 is a zero of multiplicity 2.
2. Factor or use the quadratic formula
For ax² + bx + c, try factoring first. If it does not factor, x = (−b ± √(b² − 4ac)) / 2a gives exact zeros.
3. List rational candidates
Rational root theorem: ±p/q with p | constant and q | leading coefficient. For x³ − 15x² + 71x − 105: 1, −1, 3, −3, 5, −5, 7, −7, 15, −15, ….
4. Test with synthetic division
A remainder of 0 means the candidate is a zero (factor theorem). Keep the quotient and divide again by the same zero to check multiplicity.
5. Solve what is left
Once the quotient is quadratic, factor it or use the formula. A negative discriminant gives a complex conjugate pair.
6. Use numbers when algebra stops
Cubic and higher factors with no rational zeros are solved numerically (this calculator uses Durand–Kerner with Newton polishing), e.g. (x − 1)(x³ + x² − 3x − 10) has a real zero ≈ 2.26765.
| Degree | Method | Number of zeros |
|---|---|---|
| 1 (linear)ax + b | x = −b/a | Exactly 1 real zero |
| 2 (quadratic)ax² + bx + c | Factor, complete the square or quadratic formula | 2 (real, repeated or a complex pair) |
| 3 (cubic)ax³ + bx² + cx + d | Rational root theorem + synthetic division, then a quadratic | 3; at least 1 real |
| 4 (quartic)ax⁴ + … + e | GCF, substitution u = x² if biquadratic, rational roots, then quadratics | 4; 0, 2 or 4 real |
| 5+aₙxⁿ + … + a₀ | Factor what you can; no general formula exists — use numerical methods | n; odd degree has at least 1 real |
| Discriminant | Zeros | Example |
|---|---|---|
| b² − 4ac > 0, perfect square | Two distinct rational zeros | x² − 2x − 15 = (x + 3)(x − 5) → −3, 5 |
| b² − 4ac > 0, not a square | Two distinct irrational zeros | 2x² − 10x − 3 → (5 − √31)/2, (5 + √31)/2 |
| b² − 4ac = 0 | One repeated zero (multiplicity 2) | x² − 6x + 9 = (x − 3)² → 3, 3 |
| b² − 4ac < 0 | Two complex conjugate zeros | x² − 3x + 3 → (3 − i√3)/2, (3 + i√3)/2 |
A classic exam problem: “If the zeros of x³ − 15x² + 71x − k are in arithmetic progression (AP), find k.” Let the zeros be a − d, a and a + d, then use the relationships between zeros and coefficients:
| Cubic x³ − px² + qx − k | Middle zero a = p/3 | d² = 3a² − q | Zeros | k = a(a² − d²) |
|---|---|---|---|---|
| x³ − 15x² + 71x − k | 5 | 4 | 3, 5, 7 | 105 |
| x³ − 6x² + 11x − k | 2 | 1 | 1, 2, 3 | 6 |
| x³ − 9x² + 23x − k | 3 | 4 | 1, 3, 5 | 15 |
| x³ − 12x² + 39x − k | 4 | 9 | 1, 4, 7 | 28 |
| x³ − 3x² + x − k | 1 | 2 | 1 − √2, 1, 1 + √2 | −1 |
If r is a zero, then p(r) = 0 (factor theorem). Substitute each known zero and solve for the unknowns, then use the calculator to find the remaining zeros.
p(1) = 1 − a − 13 + b = 0 → b − a = 12
p(−3) = −27 − 9a + 39 + b = 0 → b − 9a = −12
Subtract: 8a = 24 → a = 3, b = 15
x³ − 3x² − 13x + 15 = (x + 3)(x − 1)(x − 5) → zeros −3, 1, 5
6(4/3)³ − 11(4/3)² + k(4/3) − 20 = 0
128/9 − 176/9 + 4k/3 − 20 = 0 → 4k/3 = 76/3 → k = 19
6x³ − 11x² + 19x − 20 = (3x − 4)(2x² − x + 5)
Other zeros: (1 − i√39)/4 and (1 + i√39)/4
| Multiplicity | Graph at the zero | Example |
|---|---|---|
| 1 (odd) | Crosses the x-axis | x − 1 in 4(x + 3)²(x − 1) |
| 2 (even) | Touches and turns around (bounces) | (x + 3)² in 4(x + 3)²(x − 1) |
| 3 (odd) | Crosses and flattens (inflection at the axis) | x³ in x³(x + 5)² |
| Complex | No x-intercept | x² − 3x + 3 |
4x³ + 20x² + 12x − 36 = 4(x + 3)²(x − 1): zeros −3 (multiplicity 2), 1. Sum of multiplicities = degree = 3.
| Polynomial | Sum | Pairs | Product |
|---|---|---|---|
| Quadratic ax² + bx + c | α + β = −b/a | — | αβ = c/a |
| Cubic ax³ + bx² + cx + d | α + β + γ = −b/a | αβ + βγ + γα = c/a | αβγ = −d/a |
| Quartic ax⁴ + bx³ + cx² + dx + e | −b/a | c/a | e/a |
Check for 3x² + x − 10 (zeros −2, 5/3): sum of zeros −1/3 = −(1)/3 = −1/3; product of zeros −10/3 = (−10)/3 = −10/3.
Reverse the process: p(x) = a(x − r₁)(x − r₂)…(x − rₙ). Pick the leading coefficient a (often 1), multiply out, and clear fractions if you want integer coefficients. Complex zeros need their conjugate for real coefficients.
| Zeros | Factored form | Standard form |
|---|---|---|
| −2 and 5 | (x + 2)(x − 5) | x² − 3x − 10 |
| −3 and 4 | (x + 3)(x − 4) | x² − x − 12 |
| 3 and 6 | (x − 3)(x − 6) | x² − 9x + 18 |
| 3 and 5 | (x − 3)(x − 5) | x² − 8x + 15 |
| 1/4 and −1 | (x − 1/4)(x + 1) | x² + (3/4)x − 1/4integer form: 4x² + 3x − 1 |
| −2, 1, 3 (leading coefficient 2) | 2(x + 2)(x − 1)(x − 3) | 2x³ − 4x² − 10x + 12 |
| √2, −√2, 3 | (x − √2)(x + √2)(x − 3) | x³ − 3x² − 2x + 6 |
| 2 + 3i (conjugate added) | (x − (2 + 3i))(x − (2 − 3i)) | x² − 4x + 13 |
−0.3x² − 1.7x + 2
= −0.1(3x + 20)(x − 1)
Zeros: −20/3, 1
√3x² − 8x + 4√3
= √3(x − 2√3/3)(x − 2√3)
Zeros: 2√3/3, 2√3
x³ − 9x² − 36x
= x(x + 3)(x − 12)
Zeros: −3, 0, 12
x⁵ + 10x⁴ + 25x³
= x³(x + 5)²
Zeros: −5 (multiplicity 2), 0 (multiplicity 3)
25x² − 16
= (5x + 4)(5x − 4)
Zeros: −4/5, 4/5
x² − 10x + 21
= (x − 3)(x − 7)
Zeros: 3, 7
Linear: 5x − 20 → 4 · 2y + 7 → −7/2 · 3x − 4 → 4/3 · −5x + 50 → 10.
For f(x) = N(x)/D(x), factor both parts. A factor that cancels makes a hole. The remaining zeros of N are the zeros of f; the remaining zeros of D are vertical asymptotes. Compare degrees for the horizontal asymptote.
| Function | Zeros | Holes | Vertical asymptotes | Horizontal asymptote | y-intercept |
|---|---|---|---|---|---|
| (x² − 4) / (x² − x − 2) | x = −2 | (2, 4/3) | x = −1 | y = 1 | 2 |
| (2x − 5) / (2x − 1) | x = 5/2 | None | x = 1/2 | y = 1 | 5 |
x^3, x³ or x3; multiplication can be implied: 4(x+3)^2(x-1)f(x) =, p(y) =, y =, = 0√3x^2 - 8x + 4√3, sqrt(2)x, 2x^2 - 6.9x + 1(x^2-4)/(x^2-x-2); any single letter works as the variable (x, y, t)Need the division steps? Use the synthetic division calculator or the quadratic formula calculator.
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