Three Techniques for Evaluating and Finding Zeros of Polynomial Functions

HOW TO

Given a polynomial function $f$, 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 $f(x)=4 x^{3}-3 x-1$.

Solution

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

$\begin{aligned} \frac{p}{q} &=\frac{\text { factor of constant term }}{\text { factor of leading coefficient }} \\ &=\frac{\text { factor of }-1}{\text { factor of } 4} \end{aligned}$

The factors of $–1$ are ±1 and the factors of $4$ are $\pm 1, \pm 2$, and $\pm 4$. The possible values for $\frac{p}{q}$ are $\pm 1, \pm \frac{1}{2}$, and $\pm \frac{1}{4}$. 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 $1$.

Dividing by $(x−1)$ gives a remainder of $0$, so $1$ is a zero of the function. The polynomial can be written as

$(x-1)\left(4 x^{2}+4 x+1\right)$

The quadratic is a perfect square. $f(x)$ can be written as

$(x-1)(2 x+1)^{2}$

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 $0$.

$\begin{aligned} 2 x+1 &=0 \\ x &=-\frac{1}{2} \end{aligned}$

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

Analysis

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

Figure 1