We've updated our
Privacy Policy effective December 15. Please read our updated Privacy Policy and tap

Study Guides > MATH 1314: College Algebra

Find zeros of a polynomial function

The Rational Zero Theorem helps us to narrow down the list of possible rational zeros for a polynomial function. Once we have done this, we can use synthetic division repeatedly to determine all of the zeros of a polynomial function.

How To: Given a polynomial function ff\\, use synthetic division to find its zeros.

  1. Use the Rational Zero Theorem to list all possible rational zeros of the function.
  2. Use synthetic division to evaluate a given possible zero by synthetically dividing the candidate into the polynomial. If the remainder is 0, the candidate is a zero. If the remainder is not zero, discard the candidate.
  3. Repeat step two using the quotient found with synthetic division. If possible, continue until the quotient is a quadratic.
  4. Find the zeros of the quadratic function. Two possible methods for solving quadratics are factoring and using the quadratic formula.

Example 5: Finding the Zeros of a Polynomial Function with Repeated Real Zeros

Find the zeros of f(x)=4x33x1f\left(x\right)=4{x}^{3}-3x - 1\\.

Solution

The Rational Zero Theorem tells us that if pq\frac{p}{q}\\ is a zero of f(x)f\left(x\right)\\, then is a factor of –1 and q is a factor of 4.

{pq=factor of constant termfactor of leading coefficient =factor of -1factor of 4\begin{cases}\frac{p}{q}=\frac{\text{factor of constant term}}{\text{factor of leading coefficient}}\hfill \\ \text{ }=\frac{\text{factor of -1}}{\text{factor of 4}}\hfill \end{cases}\\

The factors of –1 are ±1\pm 1\\ and the factors of 4 are ±1,±2\pm 1,\pm 2\\, and ±4\pm 4\\. The possible values for pq\frac{p}{q}\\ are ±1,±12\pm 1,\pm \frac{1}{2}\\, and ±14\pm \frac{1}{4}\\. These are the possible rational zeros for the function. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let’s begin with 1.

Synthetic division with 1 as the divisor and {4, 0, -3, -1} as the quotient. Solution is {4, 4, 1, 0}

Dividing by (x1)\left(x - 1\right)\\ gives a remainder of 0, so 1 is a zero of the function. The polynomial can be written as

(x1)(4x2+4x+1)\left(x - 1\right)\left(4{x}^{2}+4x+1\right)\\.

The quadratic is a perfect square. f(x)f\left(x\right)\\ can be written as

(x1)(2x+1)2\left(x - 1\right){\left(2x+1\right)}^{2}\\.

We already know that 1 is a zero. The other zero will have a multiplicity of 2 because the factor is squared. To find the other zero, we can set the factor equal to 0.

{2x+1=0 x=12\begin{cases}2x+1=0\hfill \\ \text{ }x=-\frac{1}{2}\hfill \end{cases}\\

The zeros of the function are 1 and 12-\frac{1}{2}\\ with multiplicity 2.

Analysis of the Solution

Look at the graph of the function f in Figure 1. Notice, at x=0.5x=-0.5\\, the graph bounces off the x-axis, indicating the even multiplicity (2,4,6…) for the zero –0.5. At x=1x=1\\, the graph crosses the x-axis, indicating the odd multiplicity (1,3,5…) for the zero x=1x=1\\.
Graph of a polynomial that have its local maximum at (-0.5, 0) labeled asFigure 1

Licenses & Attributions