# Real Zeros of Polynomials

These are my notes on real zeros of polynomials.

In this section, we discuss the techniques that can be used to find the real

zeros of a polynomial function. Recall that if r is a real zero of a polynomial

function f then f(r)=0, r is an x-intercept of the graph of f, and r is a

solution of the equation f(x)=0.

For polynomial and rational functions, we have seen the importance of the real

zeros for graphing. In most cases, the real zeros of a polynomial function are

difficult to find using algebraic methods. No nice formulas like the quadratic

formula are available to help us find zeros for polynomials of degree 3 or

higher. Formulas do exist for solving any third or fourth degree polynomial

equation, but they are complicated to use. No general formulas exist for

polynomial equations of degree 5 or higher.**Remainder and Factor Theorems**

When we divide one polynomial by another, we obtain a quotient polynomial and a

remainder, the remainder being either the zero polynomial or a polynomial whose

degree is less than the degree of the divisor. To check our work, we do:

(quotient)(divisor) + remainder = dividend

This checking routine is the basis for a famous theorem called the division

algorithm for polynomials.**Remainder Theorem**

Let f be a polynomial function. If f(x) is divided by x-c, then the remainder is

f(c).**Example 1**

Using the remainder theorem

Find the remainder if \(f(x)=x^3-4x^2-5\) when divided by x-3 and x+2

Solution:

We could use long division, but who wants to do that, it is easier to use the

remainder theorem.

\(f(3)=(3)^3-4(3)^2-5 = 27-36-5 = 14\)

The remainder is 14

To find the remainder when f(x) is divided by x+2 = x-(-2) = -2

\(f(-2) = (-2)^3-4(-2)^2-5 = -8-16-5 = -29\)

The remainder is -29**Factor Theorem**

Let f be a polynomial function. Then x-c is a factor of f(x) if ands only if

f(c)=0. If f(c)=0, then x-c is a factor of f(x). If x-c is a factor of f(x),

then f(c)=0.

Suppose that f(c)=0. Then, we have \(f(x)=(x-c)q(x)\) for some polynomial q(x).

That is, x-c is a factor of f(x). Suppose that x-c is a factopr of f(x). Then

there is a polynomial function q such that \(f(x)=(x-c)q)x)\). Replacing x by c,

we find that \(f(c)=(c-c)q(c) = 0*q(c) = 0\). That completes the proof. One use

of the factor theorem is to determine whether a polynomial has a particular

factor.**Example 2**

Use the factor theorem to determine whether a function has a factor

\(f(x)=2x^3-x^2+2x-3\)

For x-1 and x+3

Solution:

That factor theorem states that if f(c)=0 then x-c is a factor

Because x-1 is of the form x-c with c=1, we find the value of f(1). We choose to

use substitution.

\(f(1) = 2(1)^3-(1)^2+2(1)-3 = 2-1+2-3 = 0\)

By the factor theorem, x-1 is a factor of x.

To test the factor x+3, we first need to write it in the form of x-c. Since

x+3=x-(-3), we find the value of f(-3). Because f(-3) =-72 and not 0, we

conclude from the factor theorem that x-(-3) or x+3 is not a factor of f(x).**Number of Real Zeros**

The next theorem concerns the number of real zeros that a polynomial function

may have. In counting the zeros of a polynomial, we count each zero as many

times as its multiplicity.

A polynomial function cannot have more real zeros than its degree.

The proof is based on the factor theorem. If r is a real zero of a polynomial

function f, then f(r)=0 and so, x-r is a factor of f(x). Each real zero

corresponds to a factor of degree 1. Because f cannot have more first degree

factors than its degree.

Descarte's rule of signs provides information about the number and location of

the real zeros of a polynomial function written in standard form. It requires

that we count the number of variations in the sign of the coefficients of f(x)

and f(-x). For example, the following polynomial function has two variations in

the signs of the coefficients.

\(f(x)=3x^7+4x^4+3x^2-2x-1 = -3x^7+0x^6+0x^5+4x^4+0x^3+3x^2-2x-1\)

Notice that we ignored the zero coefficients in \(0x^6,0x^5\) and \(0x^3\) in

counting the number of variations in the sign of f(x). Replacing x by -x, we

get:

\(f(-x)=-3(-x)^7+4(-x)^4+3(-x)^2-2(-x)-1 = 3x^7+4x^4+3x^2+2x-1\)

This has one variation in sign.

Descarte's Rule of Signs

Let f denote a polynomial function written in standard form.

The number of positive real zeros of f either equals the number of variations in

the sign of the nonzero coefficients of f(x) or else equals that number less an

even integer.

The number of negative real zeros of f either equals the number of variations in

the sign of the nonzero coefficients of f(-x) or else equals that number less an

even integer.**Example 3**

Using the number of real zeros theorem and the rule of signs.

Discuss the real zeros of \(f(x)=3x^6-4x^4+3x^3+2x^2-x-3\)

Solution:

Because the polynomial is of degree 6, by the real zeros theorem there are at

most six real zeros. Since there are three variations in the sign of the nonzero

coefficients of f(x), by the rule of signs, we expect either three or one

positive real zero. To continue, look at f(-x).

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

There are three variations in sign, so we expect either three or one negative

real zeros. Equivalently, we know now that the graph of f has either three or one

positive x-intercepts and three or one negative x-intercepts.**Rational Zeros Theorem**

The rational zeros theorem provides information about the rational zeros of a

polynomial with integer coefficients.

Let f be a polynomial function of degree 1 or higher of the form:

\(f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...a_{1}x+a_{0}\)

Each coefficient is an integer. If p/q is in lowest terms, it is a rational zero

of f, then p must be a factor of \(a_o\), and q must be a factor of \(a_n\).**Example 4**

Listing potential rational zeros

\(f(x)=2x^3+11x^2-7x-6\)

Solution:

Because f has integer coefficients, we may use the rational zero theorem. First,

we list all the integers p that are factors of the constant term a0=-6 and all

the integers q that are factors of the leading coefficient ay=2.

\(p:....\pm1,\pm2,\pm3,\pm6\)

\(q:....\pm1,\pm2\)

Now, we form all possible ratios p/q

\(\frac{p}{q}:....\pm1,\pm2,\pm3,\pm6,\pm\frac{1}{2},\pm\frac{3}{2}\)

If f has a rational zero, it will be found in this list, which contains 12

possibilities.

Be sure that you understand what the rational zeros theorem says. For a

polynomial with integer coefficients, if there is a rational zero, it is one of

those listed. It may be the case that the function does not have any rational

zeros.

Long division, synthetic division, or substitution can be used to test each

potential rational zero to determine whether it is indeed a zero. To make the

work easier, integers are usually tested first.**Example 5**

Finding the rational zeros of a polynomial function

\(f(x)=2x^3+11x^2-7x-6\)

Write f in factored form

Solution:

We gather all the information that we can about the zeros.

1. There are at most 3 zeros

2. By the rule of signs, there is one positive real zero. Also, because

\(f(-x)=-2x^3+11x^2+7x-6\)

There are two negative zeros or no negative zeros.

3. Now we use the list of potential rational zeros obtained in the previous

example.

\(\pm1,\pm2,\pm3,\pm6,\pm\frac{1}{2},\pm\frac{3}{2}\)

We choose to test the potential rational zero using substitution.

\(f(1)=2(1)^3+11(1)^2-7(1)-6 = 2+11-7-6 = 0\)

Since f(1)=0, 1 is a zero and x-1 is a factor of f. We can use long division or

synthetic division to factor f.

\(f(x)=2x^3+11x^2-7x-6 = (x-1)(2x^2+13x+6)\)

Now, any solution of the equation \(2x^2+13x+6=0\) will be a zero of f. Because

of this, we call the equation a depressed equation of f. Since the degree of the

depressed equation of f is less than that of the original polynomial, it is

easier to work with the depressed equation to find the zeros of f.

4. The depressed equation \(2x^2+13x+6=0\) is a quadratic equation with

discriminant \(b^2-4ac = 169-48 = 121 > 0\). The equation has two real

solutions, which can be found by factoring:

\(2x^2+13x+6 = (2x+1)(x+6) = 0\)

\(2x+1 = 0\) and \(x+6 = 0\)

So, \(x=-\frac{1}{2}\) or \(x = -6\)

The zeros of f are \(-6,-\frac{1}{2},1\)

We know that \(f(x)=2x^3+11x^2-7x-6\) can be written as

\(f(x)=(x-1)(2x^2+13x+6)\). Because \(2x^2+13x+6 = (2x+1)(x+6)\), we write f in

factored form as: \(f(x)=(x-1)(2x+1)(x+6)\)

Notice that the three zeros of f are found in this example are among those given

in the list of potential zeros. **Find the Real Zeros of a Polynomial Function**

1. Use the degree of the polynomial to determine the maximum number of real

zeros.

2. Use the rule of signs to determine the possible number of positive zeros and

negative zeros.

3. If the polynomial has integer coefficients, use the rational zeros theorem to

identify those rational numbers that potentially could be zeros.

Use substitution, synthetic division, or long division to test each potential

rational zero. Each time that a zero(and thus a factor) is found, repeat step 3

on the depressed equation.

4. In attempting to find the zeros, remember to use the factoring techniques

that you already know.**Example 6**

Finding the real zeros of a polynomial function

Find the real zeros of \(f(x)=x^5-5x^4+12x^3-24x^2+32x-16\)

Write f in factored form.

Solution:

We gather all the information that we can about the zeros.

There are at most 5 real zeros because of \(x^5\).

By the rule of signs, there are 5, three, or one positive zero.

There are no negative zeros because \(f(x)=-x^5-5x^4-12x^3-24x^2-32x-16\) has no

variation in sign.

Because the leading coefficient \(a_{5}=1\) and there are no negative zeros, the

potential rational zeros are the integers 1,2,4,8, and 16, the positive factors

of the constant term, 16. We test the potential rational zero 1 first, using

synthetic division.

We get a remainder of f(1)=0, so 1 is a zero and x-1 is a factor of f. Using the

entries in the bottom row of the synthetic division, we can begin to factor f.

\(f(x)=x^5-5x^4+12x^3-24x^2+32x-16 = (x-1)(x^4-4x^3+8x^2-16x+16)\)

We now work with the first depressed equation:

\(q1(x)=x^4-4x^3+8x^2-16x+16=0\)

The potential rational zeros of q1 are still 1,2,4,8, and 16. We test 1 first

since it may be a repeated zero of f.

Since the remainder is 5, 1 is not a repeated zero. We try 2 next:

The remainder is f(2)=0 so 2 is a zero and x-2 is a factor of f. Again using the

bottom row, we find:

\(f(x)=x^5-5x^4+12x^3-24x^2+32x-16 = (x-1)(x-2)(x^3-2x^2+4x-8)\)

The remaining zeros satisfy the new depressed equation:

\(q2(x)=x^3-2x^2+4x-8=0\)

Notice that q2(x) can be factored using grouping.

Since \(x^2+4=0\) has no real solutions, the real zeros of f are 1 and 2, the

latter being a zero of multiplicity 2. The factored form of f is:

\(f(x)=x^5-5x^4+12x^3-24x^2+32x-16 = (x-1)(x-2)^2(x^2+4)\)**Example 7**

Solving a polynomial equation

Find the real solutions of the equation:

\(x^5-5x^4+12x^3-24x^2+32x-16=0\)

Solution:

The real solutions of this equation are the real zeros of the polynomial

function \(f(x)=x^5-5x^4+12x^3-24x^2+32x-16\)

Using the result from previous example, the real zeros of f are 1 and 2. These

are the real solutions of the equation:

\(x^5-5x^4+12x^3-24x^2+32x-16=0\)

In example 6, the quadratic factor \(x^2+4\) that appears in the factored form

of f is called irreducible, because the polynomial \(x^2+4\) cannot be factored

over the real numbers. In general, we say that a quadratic factor \(ax^2+bx+c\)

is irreducible if it cannot be factored over the real numbers, that is, if it is

prime over the real numbers.

Refer to examples 5 and 6. The polynomial function of example 5 has three real

zeros, and its factored form contains three linear factors. The polynomial

function of example 6 has two distinct real zeros, and its factored form

contains two distinct linear factors and one irreducible quadratic factor.

Every polynomial function with real coefficients can be uniquely factored into a

product of linear factors and/or irreducible quadratic factors.

We shall prove this result and we shall draw several additional conclusions

about the zeros of a polynomial function. One conclusion is worth noting. If a

polynomial with real coefficients is of odd degree, it must contain at least one

linear factor. This means it must have at least one real zero.

A polynomial function with real coefficients of odd degree has at least one real

zero.**Theorem for Bounds on Zeros**

The search for the real zeros of a polynomial function can be reduced somewhat

if bounds on the zeros are found. A number M is a bound on the zeros of a

polynomial if every zero lies between -M and M. That is, M is a bound on the

zeros of a polynomial if \(-M \leq\text{ any real zero of f }\leq M\)**Example 8**

Using the theorem for finding bounds on zeros

Find a bound on the real zeros of each polynomial

1. \(f(x)=x^5+3x^3-9x^2+5\)

2. \(g(x)=4x^5-2x^3+2x^2+1\)

Solution:

1. The leading coefficient of f is 1.

\(f(x)=x^5+3x^3-9x^2+5 = 0,3,-9,0 = 5\)

The smaller of the two numbers, 10, is the bound. Every real zero of f lies

between -10 and 10.

2. First, we write g so that it is the product of a constant times a polynomial

whose leading coefficient is 1.

\(g(x)=4x^5-2x^3+2x^2+1 = 4(x^5-\frac{1}{2}x^3+\frac{1}{2}x^2+\frac{1}{4}\)

Next, we evaluate the two expressions in (4) with

\(a_4=0,a_3=-\frac{1}{2},a_2=\frac{1}{2},a_1=0\text{ and }a_0=\frac{1}{4}\)

The smaller of the two numbers, \(\frac{5}{4}\) is the bound. Every real zero

of g lies between \(-\frac{5}{4}\text{ and }\frac{5}{4}\)**Intermediate Value Theorem**

It is based on the fact that the graph of a polynomial function is continuous,

that is, it contains no holes or gaps. Let f denote a polynomial function. If

a<b and if f(a) and f(b) are of opposite sign, there is at least one real zero

of f between a and b.**Example 9**

Using the intermediate value theorem to locate real zeros

Show that \(f(x)=x^5-x^3-1\) has a real zero between 1 and 2.

Solution:

We evaluate f at 1 and 2

\(f(1)=-1\)

\(f(2)=23\)

Because f(1)<0 and f(2)>0, it follows from the intermediate value theorem that f

has a zero between 1 and 2.

Let us look at the polynomial f of example 9 more closely. Based on the rule of

signs, f has exactly one positive real zero. Based on the rational zeros

theorem, 1 is the only potential positive rational zero. Since f(1) does not

equal 0, we conclude that the zero between 1 and 2 is irrational. We can use the

intermediate value theorem to approximate it.

1. Find two consecutive integers a and a+1 such that f has a zero between them.

2. Divide the interval into 10 equal subintervals.

3. Evaluate f at each endpoint of the subintervals until the intermediate value

theorem applies. This endpoint then contains a zero.

4. Repeat the process until the desired accuracy is achieved.**Example 10**

Approximating the real zeros of a polynomial function

Find the positive zero of:

\(f(x)=x^5-x^3-1\) to two decimal places.

Solution:

From example 9, we know that the positive zero is between 1 and 2. We divide the

interval [1,2] into 10 equal subintervals.

[1,1.1][1.1,1.2][1.2,[1.3][1.3,1.4][1.4,1.5][1.5,1.6][1.6,1.7][1.7,1.8][1.9,1.9][1.9,2].

Now we find the value of f at each endpoint until the intermediate value theorem

applies.

\(f(x)=x^5-x^3-1\)

f(1.0)=-1 f(1.2)=-0.23968 f(1.1)=-0.72049 f(1.3)=0.51593

We can stop here and conclude that the zero is between 1.2 and 1.3. now we

divide the interval[1.2,1.3] into 10 equal subintervals and proceed to evaluate

f at each endpoint. We conclude that the zero lies between 1.23 and 1.24 and

correct to two decimal places, the zero is 1.23.