2.5 Zeros of Polynomial Functions

How many zeros does a polynomial function have?

A polynomial function of degree \(n\) has \(n\) zeros, provided multiple zeros are counted more than once and provided complex zeros are counted.

Keep in mind that any complex zeros of a function are not considered to be part of the domain of the function, since only real numbers domains are being considered. It is only for the purpose of counting zeros that we consider complex and multiple zeros.

Exercise 2.5.1

Let \(f ( x ) = x^2 + 1\).

Use the quadratic formula to show that the imaginary numbers \(i\) and \(- i\) are the two zeros of \(f\). Verify this using synthetic division.

See Solution

Exercise 2.5.2

Let \(f ( x ) = x^2 + 2 x + 1\).

Factor \(f ( x ) = x^2 + 2 x + 1\) and show that it contains the same linear factor twice. Thus it has a double zero.

See Solution

Exercise 2.5.3

Let \(f ( x ) = 2 x^3 + 3 x^2 + 8 x - 5\).

This function \(f(x)\) has one real zero and two complex zeros.

Use the rational zero principle from section 2.3 to list all possible rational zeros. Then use synthetic division from section 2.4 to find a rational zero from among the possible rational zeros. Use synthetic division to factor the function. Use the quadratic formula on the quotient to find the two complex zeros of the function.

See Solution

Complex zeros occur in pairs.

Notice that in exercises 2.5.1 and 2.5.3, the complex zeros were conjugates of each other. This always happens. We can state the following general principle:

Complex zeros of polynomial functions occur in conjugate pairs.

Descartes Rule of Signs for positive zeros

Next, we are going to investigate a general principle which tells us how many positive zeros a polynomial function may have.

Consider the polynomial function \(f ( x ) = 2 x^3 + 3 x^2 + 8 x - 5\). Notice that it is written in standard form for polynomials, that is, terms are written in decreasing order according to their exponents. Now, count the number of changes in sign of the coefficients. There is exactly one such change in sign. Thus, there can be only one positive zero.

Descartes' rule of signs for positive zeros may be stated as follows:

When a polynomial function is written in standard form, the number of changes in sign of the coefficients is the maximum number of positive zeros of the function. The actual number of positive zeros may be less than the maximum by an even amount.

Exercise 2.5.4

Consider the function \(g ( x ) = 2 x^5 - 8 x^4 + 7 x^3 - 2 x^2 + 5 x - 1\).

Use the positive zeros principle to show that \(g\) has either five, three or one positive zeros.

See Solution

Descartes rule of signs for negative zeros.

We can use we can use Descartes' Rule of Signs for positive zeros to find a Descartes' rule of signs for negative zeros..

Consider the graph of y\( = g ( - x )\). It will look like the graph of \(y = g ( x )\) reflected about the \(y\)-axis. Thus there will be a one-to-one correspondence with the negative zeros of \(y = g ( x )\) and the positive zeros of \(y = g ( - x )\). That is to say, the number of positive zeros of

\[ y = g ( - x ) = - 2 x^5 - 8 x^4 - 7 x^3 - 2 x^2 - 5 x - 1 \]

equals the number of negative zeros of

\[ y = g ( x ) = 2 x^ - 8 x^4 + 7 x^3 - 2 x^2 + 5 x - 1\]

Because there are no sign changes in the cofficients of\( y = g ( - x )\), it follows that \(y = g ( x )\) has no negative zeros.

The following table shows the possible distribution of the five zeros of \(g ( x )\) :

Positive\(5\)\(3\)\(1\)
Negative\(0\)\(0\)\(0\)
Complex\(0\)\(2\)\(4\)

In other words, there are only three possible arrangements of the zeros of \(g\). Either \(g\) has five positive zeros, no negative and no complex zeros, or it has three positive zeros, no negative and two complex zeros, or it has one positive zero, no negative and four complex zeros.

Exercise 2.5.5

Draw a chart showing the possible distribution of the zeros of the function

\(p ( x ) = 2 x^5 + 7 x^3 - 2 x^2 + 5 x - 1\)

See Solution

Bounds on sets of zeros.

Next, let us consider the polynomial function \(f ( x ) = x^4 + 5 x^3 + 3 x^2 + 2 x + 1\). Since there are no changes of sign in the coefficients, there are no positive zeros.

Now, consider the function \(g ( x ) = ( x - 1 ) ( x^4 + 5 x^3 + 3 x^2 + 2 x + 1 )\).

Its only positive zero is 1, since ( x - 1 ) is a factor and since \(x^4 + 5 x^3 + 3 x^2 + 2 x + 1\) has no positive zeros. Furthermore, since the leading coefficient is positive, the graph of \(y = g ( x )\) to the right of \(x =1\) lies entirely above the \(x\)-axis.

Now, consider the polynomial function \(h ( x ) = ( x - 1 ) ( x^4 + 5 x^3 + 3 x^2 + 2 x + 1 ) + 3\).

Its graph is just the graph of \(y = g ( x )\) raised upwards by three units. Thus \(y = h ( x )\) lies entirely above the \(x\)-axis for \(x\ge1\). So \(1\) is an upper bound on the set of zeros of the polynomial function \(y = h ( x )\).

Next, consider the polynomial function \(f ( x ) = - x^4 - 2 x^3 - x^2 - 5 x - 2\). Since it has no change in the signs of its coefficients, it has no positive zeros. Thus the function \(g ( x ) = ( x - 1 ) ( - x^4 - 2 x^3 - x^2 - 5 x - 2 )\) has no zeros greater than \(1\). And since the leading coefficient of \(g\) is negative, all the graph of \(g\) to the right of \(1\) lies below the \(x\)-axis. Thus the graph of a function such as \(h ( x ) = ( x - 1 ) ( - x^4 - 2 x^3 - x^2 - 5 x - 2 ) - 3\), for example, will also lie entirely below the \(x\)-axis to the right of \(1\). Thus \(1\) would be an upper bound on the zeros of \(h\).

Now what is the point of all this? How can this example be generalized?

In general, if \(f ( x ) = ( x - c ) q ( x ) + r\) and if \(r\) and all the coefficients of \(q ( x )\) are of the same sign, then \(c\) is an upper bound on the set of zeros of \(f ( x )\).

Let us see how to use this principle to find an upper bound on the set of zeros of the function \(f ( x ) = x^5 - 2 x^4 + 7 x^3 - x^2 - 12 x - 3\).

Use synthetic division to rewrite the function in the following form: \(f ( x ) = ( x - 2 ) ( x^4 + 7 x^2 + 13 x + 14 ) + 25\).

Then, using the principle above, \(2\) is an upper bound on the set of zeros of \(f\).

If we are searching for the zeros of a polynomial function by means of synthetic division and find that, when evaluating\(f ( c )\), for some \(c > 0\), we get a row of all non-negative (or non-positive) numbers from the synthetic division process, then we know that \(c\) is an upper bound on the set of zeros of \(f\).

Exercise 2.5.6

Use synthetic division to find the smallest positive integer which is an upper bound on the zeros of the function \(f ( x ) = x^3 - 3 x^2 - 3 x + 1\)

See Solution

Exercise 2.5.7

Find the smallest positive integer which is an upper bound on the zeros of the function \(g ( x ) = - x^3 + 3 x^2 + 3 x - 1\).

See Solution

Greatest integral lower bounds on the set of negative zeros

This procedure can be adapted to find negative numbers which are lower bounds on the set of zeros of a polynomial function. Consider the following fact: The number \(- c\) is a zero of \(y = f ( x )\) if and only if \(c\) is a zero of \(y = f ( - x )\). [Remember that the graph of \(y = f ( - x )\) is the reflection of the graph of \(y = f ( x )\) in the \(y\)-axis.] Thus, we can find the negative lower bounds of \(y = f ( x )\) by finding the positive upper bounds of \(y = f ( - x )\) and changing the sign.

Exercise 2.5.8

Find the largest negative integer which is a lower bound on the zeros of the function \(f ( x ) = x^3 - 5 x + 2\).

See Solution

Exercise 2.5.9

Find the largest negative integer which is a lower bound on the zeros of the function \(g( x ) = - x^3 + 5 x - 2\).

See Solution