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

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

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

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

The factors of –1 are $\pm 1$ 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 $\left(x - 1\right)$ gives a remainder of 0, so 1 is a zero of the function. The polynomial can be written as

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

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

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

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.

$\begin{array}{l}2x+1=0\hfill \\ \text{ }x=-\frac{1}{2}\hfill \end{array}$

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

Analysis of the Solution

Look at the graph of the function f. 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 $x=1$.

Example: Finding the Zeros of a Polynomial Function with Complex Zeros

Find the zeros of $f\left(x\right)=3{x}^{3}+9{x}^{2}+x+3$.

Answer: The Rational Zero Theorem tells us that if $\frac{p}{q}$ is a zero of $f\left(x\right)$, then p is a factor of 3 and q is a factor of 3.

$\begin{array}{l}\frac{p}{q}=\frac{\text{factor of constant term}}{\text{factor of leading coefficient}}\hfill \\ \text{ }=\frac{\text{factor of 3}}{\text{factor of 3}}\hfill \end{array}$

The factors of 3 are $\pm 1$ and $\pm 3$. The possible values for $\frac{p}{q}$, and therefore the possible rational zeros for the function, are $\pm 3,\text{\pm 1, and }\pm \frac{1}{3}$. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let’s begin with –3. Dividing by $\left(x+3\right)$ gives a remainder of 0, so –3 is a zero of the function. The polynomial can be written as $$\left(x+3\right)\left(3{x}^{2}+1\right)$$ We can then set the quadratic equal to 0 and solve to find the other zeros of the function.

$\begin{array}{l}3{x}^{2}+1=0\hfill \\ \text{ }{x}^{2}=-\frac{1}{3}\hfill \\ \text{ }x=\pm \sqrt{-\frac{1}{3}}=\pm \frac{i\sqrt{3}}{3}\hfill \end{array}$

The zeros of $f\left(x\right)$ are –3 and $\pm \frac{i\sqrt{3}}{3}$.

Analysis of the Solution

Look at the graph of the function f. Notice that, at $x=-3$, the graph crosses the x-axis, indicating an odd multiplicity (1) for the zero $x=-3$. Also note the presence of the two turning points. This means that, since there is a 3^rd degree polynomial, we are looking at the maximum number of turning points. So, the end behavior of increasing without bound to the right and decreasing without bound to the left will continue. Thus, all the x-intercepts for the function are shown. So either the multiplicity of $x=-3$ is 1 and there are two complex solutions, which is what we found, or the multiplicity at $x=-3$ is three. Either way, our result is correct.