Lecture 10 The Fundamental Theorem of Algebra

Lecture 10

The Fundamental Theorem of Algebra

In this chapter, we continue our study of roots of polynomials. In particular, we will learn how to find all rational roots of polynomials, and study complex numbers, which will allow us to state the Fundamental Theorem of Algebra.

Material in this section comes from Section 5.5 of the textbook.

10.1 The Rational Roots Test

The Rational Roots Test (called the Rational Zero Theorem by the book) is a tool that can help us to find the x-intercepts, or roots of any polynomial. The test will allow us to list the possibilities of x-intercepts if they happen to be rational, hence the name.

We recall that a rational number is a number that can be written as a fraction a/b, where both a and b are integers. For example 1/2, 2/3, and 107/259 are rational numbers. Any integer n is a rational number, since it can be written as n/1. Examples of numbers that are not rational are 2 and (although we will not prove that these numbers are not rational).

Theorem 10.1 (The Rational Roots Test). Given a polynomial f (x) = anxn +

an-1xn-1 + . . . + a1x + a0 that has integer coefficients, every rational x-root of

f (x)

has

the

form

p q

where

p

is

any

factor

of

the

constant

term

a0

and

q

is

any

factor of the leading coefficient an.

1

LECTURE 10. THE FUNDAMENTAL THEOREM OF ALGEBRA

2

Example 10.2. List the possible rational x-intercepts given by the rational roots test for the polynomial g(x) = 6x3 + 5x2 - 13x - 2.

We will use the Rational Roots Test to list all possible x-intercepts. We know

that

any

rational

root

will

have

the

form

p q

with

p

and

q

factors

of

the

constant

term

and the leading coefficient, respectively. Thus, we first list the factors of the constant

term 2. They are ?1, and ?2. Then we list the factors of the leading coefficient 6,

they are ?1, ?2, ?3, and ?6. Combining these, we get the following list of possible

rational roots:

11112222 ? ,? ,? ,? ,? ,? ,? ,? .

12361236

Simplifying and removing repetition, this list becomes:

111

2

?1, ? , ? , ? , ?2, ? .

236

3

Note that not all of these possibilities will be rational roots, but all rational roots will be listed among these possibilities. In this case, plugging each of the possible rational roots into g(x), we find that none of them result in zero, so there are no rational roots of g(x).

Advice 10.3. As in this example, we note that the Rational Roots Theorem gives a list of possible rational roots of a polynomial. Every rational root will be in this list, but not every number in the list will be a rational root. In order to determine the actual rational roots, we need to plug each number on the list into the polynomial, and find which ones result in a value of 0.

Example 10.4. Find the roots of the polynomial f (x) = 2x3 - x2 - 2x + 1 using the Rational Roots Test to find possible rational roots.

First, list the factors of 1, which are ?1. Then list factors of 2, which are ?1 and ?2. Then the possible rational roots are given by

?1 ?1 ,

?1 ?2

By examining each possible choice, simplifying, and removing repeats, we find that the list of all possible rational roots is

1 ?1, ? .

2

There are only four possibilities, and we can use synthetic division to check each of them (note that instead of using synthetic division, we could plug the values into the polynomial). As our first guess, we will check to see if -1/2 is a root. Recall that the remainder theorem (from Lecture 8) states that -1/2 is a root of f (x) if x + 1/2 is a factor. Thus we use synthetic division to divide f (x) by x + 1/2, which means we place a -1/2 outside the box.

LECTURE 10. THE FUNDAMENTAL THEOREM OF ALGEBRA

3

-1/2 2 2

-1 -2 1 -1 1 1/2 -2 -1 3/2

In the bottom row, the last entry tells us that the remainder, which is the value

of

f (-1/2)

is

3 2

.

The

remaining

numbers

give

us

that

the

quotient

is

2x2 - 2x - 1,

which we do not need for this problem.

Since f (-1/2) = 3/2 we see that-1/2 is not a root.

As our second choice, we will check whether 1 is a root. Recall that the remainder theorem states that 1 is a root of f (x) if x - 1 is a factor. Thus, we divide f (x) by x - 1 which means we place 1 outside the box.

12 2

-1 -2 1

2

1 -1

1 -1 0

We have remainder 0, so we know that f (1) = 0, so that 1 is a root. We could

keep guessing to see if the other possible roots also work, but a quicker strategy is

to continue to factor f (x). We learned from the synthetic division and the division algorithm that f (x) = (x-1)(2x2+x-1). Now we can factor the remaining quadratic: 2x2 + x - 1 = (2x - 1)(x + 1). Hence,

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

Setting each of these factors equal to zero gives the roots of f (x). There are three

roots,

namely

1,

1 2

,

and

-1.

Remark 10.5. Note that instead of using the rational roots test, we could have factored f (x) = 2x3 - x2 - 2x + 1 by grouping in order to find its roots.

10.2 Introduction to Complex Numbers

Before we can fully grasp the Fundamental Theorem of Algebra, the main subject of this section, we must first introduce the concept of complex numbers. Suppose that you are asked to solve the following equation for x.

x2 + 4 = 0

LECTURE 10. THE FUNDAMENTAL THEOREM OF ALGEBRA

4

You can't see an obvious way to factor it into two linear terms, so you proceed by solving it by isolating x2, then square rooting both sides.

x2 + 4 = 0 x2 = -4

x = ? -4

But, then we stop because the square root of a negative number is not allowed! Mathematicians got to this point too, but eventually were able to come up with a way to describe these numbers even though they are not real numbers. They defined

i = -1

and called i an imaginary number. Thus the answer to our equation would be

x = ? -4

x = ? -1 4 x = ?i ? 2 x = ?2i

Since i is a square root of -1, we see that i2 = -1. Then i3 = -i and i4 = (i2)2 = (-1)2 = 1.

Definition 10.6. Complex Numbers are any numbers that are in the form a + bi where i = -1.

? a is called the real part of the complex number. ? b is called the imaginary part of the complex number.

If b = 0, then a + bi = a + 0i = a is a real number. If a = 0 and b = 0, then the complex number a + bi = bi is said to be an imaginary number.

Remark 10.7. We note that real and imaginary numbers are special cases of complex numbers. A real number like 2 can be considered to be the complex number 2 + 0i; a real number is just a complex number with imaginary part equal to 0. An imaginary number is a nonzero complex number with real part equal to zero, like 2i = 0 + 2i or -7i = 0 + (-7i).

Advice 10.8. Note that for a positive real number a, we can write

-a = -1 a = i a.

LECTURE 10. THE FUNDAMENTAL THEOREM OF ALGEBRA

5

Complex numbers are similar to real numbers; we can add, subtract, multiply and divide complex numbers. Below we explain how these operations work.

Adding and subtracting complex numbers are the easiest operations to calculate. Below we add two complex numbers below by adding the real part and the complex part together separately to get a new complex number.

(1 + 3i) + (4 - 5i) = (1 + 4) + (3 + -5)i = 5 - 2i

Subtraction is the exact same except that you change the sign to subtract instead of add. If it helps, when adding and subtracting complex numbers, we can think of i as if it were a variable.

Example 10.9. Add and subtract -3 + 2i and -4 - 3i.

First, we add the complex numbers. (-3 + 2i) + (-4 - 3i) = (-3 - 4) + (2 - 3)i = -7 - i

Next, we subtract the complex numbers. (-3 + 2i) - (-4 - 3i) = (-3 - (-4)) + (2 - (-3))i = (-3 + 4) + (2 + 3)i = 1 + 5i

We can also multiply complex numbers. We multiply complex numbers exactly as if they were polynomials with i as a variable, except that when we are done, we can replace i2 by -1. Example 10.10. Multiply -3 + 2i and -4 - 3i.

We begin by expanding just like expanding a polynomial. (-3 + 2i)(-4 - 3i) = 12 + 9i - 8i - 6i2 = 12 + i - 6i2

Using the fact that i2 = -1, we can simplify this to 12 + i - 6i2 = 12 + i - 6(-1) = 12 + i + 6 = 18 + i.

Dividing complex numbers takes a little more care than the other operations. First, we must have the definition of a complex conjugate.

LECTURE 10. THE FUNDAMENTAL THEOREM OF ALGEBRA

6

Definition 10.11. The complex conjugate of a + bi is a - bi.

We note the following relations between a complex number and its complex conjugate.

? When a complex number is multiplied by its conjugate, then the result is a real number.

? When a complex number is added to its conjugate, then the result is also a real number.

Example 10.12. The complex conjugate of 2+3i is 2-3i. The complex conjugate of 2 - 3i is 2 + 3i, since 2 - 3i = 2 + (-3)i, so its complex conjugate is 2 - (-3i) = 2 + 3i. Similarly, the complex conjugate of -7 - 5i is -7 + 5i.

Example

10.13.

Calculate

-3+2i -4-3i

.

The object of this problem is simplify the expression to be in the form a + bi. To do this we will get rid of the complex numbers on the denominator. The strategy is to multiply the expression by "1" in a special form:

-4 + 3i = 1,

-4 + 3i

where -4 + 3i is the complex conjugate of -4 - 3i. When we multiply the fraction by this special "1" then we will get a real number on the bottom (see definition of complex conjugate) which will allow us to simplify it to the form a + bi.

-3 + 2i -4 + 3i (-3 + 2i)(-4 + 3i) 12 - 9i - 8i + 6i2

?

=

=

-4 - 3i -4 + 3i (-4 - 3i)(-4 + 3i) 16 - 12i + 12i - 9i2

12 - 17i + 6(-1) 6 - 17i 6 17

=

=

= - i.

16 - 9(-1)

25 25 25

10.3 The Fundamental Theorem of Algebra

The beginning of this lecture discussed the method for finding all the rational roots of a polynomial. The Fundamental Theorem of Algebra discusses all the roots of the polynomial.

Theorem 10.14 (The Fundamental Theorem of Algebra). Any polynomial f (x) of degree n > 0 with real coefficients can be factored as

f (x) = a(x - c1)(x - c2) . . . (x - cn)

where a is a real number and c1, c2, . . . , cn are complex numbers. Therefore, f (x) has n roots, allowing for multiplicities.

LECTURE 10. THE FUNDAMENTAL THEOREM OF ALGEBRA

7

Example 10.15. Find all the roots of the polynomial g(x) = x3 + x2 + x + 1.

First, we will use the Rational Zero Theorem to see if there are any rational roots. We calculate the factors of the constant term 1 and the leading coefficient 1 and come up with the possible rational roots.

?1 = ?1

?1

The value 1 is easily seen not to be a root, since g(1) = 14 + 12 + 1 + 1 = 0. Next, we check whether -1 is a root by using synthetic division to see if (x + 1) is a factor of g(x).

-1 1

1

1

1

-1 0 -1

1010

The last entry of the bottom row is the remainder (and the value of g(-1)). Since the remainder is 0, (x - 1) is a factor of g(x) and the other factor is x2 + 1 (which we read off from the earlier entries in the bottom row). We can solve for the other roots by setting x2 + 1 = 0.

x2 + 1 = 0 x2 = -1 x = ? -1 x = ?i.

Thus, g(x) can be written g(x) = (x - 1)(x + i)(x - i) and the roots of g(x) are 1, i, and -i.

Example 10.16. Find all roots of the polynomial f (x) = x2 + 2x + 3.

This is a quadratic polynomial, so we can use the quadratic formula to find its roots. It has the form f (x) = ax2 + bx + c with a = 1, b = 2, and c = 3. Plugging

LECTURE 10. THE FUNDAMENTAL THEOREM OF ALGEBRA

8

these values into the quadratic formula, we find that the roots are

-b ? b2 - 4ac x=

2a -2 ? 22 - 4(1)(3) =

2(1) -2 ? -8 = 2 -2 ? -1 8 = 2 -2 ? i 8 = 2 -2 ? 2i 2 = 2 = -1 ? i 2, where we use the fact that 8 = 4 2 = 2 2. Hence, we seethat the two roots guaranteed by the Fundamental Theorem of Algebra are -1 + i 2 and -1 - i 2.

We could also have used a calculator to simplify the roots into the form -1+1.414i and -1 - 1.414i.

The following is a useful theorem that corresponds to the Fundamental Theorem of Algebra. Once we know that one complex number is a root, then its conjugate is also a root.

Theorem 10.17 (The Complex Conjugate Theorem). If a polynomial f (x) has real coefficients and the complex number a + bi is a root of of f (x), then a - bi is also a root of f (x).

Example 10.18. Find a polynomial of degree 4 with real coefficients that has -1, 2, and 2i as roots, and satisfies f (1) = 20.

By the Fundamental Theorem of Algebra we know that the polynomial f (x) can be written in the form

f (x) = a(x - c1)(x - c2)(x - c3)(x - c4)

where there are 4 linear factors since we want the degree of f (x) to be 4. Three of

the roots are given in the statement of the problem, so we set c1 = -1, c2 = 2, and

c3 = 2i. Hence,

f (x) = a(x - (-1))(x - 2)(x - 2i)(x - c4).

We still need to find the last root c4 and the value of a. Since f (x) has real coefficients and has 2i as a root, then we also know that -2i is a root of f (x) by the Complex

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download