Identify the x-Intercepts of Polynomial Functions whose Equations are Factorable

In this section, you will:

• Recognize characteristics of graphs of polynomial functions.
• Use factoring to find zeros of polynomial functions.
• Identify zeros and their multiplicities.
• Determine end behavior.
• Understand the relationship between degree and turning points.
• Graph polynomial functions.
• Use the Intermediate Value Theorem.

The revenue in millions of dollars for a fictional cable company from 2006 through 2013 is shown in Table 1.

Table 1

The revenue can be modeled by the polynomial function

$R(t)=-0.037 t^{4}+1.414 t^{3}-19.777 t^{2}+118.696 t-205.332$

where $R$ represents the revenue in millions of dollars and $t$ represents the year, with $t=6$ corresponding to $2006$. Over which intervals is the revenue for the company increasing? Over which intervals is the revenue for the company decreasing? These questions, along with many others, can be answered by examining the graph of the polynomial function. We have already explored the local behavior of quadratics, a special case of polynomials. In this section we will explore the local behavior of polynomials in general.

Source: Rice University, https://openstax.org/books/college-algebra/pages/5-3-graphs-of-polynomial-functions
This work is licensed under a Creative Commons Attribution 4.0 License.

Polynomial functions of degree $2$ or more have graphs that do not have sharp corners; recall that these types of graphs are called smooth curves. Polynomial functions also display graphs that have no breaks. Curves with no breaks are called continuous. Figure 1 shows a graph that represents a polynomial function and a graph that represents a function that is not a polynomial.

Figure 1

EXAMPLE 1

Recognizing Polynomial Functions

Which of the graphs in Figure 2 represents a polynomial function?

Figure 2

Solution

The graphs of $f$ and $h$ are graphs of polynomial functions. They are smooth and continuous.

The graphs of $g$ and $k$ are graphs of functions that are not polynomials. The graph of function $g$ has a sharp corner. The graph of function $k$ is not continuous.

Q&A

Do all polynomial functions have as their domain all real numbers?

Yes. Any real number is a valid input for a polynomial function.

Recall that if $f$ is a polynomial function, the values of $x$ for which $f(x)=0$ are called zeros of f. If the equation of the polynomial function can be factored, we can set each factor equal to zero and solve for the zeros.

We can use this method to find $x$- intercepts because at the $x$- intercepts we find the input values when the output value is zero. For general polynomials, this can be a challenging prospect. While quadratics can be solved using the relatively simple quadratic formula, the corresponding formulas for cubic and fourth-degree polynomials are not simple enough to remember, and formulas do not exist for general higher-degree polynomials. Consequently, we will limit ourselves to three cases:

HOW TO

Given a polynomial function $f$, find the $x$-intercepts by factoring.

EXAMPLE 2

Finding the $x$-Intercepts of a Polynomial Function by Factoring

Find the x-intercepts of $f(x)=x^{6}-3 x^{4}+2 x^{2}$.

Solution

We can attempt to factor this polynomial to find solutions for $f(x)=0$.

$\begin{aligned} x^{6}-3 x^{4}+2 x^{2} &=0 & & \text { Factor out the greatest } \\ & & & \text { common factor. } \\ x^{2}\left(x^{4}-3 x^{2}+2\right) &=0 & & \text { Factor the trinomial. } \\ x^{2}\left(x^{2}-1\right)\left(x^{2}-2\right) &=0 & & \text { Set each factor equal to zero. } \end{aligned}$

$\left(x^{2}-1\right)=0 \quad\left(x^{2}-2\right)=0$

$x^{2}=0 \quad \text { or } \quad x^{2}=1 \quad \text { or } \quad x^{2}=2$

$x=0 \quad x=\pm 1 \quad x=\pm \sqrt{2}$

This gives us five x-intercepts: $(0,0)$, $(1,0)$, $(−1,0)$, $(\sqrt{2}, 0)$, and $(-\sqrt{2}, 0)$. See Figure 3. We can see that this is an even function because it is symmetric about the $y$-axis.

Figure 3

EXAMPLE 3

Finding the $x$-Intercepts of a Polynomial Function by Factoring

Find the x-intercepts of $f(x)=x^{5}-5 x^{2}-x+5$.

Solution

Find solutions for $f(x)=0$ by factoring.

$\begin{aligned} x^{3}-5 x^{2}-x+5 &=0 & & \text { Factor by grouping. } \\ x^{2}(x-5)-(x-5) &=0 & & \text { Factor out the common factor. } \\ \left(x^{2}-1\right)(x-5) &=0 & & \text { Factor the difference of squares. } \\ (x+1)(x-1)(x-5) &=0 & & \text { Set each factor equal to zero. } \end{aligned}$

$x+1=0 \quad \text { or } \quad x-1=0 \quad \text { or } \quad x-5=0$

$x= -1 \quad \text { or } \quad x=1 \quad \text { or } \quad x=5$

There are three x-intercepts: $(−1,0)$, $(1,0)$, and $(5,0)$. See Figure 4.

Figure 4

EXAMPLE 4

Finding the $y$- and $x$-Intercepts of a Polynomial in Factored Form

Find the $y$- and $x$-intercepts of $g(x)=(x-2)^{2}(2 x+3)$.

Solution

The $y$-intercept can be found by evaluating $g(0)$

$\begin{aligned} g(0) &=(0-2)^{2}(2(0)+3) \\ &=12 \end{aligned}$

So the $y$-intercept is $(0,12)$.

The $x$-intercepts can be found by solving $g(x)=0$.

$(x-2)^{2}(2 x+3)=0$

$\begin{aligned} (x-2)^{2} &=0 & & &(2 x+3) &=0 \\ x-2 &=0 & \text { or } & & x &=-\frac{3}{2} \\ x &=2 & & & \end{aligned}$

So the $x$-intercepts are $(2,0)$ and $(-\frac{3}{2},0)$.

Analysis

We can always check that our answers are reasonable by using a graphing calculator to graph the polynomial as shown in Figure 5.

Figure 5

EXAMPLE 5

Finding the $x$-Intercepts of a Polynomial Function Using a Graph

Find the $x$-intercepts of $h(x)=x^{3}+4 x^{2}+x-6$.

Solution

This polynomial is not in factored form, has no common factors, and does not appear to be factorable using techniques previously discussed. Fortunately, we can use technology to find the intercepts. Keep in mind that some values make graphing difficult by hand. In these cases, we can take advantage of graphing utilities.

Looking at the graph of this function, as shown in Figure 6, it appears that there are $x$-intercepts at $x=−3,−2$, and $1$.

Figure 6

We can check whether these are correct by substituting these values for $x$ and verifying that

$h(-3)=h(-2)=h(1)=0$

Since $h(x)=x^{5}+4 x^{2}+x-6$, we have:

$\begin{aligned} h(-3) &=(-3)^{3}+4(-3)^{2}+(-3)-6=-27+36-3-6=0 \\ h(-2) &=(-2)^{3}+4(-2)^{2}+(-2)-6=-8+16-2-6=0 \\ h(1) &=(1)^{3}+4(1)^{2}+(1)-6=1+4+1-6=0 \end{aligned}$

Each $x$-intercept corresponds to a zero of the polynomial function and each zero yields a factor, so we can now write the polynomial in factored form.

$\begin{aligned} h(x) &=x^{3}+4 x^{2}+x-6 \\ &=(x+3)(x+2)(x-1) \end{aligned}$

TRY IT #1

Find the $y-$ and $x$-intercepts of the function $f(x)=x^{4}-19 x^{2}+30 x$.

Graphs behave differently at various $x$-intercepts. Sometimes, the graph will cross over the horizontal axis at an intercept. Other times, the graph will touch the horizontal axis and "bounce" off.

Suppose, for example, we graph the function shown.

$f(x)=(x+3)(x-2)^{2}(x+1)^{3}$

Notice in Figure 7 that the behavior of the function at each of the $x$-intercepts is different.

Figure 7 Identifying the behavior of the graph at an $x$-intercept by examining the multiplicity of the zero.

The $x$-intercept $x=−3$ is the solution of equation $(x+3)=0$. The graph passes directly through the $x$-intercept at $x=−3$. The factor is linear (has a degree of $1$), so the behavior near the intercept is like that of a line—it passes directly through the intercept. We call this a single zero because the zero corresponds to a single factor of the function.

The $x$-intercept $x=2$ is the repeated solution of equation $(x-2)^{2}=0$. The graph touches the axis at the intercept and changes direction. The factor is quadratic (degree $2$), so the behavior near the intercept is like that of a quadratic—it bounces off of the horizontal axis at the intercept.

$(x-2)^{2}=(x-2)(x-2)$

The factor is repeated, that is, the factor $(x−2)$ appears twice. The number of times a given factor appears in the factored form of the equation of a polynomial is called the multiplicity. The zero associated with this factor, $x=2$, has multiplicity $2$ because the factor $(x−2)$ occurs twice.

The $x$-intercept $x=−1$ is the repeated solution of factor $(x+1)^{3}=0$. The graph passes through the axis at the intercept, but flattens out a bit first. This factor is cubic (degree $3$), so the behavior near the intercept is like that of a cubic—with the same S-shape near the intercept as the toolkit function $f(x)=x^{3}$. We call this a triple zero, or a zero with multiplicity $3$.

For zeros with even multiplicities, the graphs touch or are tangent to the $x$-axis. For zeros with odd multiplicities, the graphs cross or intersect the $x$-axis. See Figure 8 for examples of graphs of polynomial functions with multiplicity $1$, $2$, and $3$.

Figure 8

For higher even powers, such as $4$, $6$, and $8$, the graph will still touch and bounce off of the horizontal axis but, for each increasing even power, the graph will appear flatter as it approaches and leaves the $x$-axis.

For higher odd powers, such as $5$, $7$, and $9$, the graph will still cross through the horizontal axis, but for each increasing odd power, the graph will appear flatter as it approaches and leaves the $x$-axis.

GRAPHICAL BEHAVIOR OF POLYNOMIALS AT $X$-INTERCEPTS

If a polynomial contains a factor of the form $(x-h)^{p}$, the behavior near the $x$- intercept $h$ is determined by the power $p$. We say that $x=h$ is a zero of multiplicity $p$.

The graph of a polynomial function will touch the $x$-axis at zeros with even multiplicities. The graph will cross the $x$-axis at zeros with odd multiplicities.

The sum of the multiplicities is the degree of the polynomial function.

HOW TO

Given a graph of a polynomial function of degree $n$, identify the zeros and their multiplicities.

1. If the graph crosses the $x$-axis and appears almost linear at the intercept, it is a single zero.

2. If the graph touches the $x$-axis and bounces off of the axis, it is a zero with even multiplicity.

3. If the graph crosses the $x$-axis at a zero, it is a zero with odd multiplicity.

4. The sum of the multiplicities is $n$.

EXAMPLE 6

Identifying Zeros and Their Multiplicities

Use the graph of the function of degree $6$ in Figure 9 to identify the zeros of the function and their possible multiplicities.

Figure 9

Solution

The polynomial function is of degree $6$. The sum of the multiplicities must be 6.

Starting from the left, the first zero occurs at $x=−3$. The graph touches the $x$-axis, so the multiplicity of the zero must be even. The zero of $−3$ most likely has multiplicity $2$.

The next zero occurs at $x=−1$. The graph looks almost linear at this point. This is a single zero of multiplicity $1$.

The last zero occurs at $x=4$. The graph crosses the $x$-axis, so the multiplicity of the zero must be odd. We know that the multiplicity is likely $3$ and that the sum of the multiplicities is 6.

TRY IT #2

Use the graph of the function of degree $9$ in Figure 10 to identify the zeros of the function and their multiplicities.

Figure 10