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College Algebra

Zeros of Polynomials

Lesson

A zero of a polynomial is a value of xxthat makes the polynomial equal zero. For polynomials beyond degree 2, factoring isn’t always obvious — but the Rational Root Theorem gives you a short list of candidates to test.

For a polynomial with integer coefficients, any rational zero pq\tfrac{p}{q} (in lowest terms) must satisfy:

  • pp divides the constant term.
  • qq divides the leading coefficient.

Practical version: when the leading coefficient is 1, just try the integer factors of the constant term (positive and negative).

The full strategy:

  1. List candidate rational zeros.
  2. Test them by substituting (look for f(c)=0f(c) = 0).
  3. Once you find one, use synthetic division to factor out (x−c)(x - c).
  4. Solve the smaller polynomial that’s left — factor it or use the quadratic formula.

Worked example 1

f(x)=x3−6x2+11x−6f(x) = x^3 - 6x^2 + 11x - 6

Constant term is −6, leading coefficient is 1, so candidates are ±1,±2,±3,±6\pm 1, \pm 2, \pm 3, \pm 6. Try x=1x = 1:

1−6+11−6=0✓1 - 6 + 11 - 6 = 0 \checkmark

Synthetic division by (x−1)(x - 1) gives quotient x2−5x+6x^2 - 5x + 6, which factors as (x−2)(x−3)(x - 2)(x - 3).

Zeros: x=1, 2, 3\text{Zeros: } x = 1,\ 2,\ 3

Worked example 2

f(x)=2x3−3x2−8x−3f(x) = 2x^3 - 3x^2 - 8x - 3

Constant −3, leading coefficient 2, so candidates are ±1,±3,±12,±32\pm 1, \pm 3, \pm \tfrac{1}{2}, \pm \tfrac{3}{2}. Try x=3x = 3:

2(27)−3(9)−8(3)−3=54−27−24−3=0✓2(27) - 3(9) - 8(3) - 3 = 54 - 27 - 24 - 3 = 0 \checkmark

Synthetic division by (x−3)(x - 3) gives quotient 2x2+3x+1=(2x+1)(x+1)2x^2 + 3x + 1 = (2x + 1)(x + 1).

Zeros: x=3, −12, −1\text{Zeros: } x = 3,\ -\tfrac{1}{2},\ -1

How to type your answer

List every zero, separated by commas — order doesn’t matter. Use fractions where helpful. Examples: 1,2,3, 0,3,-3, 3,-1/2,-1.

Practice

Work through these. Stuck? Click Get a hint.

Warm-Up

Quick problems to get going.

Problem 1

x3−x=0x^3 - x = 0

Problem 2

x3−4x=0x^3 - 4x = 0

Problem 3

x3−9x=0x^3 - 9x = 0

Problem 4

x3−x2−6x=0x^3 - x^2 - 6x = 0

Practice

Standard problems matching the lesson.

Problem 5

x3−6x2+11x−6=0x^3 - 6x^2 + 11x - 6 = 0

Problem 6

x3−x2−4x+4=0x^3 - x^2 - 4x + 4 = 0

Problem 7

x3+2x2−x−2=0x^3 + 2x^2 - x - 2 = 0

Problem 8

x3−7x+6=0x^3 - 7x + 6 = 0

Problem 9

x3−4x2+x+6=0x^3 - 4x^2 + x + 6 = 0

Problem 10

x3+3x2−4x−12=0x^3 + 3x^2 - 4x - 12 = 0

Problem 11

x3−2x2−5x+6=0x^3 - 2x^2 - 5x + 6 = 0

Problem 12

x3+x2−4x−4=0x^3 + x^2 - 4x - 4 = 0

Problem 13

x3−3x2−4x+12=0x^3 - 3x^2 - 4x + 12 = 0

Problem 14

x3+4x2+x−6=0x^3 + 4x^2 + x - 6 = 0

Challenge

Harder problems — edge cases, trickier numbers, multiple steps.

Problem 15

2x3−3x2−8x−3=02x^3 - 3x^2 - 8x - 3 = 0

Problem 16

2x3+x2−5x+2=02x^3 + x^2 - 5x + 2 = 0

Problem 17

3x3−x2−8x−4=03x^3 - x^2 - 8x - 4 = 0

Problem 18

x4−5x2+4=0x^4 - 5x^2 + 4 = 0

Problem 19

x4−13x2+36=0x^4 - 13x^2 + 36 = 0

Problem 20

x3−19x+30=0x^3 - 19x + 30 = 0

Problem 21

2x3−7x2+7x−2=02x^3 - 7x^2 + 7x - 2 = 0

Problem 22

2x3+5x2−4x−3=02x^3 + 5x^2 - 4x - 3 = 0

Practice

Standard problems matching the lesson.

Problem 23

Population P(t) = (t − 2)(t + 1)(t − 5). Find all zeros.

Problem 24

Rational roots of x³ − 6x² + 11x − 6 = 0.

Challenge

Harder problems — edge cases, trickier numbers, multiple steps.

Problem 25

Rational roots of 2x³ − x² − 8x + 4 = 0.

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