How to Find Zeros of Polynomials: Steps and Examples

How to find polynomial zeros step by step

1. Set the output equal to zero. A zero is an input \(c\) satisfying \(f(c)=0\). The corresponding real-coordinate intercept is \((c,0)\).
2. Look for a common factor or familiar pattern. Try a greatest common factor, a difference of squares or factoring by grouping.
3. Solve every factor equation. If a product is zero, at least one factor must equal zero. Keep repeated factors when describing multiplicity.
4. Use another method when factoring is difficult. For \(ax^2+bx+c=0\), with \(a\ne0\), the quadratic formula is \(x=(-b\pm\sqrt{b^2-4ac})/(2a)\). Graphing or numerical methods can estimate real roots.
5. Check in the original polynomial. Substitute each proposed zero. A rounded numerical approximation may leave a small nonzero residual.

For integer-coefficient polynomials, the Rational Zero Theorem supplies possible rational roots from factors of the constant and leading coefficients. Each candidate must be tested. Synthetic division by \(x-c\) gives remainder \(f(c)\); a zero remainder confirms the factor. Neither procedure guarantees that every root is rational.

A nonzero polynomial of degree \(n\) has at most \(n\) distinct real zeros. Over the complex numbers it has exactly \(n\) zeros counted with multiplicity. For real coefficients, nonreal roots occur in conjugate pairs; that fact helps check a result but does not find all roots by itself.

Cubic example: factor before solving

Find the zeros of \(f(x)=x^3-12x^2+20x\).

1. Take out the common factor: \(f(x)=x(x^2-12x+20)\).
2. Find two numbers with product 20 and sum −12: −2 and −10. Thus \(f(x)=x(x-2)(x-10)\).
3. Set the factors equal to zero: \(x=0\), \(x-2=0\), or \(x-10=0\).
4. The zeros are \(0,2,10\). Substituting any of these into the original polynomial gives 0.

Do not divide the equation by \(x\) without first considering \(x=0\). Doing so would discard one of the solutions.

Real intercepts, repeated zeros and complex roots

The graph of a real polynomial crosses the x-axis at a real zero of odd multiplicity and touches it without crossing at a real zero of even multiplicity. For example, \((x-5)^2\) has one distinct zero, 5, with multiplicity 2. Its graph touches the axis at \((5,0)\).

A graph can help estimate real roots, but a viewing window can miss an intercept or hide a repeated zero. Nonreal roots are not intercepts on an ordinary real x-y graph: \(x^2+1=0\) has roots \(i\) and \(-i\), while the real graph has no x-intercepts.

Practice: find the zeros

Factor each polynomial and identify its distinct zeros. Note any repeated root.

1. \(3x^3-19x^2+33x-9\)
2. \(x^2-10x+25\)
3. \(x^3+2x^2-25x-50\)
4. \(x^4+2x^3-16x^2-32x\)

Answers and factor checks

1. \((x-3)^2(3x-1)\): zeros 3 and \(1/3\); 3 has multiplicity 2.
2. \((x-5)^2\): zero 5 has multiplicity 2.
3. \((x+2)(x-5)(x+5)\): zeros −2, 5, −5.
4. \(x(x+2)(x-4)(x+4)\): zeros 0, −2, 4, −4.

Expand the factors to check each polynomial, then substitute the roots to check the zero equations.

Algebra II for Beginners 2026 The Ultimate Step by Step Guide to Acing Algebra II

Download

$27.99 Original price was: $27.99.$17.99Current price is: $17.99.

Rated 5.00 out of 5 based on 1 customer rating

Satisfied 1 Students

Continue with the Algebra 2 learning center. See the OpenStax polynomial-zero lesson and graph and multiplicity guidance for further explanation.

Worked example: rational root theorem and synthetic division

Find the zeros of f(x) = x³ − 2x² − 5x + 6.

1. List candidates. The constant is 6 and the leading coefficient is 1, so possible rational zeros are ±1, ±2, ±3, ±6.
2. Test one. f(1) = 1 − 2 − 5 + 6 = 0, so x = 1 is a zero and (x − 1) is a factor.
3. Divide. Synthetic division by 1 on the coefficients 1, −2, −5, 6: bring down 1; 1 × 1 = 1, −2 + 1 = −1; −1 × 1 = −1, −5 − 1 = −6; −6 × 1 = −6, 6 − 6 = 0. The remainder is 0 and the quotient is x² − x − 6.
4. Finish. x² − x − 6 = (x − 3)(x + 2).

So f(x) = (x − 1)(x − 3)(x + 2) and the zeros are 1, 3 and −2. You can preview this with Descartes’ rule of signs: the coefficients +, −, −, + change sign twice, so there are 2 or 0 positive zeros, and f(−x) = −x³ − 2x² + 5x + 6 changes sign once, so there is exactly 1 negative zero.

How many zeros can a polynomial have?

A polynomial of degree n has exactly n zeros if you count complex zeros and repeated zeros (multiplicity). That is the Fundamental Theorem of Algebra. It can have at most n real zeros, which are its x-intercepts.

4th degree tip: if only even powers appear, substitute u = x². Then x⁴ − 5x² + 4 becomes u² − 5u + 4 = (u − 1)(u − 4), so x² = 1 or x² = 4, giving x = ±1 and x = ±2. Otherwise, use the rational root theorem twice to bring it down to a quadratic.

Frequently Asked Questions

How do you find the zeros of a polynomial step by step?

Set the polynomial equal to zero, then factor. Take out any common factor, look for patterns such as a difference of squares, and use the rational root theorem with synthetic division if nothing else works. Set each factor equal to zero and solve. For x³ − x² − 6x = x(x − 3)(x + 2), the zeros are 0, 3 and −2. Plug each one back in to check.

How do you find the number of zeros of a polynomial?

The degree tells you. A polynomial of degree n has exactly n zeros when complex and repeated zeros are counted, though it can have fewer real ones. For example, x⁴ − 1 = (x² + 1)(x − 1)(x + 1) has degree 4, with real zeros 1 and −1 plus the complex pair i and −i. A polynomial of odd degree always has at least one real zero.

How do you find zeros from standard form?

With the terms written from highest power to lowest, factor or use a formula. For a quadratic ax² + bx + c, the quadratic formula x = (−b ± √(b² − 4ac)) ÷ 2a always works. With 2x² + 3x − 5, factoring gives (2x + 5)(x − 1), so the zeros are x = 1 and x = −5/2. The constant term and leading coefficient in standard form also give the rational root candidates.

How do you find the real zeros of a polynomial?

Real zeros are the x-values where the graph meets the x-axis. Factor completely, then keep only the solutions that are real numbers. In x³ + 4x = x(x² + 4), x = 0 is the only real zero, because x² + 4 = 0 gives x = ±2i, which are complex. A quick graph confirms it: the curve crosses the x-axis just once, at the origin.

How do you find the zeros of a function?

Solve f(x) = 0. The answers are the inputs that make the output zero, and on a graph they are the x-intercepts. For f(x) = x² − 7x + 12, factoring gives (x − 3)(x − 4), so the zeros are 3 and 4. When a function does not factor nicely, the quadratic formula or a graphing calculator can find or estimate the zeros.