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Course: Algebra (all content) > Unit 10

Lesson 32: Zeros of polynomials and their graphs

Positive & negative intervals of polynomials

Learn about the relationship between the zeros of polynomials and the intervals over which they are positive or negative.

What you should be familiar with before taking this lesson

The zeros of a polynomial

f

correspond to the

x

-intercepts of the graph of

y = f (x)

For example, let's suppose

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

. Since the zeros of function

f

are

- 3

and

1

, the graph of

y = f (x)

will have

x

-intercepts at

(- 3, 0)

and

(1, 0)

If this is new to you, we recommend that you check out our zeros of polynomials article.

What you will learn in this lesson

While the

x

-intercepts are an important characteristic of the graph of a function, we need more in order to produce a good sketch.

Knowing the sign of a polynomial function between two zeros can help us fill in some of the gaps.

In this article, we'll learn how to determine the intervals over which a polynomial is positive or negative and connect this back to the graph.

Positive and negative intervals

The sign of a polynomial between any two consecutive zeros is either always positive or always negative.

For example, consider the graphed function

f (x) = (x + 1) (x - 1) (x - 3)

From the graph, we see that

f (x)

is always ...

...negative when $- \infty < x < - 1$ ‍.
...positive when $- 1 < x < 1$ ‍.
...negative when $1 < x < 3$ ‍.
...positive when $3 < x < \infty$ ‍.

It is not necessary, however, for a polynomial function to change signs between zeros.

For example, consider the graphed function

g (x) = x (x + 2)^{2}

From the graph, we see that

g (x)

is always...

...negative when $- \infty < x < - 2$ ‍.
...negative when $- 2 < x < 0$ ‍.
...positive when $0 < x < \infty$ ‍.

Notice that

g (x)

does not change sign around

x = - 2

Determining the positive and negative intervals of polynomials

Let's find the intervals for which the polynomial

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

is positive and the intervals for which it is negative.

The zeros of

f

are

- 3

and

1

. This creates three intervals over which the sign of

f

is constant:

Let’s find the sign of

f

for

- \infty < x < - 3

We know that

f

will either be always positive or always negative on this interval. We can determine which is the case by evaluating

f

for one value in this interval. Since

- 4

is in this interval, let's find

f (- 4)

Because we are only interested in the sign of the polynomial here, we don't have to completely evaluate it:

\begin{aligned} f (x) & = (x + 3) (x - 1)^{2} \\ f (- 4) & = (- 4 + 3) (- 4 - 1)^{2} \\ = (-) (-)^{2} & Evaluate only the sign of the answer. \\ = (-) (+) & A negative squared is a positive. \\ = - & A negative times a positive is a negative. \end{aligned}

Here we see that

f (- 4)

is negative, and so

f (x)

will always be negative for

- \infty < x < - 3

We can repeat the process for the remaining intervals.

Interval: $- 3 < x < 1$ ‍

Since

0

is in this interval, let's find the sign of

f (0)

$\begin{aligned} f (x) & = (x + 3) (x - 1)^{2} \\ f (0) & = (0 + 3) (0 - 1)^{2} \\ = (+) (-)^{2} & Evaluate only the sign of the answer. \\ = (+) (+) & A negative squared is a positive. \\ = + & A positive times a positive is a positive. \end{aligned}$ ‍

Because

f (0)

is positive,

f (x)

will be positive for

- 3 < x < 1

Interval: $1 < x < \infty$ ‍

Since

2

is in this interval, let's find the sign of

f (2)

$\begin{aligned} f (x) & = (x + 3) (x - 1)^{2} \\ f (2) & = (2 + 3) (2 - 1)^{2} \\ = (+) (+)^{2} & Evaluate only the sign of the answer. \\ = (+) (+) & A positive squared is a positive \\ = + & A positive times a positive is a positive. \end{aligned}$ ‍

Because

f (2)

is positive,

f (x)

will be positive for

1 < x < \infty

The results are summarized in the table below.

Interval	The value of a specific $f (x)$ ‍ within the interval	Sign of $f$ ‍ on interval	Connection to graph of $f$ ‍
$- \infty < x < - 3$ ‍	$f (- 4) < 0$ ‍	negative	Below the $x$ ‍-axis
$- 3 < x < 1$ ‍	$f (0) > 0$ ‍	positive	Above the $x$ ‍-axis
$1 < x < \infty$ ‍	$f (2) > 0$ ‍	positive	Above the $x$ ‍-axis

This is consistent with the graph of

y = f (x)

Check your understanding

g (x) = (x + 1)^{2} (x + 6)

has zeros at

x = - 6

and

x = - 1

What is the sign of $g$ ‍ on the interval $- 6 < x < - 1$ ‍?

h (x) = (3 - x) (x + 5) (x - 2)

has zeros at

x = - 5

x = 2

, and

x = 3

What is the sign of $h (x)$ ‍ on the interval $- 5 < x < 2$ ‍?

Challenge problem

3*) Which of the following could be the graph of $g (x) = (x - 2)^{2} (x + 1)^{3}$ ‍?

Determining positive & negative intervals from a sketch of the graph

Another way to determine the intervals over which a polynomial is positive or negative is to draw a sketch of its graph, based on the polynomial's end behavior and the multiplicities of its zeros.

Check out our graphs of polynomials article for further details.

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