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 [latex]f[/latex], 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 [latex]f\left(x\right)=4{x}^{3}-3x - 1[/latex].

Solution

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

[latex]\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}[/latex]

The factors of –1 are [latex]\pm 1[/latex] and the factors of 4 are [latex]\pm 1,\pm 2[/latex], and [latex]\pm 4[/latex]. The possible values for [latex]\frac{p}{q}[/latex] are [latex]\pm 1,\pm \frac{1}{2}[/latex], and [latex]\pm \frac{1}{4}[/latex].
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 [latex]\left(x - 1\right)[/latex] gives a remainder of 0, so 1 is a zero of the function. The polynomial can be written as

[latex]\left(x - 1\right)\left(4{x}^{2}+4x+1\right)[/latex].

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

[latex]\left(x - 1\right){\left(2x+1\right)}^{2}[/latex].

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.

[latex]\begin{cases}2x+1=0\hfill \\ \text{ }x=-\frac{1}{2}\hfill \end{cases}[/latex]

The zeros of the function are 1 and [latex]-\frac{1}{2}[/latex] with multiplicity 2.

Analysis of the Solution

Look at the graph of the function f in Figure 1. Notice, at [latex]x=-0.5[/latex], the graph bounces off the x-axis, indicating the even multiplicity (2,4,6…) for the zero –0.5. At [latex]x=1[/latex], the graph crosses the x-axis, indicating the odd multiplicity (1,3,5…) for the zero [latex]x=1[/latex].

Graph of a polynomial that have its local maximum at (-0.5, 0) labeled as

Figure 1