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## Degree higher than three

Polynomial functions of degree 0 and 1 are called linear functions.

Polynomial functions of degree 2 are called quadratic functions.

In this section we will study polynomial functions of degree three or greater.

What do we want to know about such functions?Generally we want to know two things:

1. What are the zeros of the function?
2. What does the graph of the function look like?

The two issues are related.

If we know what the zeros are, then we know where the $$x$$-intercepts are, so we know something about the graph.

And suppose, for a particular function, we know that the graph lies above the $$x$$-axis at $$1$$ and below it at $$2$$, then we know that there must be a zero somewhere between $$1$$ and $$2$$. This follows from the fact that the graphs of polynomial functions are of one piece--they are continuous, without any gaps or breaks. So the graph cannot pass from one side of the $$x$$-axis to the other without intersecting the $$x$$-axis.

## Exercise 2.3.1

Consider the following third degree polynomial function:$$f ( x ) = x^3 + 2 x^2 - x - 2$$

Find $$f ( 1 ), f ( -1 )$$ and $$f ( -2 )$$ .

See Solution

## Zeros and factors

Notice that the polynomial can be factored by grouping.

$f ( x ) = x^2 ( x + 2 ) - ( x + 2 ) = ( x^2 - 1 ) ( x + 2 ) = ( x - 1 ) ( x + 1 ) ( x + 2 )$

Written it this form, is it easier to see why you got the results you got in exercise 2.3.1?

This example illustrates an important principle about polynomial functions:

The number $$c$$ is a zero of a polynomial function if and only if $$( x - c )$$ is a factor of the polynomial.

So, in principle, the problem of finding the zeros of a polynomial function and the problem of factoring the polynomial are the same problem.One of the main reasons we try to find the zeros of a polynomial function is so that we may factor the polynomial.

## Exercise 2.3.2

Factor the third degree polynomial by finding the zeros of the function:

$$f ( x ) = x^3 - 4 x^2 + x + 6$$

See Solution

## The rational zero principle

How did you go about finding the zeros? By trial and error?

At this point, that is the only method at your disposal--you must try substituting different numbers into the function and see which ones have an output value of zero.

We will now learn some principles which allow us to make educated guesses about what the zeros might be.

One of the most important is the Rational Zero Principle.

Remember, a rational number is a ratio of integers, numbers like $$\frac{2}{3}, \frac{3}{4}$$, etc. Integers are also rational numbers, since they can be identified with a ratio in which the denominator is $$1$$.

The Rational Zero Principle states that if $$\frac{m}{n}$$ is a rational zero of a polynomial function and is reduced to lowest terms, then $$m$$ is a factor of the constant term of the polynomial and $$n$$ is a factor of the leading coefficient of the polynomial.

For example, consider the polynomial function from exercise 2.3.2. The constant term is $$6$$ and the leading coefficient (the coefficient of the highest degree term) is $$1$$. The set of all factors of $$6$$ is $$\{1, -1, 2, -2, 3, -3, 6, -6\}$$. The only factors of $$1$$ are $$1$$ and $$-1$$. The set of all ratios of these numbers is $$\{1, -1, 2, -2, 3, -3, 6, -6\}$$. If the function has any rational zeros, they must be in this list. There is no need to check any other rational number. If a rational number is not in this list, it cannot be a zero of this function. Look again at the solution to exercise 2.3.2 and you will see that all three of the zeros are in this list.

## Exercise 2.3.3

Use the rational zero principle to find the zeros of the function. Then factor the function.

$$f ( x ) = 2 x^3 - 3 x^2 - 11 x + 6$$

See Solution

## The multiplicity of zeros

Consider the polynomial function $$f ( x ) = x^3 - 3 x^2 + 3 x - 1 = ( x - 1 ) ( x - 1 ) ( x - 1 )$$.

Then $$( x - 1 )$$ is a factor of the polynomial three times and $$1$$ is a zero of the function three times. We say that $$1$$ is a multiple zero of the function. Specifically, it is a zero of multiplicity three.

There are principles relating the multiplicity of a zero of a function to the graph of the function.

Remember, each zero specifies an $$x$$-intercept of the graph. The point $$( c, 0 )$$ is an $$x$$-intercept of the graph of a function if and only if $$c$$ is a zero of the function. There are different kinds of $$x$$-intercepts, however.

## Transitive versus intransitive $$x$$-intercepts

The graph may, or may not, actually cross the $$x$$-axis at an intercept. When the graph crosses the $$x$$-axis at an intercept, we say that it is a transitive $$x$$-intercept. The word 'transitive' is from the Latin language and means 'goes across.' When the graph does not cross the $$x$$-axis at the intercept, we say that it is an intransitive $$x$$-intercept.

The Transitivity Principle states that zeros of odd multiplicity correspond to transitive $$x$$-intercepts and zeros of even multiplicity correspond to intransitive $$x$$-intercepts.

The Tangent Principle states that the graph is tangent to the $$x$$-axis at any zero of multiplicity greater than one.

## Exercise 2.3.4

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

Find all the $$x$$-intercepts of the graph of $$f$$ and specify their transitivity.

See Solution

## The leading coefficient principle

We need a couple of more principles, and then we will be able to sketch the basic shape of any polynomial function, provided we can factor it completely.

The Leading Coefficient Principle states that for polynomial functions, the output corresponding to any input greater than the largest zero has the same sign as the leading coefficient of the polynomial.

Thus, if the leading coefficient is negative, then the polynomial function will eventually produce only negative outputs. By 'eventually' we mean, that once we begin picking inputs 'larger than the largest zero', then all outputs will be negative. So if the leading coefficient of a polynomial function is negative, then the rightmost part of the graph will lie below the $$x$$-axis, and if the leading coefficient is positive, then the rightmost part of the graph will lie above the $$x$$-axis.

## Exercise 2.3.5

What is the leading coefficient of the polynomial function in exercise 2.3.4? Use the leading coefficient principle, the tangent principle and the transitivity principle to sketch the basic shape of the graph.

See Solution

## How many bends can the graph have?

There is an additional graphing principle which will help us make our graphs more authentic.

Notice that there are no bends in a polynomial function of degree one, since its graph is a straight line. There is only one bend in the graph of a polynomial function of degree two. In general, the maximum number of bends in the graph of a polynomial function is equal to the degree minus one. Now the graph of a polynomial needn't have all the bends it is allowed. It may have fewer bends than that. However, if it does have fewer bends, the difference between the maximum number of bends it is allowed and the number of bends it actually has must be an even number. Thus a degree $$7$$ polynomial function may have $$6$$, $$4$$ or $$2$$ bends in its graph. A degree $$6$$ polynomial function may have $$5$$, $$3$$ or $$1$$ bend in its graph.

## Exercise 2.3.6

Sketch the basic shape of the graph of the function:

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

See Solution