# 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, equals, f, left parenthesis, x, right parenthesis.
For example, let's suppose f, left parenthesis, x, right parenthesis, equals, left parenthesis, x, plus, 3, right parenthesis, left parenthesis, x, minus, 1, right parenthesis, start superscript, 2, end superscript. Since the zeros of function f are minus, 3 and 1, the graph of y, equals, f, left parenthesis, x, right parenthesis will have x-intercepts at left parenthesis, minus, 3, comma, 0, right parenthesis and left parenthesis, 1, comma, 0, right parenthesis.
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.
This is because polynomial functions are continuous functions (no breaks in the graph), which means that the only way to change signs is to cross the x-axis. But if this happened, the given zeros would not be consecutive!
For example, consider the graphed function f, left parenthesis, x, right parenthesis, equals, left parenthesis, x, plus, 1, right parenthesis, left parenthesis, x, minus, 1, right parenthesis, left parenthesis, x, minus, 3, right parenthesis.
From the graph, we see that f, left parenthesis, x, right parenthesis is always ...
• ...negative when minus, infinity, is less than, x, is less than, minus, 1.
• ...positive when minus, 1, is less than, x, is less than, 1.
• ...negative when 1, is less than, x, is less than, 3.
• ...positive when 3, is less than, x, is less than, infinity.
It is not necessary, however, for a polynomial function to change signs between zeros.
For example, consider the graphed function g, left parenthesis, x, right parenthesis, equals, x, left parenthesis, x, plus, 2, right parenthesis, start superscript, 2, end superscript.
From the graph, we see that g, left parenthesis, x, right parenthesis is always...
• ...negative when minus, infinity, is less than, x, is less than, minus, 2.
• ...negative when minus, 2, is less than, x, is less than, 0.
• ...positive when 0, is less than, x, is less than, infinity.
Notice that g, left parenthesis, x, right parenthesis does not change sign around x, equals, minus, 2.

# Determining the positive and negative intervals of polynomials

Let's find the intervals for which the polynomial f, left parenthesis, x, right parenthesis, equals, left parenthesis, x, plus, 3, right parenthesis, left parenthesis, x, minus, 1, right parenthesis, start superscript, 2, end superscript is positive and the intervals for which it is negative.
The zeros of f are minus, 3 and 1. This creates three intervals over which the sign of f is constant:
Let’s find the sign of f for minus, infinity, is less than, x, is less than, minus, 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 minus, 4 is in this interval, let's find f, left parenthesis, minus, 4, right parenthesis.
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 &&\small{\gray{\text{Evaluate only the sign of the answer.}}}\\\\ &=(-)(+)&&\small{\gray{\text{A negative squared is a positive.}}}\\ \\ &=-&&\small{\gray{\text{A negative times a positive is a negative.}}}\end{aligned}
Here we see that f, left parenthesis, minus, 4, right parenthesis is negative, and so f, left parenthesis, x, right parenthesis will always be negative for minus, infinity, is less than, x, is less than, minus, 3.
We can repeat the process for the remaining intervals.
Interval: minus, 3, is less than, x, is less than, 1
Since 0 is in this interval, let's find the sign of f, left parenthesis, 0, right parenthesis.
\begin{aligned} f(x) &= (x+3)(x-1)^2\\ \\ f(0) &= ({0+3})({0-1})^2 \\\\ &= ( +)(-)^2 &&\small{\gray{\text{Evaluate only the sign of the answer.}}}\\\\ &=(+)(+)&&\small{\gray{\text{A negative squared is a positive.}}}\\ \\ &=+&&\small{\gray{\text{A positive times a positive is a positive.}}}\end{aligned}
Because f, left parenthesis, 0, right parenthesis is positive, f, left parenthesis, x, right parenthesis will be positive for minus, 3, is less than, x, is less than, 1.
Interval: 1, is less than, x, is less than, infinity
Since 2 is in this interval, let's find the sign of f, left parenthesis, 2, right parenthesis.
\begin{aligned} f(x) &= (x+3)(x-1)^2\\ \\ f(2) &= ({2+3})({2-1})^2 \\\\ &= ( +)(+)^2 &&\small{\gray{\text{Evaluate only the sign of the answer.}}}\\\\ &=(+)(+)&&\small{\gray{\text{A positive squared is a positive}}}\\\\ &=+&&\small{\gray{\text{A positive times a positive is a positive.}}}\end{aligned}
Because f, left parenthesis, 2, right parenthesis is positive, f, left parenthesis, x, right parenthesis will be positive for 1, is less than, x, is less than, infinity.
The results are summarized in the table below.
IntervalThe value of a specific f, left parenthesis, x, right parenthesis within the intervalSign of f on intervalConnection to graph of f
minus, infinity, is less than, x, is less than, minus, 3f, left parenthesis, minus, 4, right parenthesis, is less than, 0negativeBelow the x-axis
minus, 3, is less than, x, is less than, 1f, left parenthesis, 0, right parenthesis, is greater than, 0positiveAbove the x-axis
1, is less than, x, is less than, infinityf, left parenthesis, 2, right parenthesis, is greater than, 0positiveAbove the x-axis
This is consistent with the graph of y, equals, f, left parenthesis, x, right parenthesis.

### Check your understanding

1) g, left parenthesis, x, right parenthesis, equals, left parenthesis, x, plus, 1, right parenthesis, start superscript, 2, end superscript, left parenthesis, x, plus, 6, right parenthesis has zeros at x, equals, minus, 6 and x, equals, minus, 1.
What is the sign of g on the interval minus, 6, is less than, x, is less than, minus, 1?
Please choose from one of the following options.

We can determine which is the case by finding the sign of g, left parenthesis, x, right parenthesis for one value in this interval.
Since minus, 6, is less than, minus, 2, is less than, minus, 1, let's find the sign of g, left parenthesis, minus, 2, right parenthesis.
\begin{aligned} g(x) &= (x+1)^2(x+6)\\ \\ g(-2) &= ({-2+1})^2({-2+6}) \\\\ &= ( -)^2(+)&&\small{\gray{\text{Evaluate only the sign of the answer.}}}\\\\ &=(+)(+)&&\small{\gray{\text{A negative square is a positive.}}}\\\\ &=+\end{aligned}
Because g, left parenthesis, minus, 2, right parenthesis is positive, g, left parenthesis, x, right parenthesis is always positive on the interval minus, 6, is less than, x, is less than, minus, 1.
2) h, left parenthesis, x, right parenthesis, equals, left parenthesis, 3, minus, x, right parenthesis, left parenthesis, x, plus, 5, right parenthesis, left parenthesis, x, minus, 2, right parenthesis has zeros at x, equals, minus, 5, x, equals, 2, and x, equals, 3.
What is the sign of h, left parenthesis, x, right parenthesis on the interval minus, 5, is less than, x, is less than, 2?
Please choose from one of the following options.

We can determine which is the case by finding the sign of h, left parenthesis, x, right parenthesis for one value in this interval.
Since minus, 5, is less than, 0, is less than, 2, let's find the sign of h, left parenthesis, 0, right parenthesis.
\begin{aligned} h(x) &= (3-x)(x+5)(x-2)\\ \\ h(0) &= (3-0)(0+5)(0-2) \\\\ &= ( +)(+)(-) &\small{\gray{\text{Evaluate only the sign of the answer.}}}\\\\ &=-\end{aligned}
Because h, left parenthesis, 0, right parenthesis is negative, h, left parenthesis, x, right parenthesis is always negative on the interval minus, 5, is less than, x, is less than, 2.

### Challenge problem

3*) Which of the following could be the graph of g, left parenthesis, x, right parenthesis, equals, left parenthesis, x, minus, 2, right parenthesis, start superscript, 2, end superscript, left parenthesis, x, plus, 1, right parenthesis, start superscript, 3, end superscript?
Please choose from one of the following options.

We are given that g, left parenthesis, x, right parenthesis, equals, left parenthesis, x, minus, 2, right parenthesis, start superscript, 2, end superscript, left parenthesis, x, plus, 1, right parenthesis, start superscript, 3, end superscript. Since the zeros of function g are 2 and minus, 1, the graph of y, equals, g, left parenthesis, x, right parenthesis will have x-intercepts at left parenthesis, 2, comma, 0, right parenthesis and left parenthesis, minus, 1, comma, 0, right parenthesis.
This is true of all of the graphs, so let's use what we know about positive and negative intervals to determine the correct option.
The zeros of function g create the following three intervals over which the sign of g, left parenthesis, x, right parenthesis is constant: minus, infinity, is less than, x, is less than, minus, 1, minus, 1, is less than, x, is less than, 2, and 2, is less than, x, is less than, plus, infinity.
Let's find the sign of g, left parenthesis, x, right parenthesis for one specific x-value in each interval. Then we will know the sign of all g, left parenthesis, x, right parenthesis values for any x in that interval.
IntervalThe value of a specific g, left parenthesis, x, right parenthesis within the intervalSign of g on intervalConnection to graph of g
minus, infinity, is less than, x, is less than, minus, 1g, left parenthesis, minus, 2, right parenthesis, is less than, 0negativeBelow the x-axis
minus, 1, is less than, x, is less than, 2g, left parenthesis, 0, right parenthesis, is greater than, 0positiveAbove the x-axis
2, is less than, x, is less than, plus, infinityg, left parenthesis, 3, right parenthesis, is greater than, 0positiveAbove the x-axis
The only graph to reflect this is A.
Alternatively, we could have used information about end behavior and the multiplicities of the zeros to select the correct graph.

### 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.