precalculus complex zeros v. j. motto. introduction we have already seen that an nth-degree...

PrecalculusComplex Zeros

V. J. Motto

Introduction

We have already seen that an nth-degree

polynomial can have at most n real zeros.

• In the complex number system, an nth-degree polynomial has exactly n zeros.

• Thus, it can be factored into exactly n linear factors.

• This fact is a consequence of the Fundamental Theorem of Algebra, which was proved by the German mathematician C. F. Gauss in 1799.

Fundamental Theorem of Algebra

The following theorem is the basis

for much of our work in:

• Factoring polynomials.

• Solving polynomial equations.

The Fundamental Theorem of Algebra

Every polynomial

P(x) = anxn + an-1xn-1 + . . . + a1x + a0

(n ≥ 0, an ≠ 0)

with complex coefficients has at least

one complex zero.

• As any real number is also a complex number, it applies to polynomials with real coefficients too.

Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra

and the Factor Theorem together show that

a polynomial can be factored completely

into linear factors—as we now prove.

Complete Factorization Theorem

If P(x) is a polynomial of degree n ≥ 1, then

there exist complex numbers a, c1, c2, . . . , cn

(with a ≠ 0) such that

P(x) = a(x – c1) (x – c2 ) . . . (x – cn)

Complete Factorization Theorem—Proof

By the Fundamental Theorem of Algebra,

P has at least one zero—which we will call c1.

By the Factor Theorem, P(x) can be factored

as:

P(x) = (x – c1) · Q1(x)

where Q1(x) is of degree n – 1.

Complete Factorization Theorem—Proof

Applying the Fundamental Theorem to

the quotient Q1(x) gives us the factorization

P(x) = (x – c1) · (x – c2) · Q2(x)

where:• Q2(x) is of degree n – 2.

• c2 is a zero of Q1(x).

Complete Factorization Theorem—Proof

Continuing this process for n steps,

we get a final quotient Qn(x) of degree 0—a

nonzero constant that we will call a.

• This means that P has been factored as:

P(x) = a(x – c1)(x – c2) ··· (x – cn)

Complex Zeros

To actually find the complex zeros of

an nth-degree polynomial, we usually:

• First, factor as much as possible.

• Then, use the quadratic formula on parts that we can’t factor further.

E.g. 1—Factoring a Polynomial Completely

Let P(x) = x3 – 3x2 + x – 3

(a) Find all the zeros of P.

(b) Find the complete factorization of P.

E.g. 1—Factoring Completely

We first factor P as follows.

3 2

2

2

3 3

3 3

3 1

P x x x x

x x x

x x

Example (a)

E.g. 1—Factoring Completely

We find the zeros of P by setting each

factor equal to 0:

P(x) = (x – 3)(x2 + 1)

• Setting x – 3 = 0, we see that x = 3 is a zero.

• Setting x2 + 1 = 0, we get x2 = –1; so, x = ±i.

• Thus, the zeros of P are 3, i, and –i.

Example (a)

E.g. 1—Factoring Completely

Since the zeros are 3, i, and i,

by the Complete Factorization Theorem,

P factors as:

3

3

P x x x i x i

x x i x i

Example (b)

E.g. 2—Factoring a Polynomial Completely

Let P(x) = x3 – 2x + 4.

(a) Find all the zeros of P.

(b) Find the complete factorization of P.

E.g. 2—Factoring Completely

The possible rational zeros are

the factors of 4: ±1, ±2, and ±4.

• Using synthetic division, we find that –2 is a zero, and the polynomial factors as:

P(x) = (x + 2) (x2 – 2x + 2)

Example (a)

E.g. 2—Factoring Completely

To find the zeros, we set each factor

equal to 0.

• Of course, x + 2 = 0 means x = –2.

• We use the quadratic formula to find when the other factor is 0.

Example (a)

E.g. 2—Factoring Completely

• So, the zeros of P are –2, 1 + i, and 1 – i.

2 2 2 0

2 4 8

22 2

21

x x

x

ix

x i

Example (a)

E.g. 2—Factoring Completely

Since the zeros are – 2, 1 + i, and 1 – i,

by the Complete Factorization Theorem,

P factors as:

2 1 1

2 1 1

P x x x i x i

x x i x i

Example (b)

Zeros and Their Multiplicities

Zeros and Their Multiplicities

In the Complete Factorization Theorem,

the numbers c1, c2, . . . , cn are the zeros

of P.

• These zeros need not all be different.

• If the factor x – c appears k times in the complete factorization of P(x), we say that c is a zero of multiplicity k.

Zeros and Their Multiplicities

For example, the polynomial

P(x) = (x – 1)3(x + 2)2(x + 3)5

has the following zeros:

• 1(multiplicity 3)

• –2(multiplicity 2)

• –3(multiplicity 5)

Zeros and Their Multiplicities

The polynomial P has the same

number of zeros as its degree.

• It has degree 10 and has 10 zeros—provided we count multiplicities.

• This is true for all polynomials—as we prove in the following theorem.

Zeros Theorem

Every polynomial of degree n ≥ 1

has exactly n zeros—provided a zero

of multiplicity k is counted k times.

Zeros Theorem—Proof

Let P be a polynomial of degree n.

• By the Complete Factorization Theorem,

P(x) = a(x – c1)(x – c2) ··· (x – cn)

Zeros Theorem—Proof

Now, suppose that c is a zero of P

other than c1, c2, . . . , cn.

• Then,

P(c) = a(c – c1)(c – c2) ··· (c – cn) = 0

Zeros Theorem—Proof

Thus, by the Zero-Product Property,

one of the factors c – ci must be 0.

• So, c = ci for some i.

• It follows that P has exactly the n zeros c1, c2, . . . , cn.

E.g. 3—Factoring a Polynomial with Complex Zeros

Find the complete factorization and

all five zeros of the polynomial

P(x) = 3x5 + 24x3 + 48x

E.g. 3—Factoring a Polynomial with Complex Zeros

Since 3x is a common factor,

we have:

• To factor x2 + 4, note that 2i and –2i are zeros of this polynomial.

4 2

22

3 8 16

3 4

P x x x x

x x

E.g. 3—Factoring a Polynomial with Complex Zeros

Thus, x2 + 4 = (x – 2i )(x + 2i ).

Therefore,

• The zeros of P are 0, 2i, and –2i. • Since the factors x – 2i and x + 2i each occur twice

in the complete factorization, the zeros 2i and –2i are of multiplicity 2 (or double zeros).

• Thus, we have found all five zeros.

2

2 2

3 2 2

3 2 2

P x x x i x i

x x i x i

Factoring a Polynomial with Complex Zeros

The table gives further examples of

polynomials with their complete factorizations

and zeros.

E.g. 4—Finding Polynomials with Specified Zeros

(a) Find a polynomial P(x) of degree 4,

with zeros i, –i, 2, and –2 and with

P(3) = 25.

(b) Find a polynomial of degree 4,

with zeros –2 and 0, where –2 is a zero

of multiplicity 3.

E.g. 4—Specified Zeros

The required polynomial has the form

• We know that P(3) = a(34 – 3 · 32 – 4) = 50a = 25.

• Thus, a = ½ .

• So, P(x) = ½x4 – 3/2x2 – 2

2 2

4 2

2 2

1 4

3 4

P x a x i x i x x

a x x

a x x

Example (a)

E.g. 4—Specified Zeros

We require:

3

3

3 2

4 3 2

(SpecialProduct Formula 4, Section1.4)

2 0

2

6 12 8

6 12 8

Q x a x x

a x x

a x x x x

a x x x x

Example (b)

E.g. 4—Specified Zeros

We are given no information about Q other

than its zeros and their multiplicity.

So, we can choose any number for a.

• If we use a = 1, we get:

Q(x) = x4 + 6x3 + 12x2 + 8x

Example (b)

E.g. 5—Finding All the Zeros of a Polynomial

Find all four zeros of

P(x) = 3x4 – 2x3 – x2 – 12x – 4

• Using the Rational Zeros Theorem from Section 3-3, we obtain this list of possible rational zeros:

±1, ±2, ±4, ±1/3, ±2/3, ±4/3

E.g. 5—Finding All the Zeros of a Polynomial

Checking them using synthetic division,

we find that 2 and -1/3 are zeros, and we

get the following factorization.

4 3 2

3 2

213

213

3 2 12 4

2 3 4 7 2

2 3 3 6

3 2 2

P x x x x x

x x x x

x x x x

x x x x

E.g. 5—Finding All the Zeros of a Polynomial

The zeros of the quadratic factor

are:

• So, the zeros of P(x) are:

1 1 8 1 7

2 2 2x i

1 1 7 1 72, , ,

3 2 2 2 2i i

Finding All the Zeros of a Polynomial

The figure shows the graph of the

polynomial P in Example 5.

• The x-intercepts correspond to the real zeros of P.

• The imaginary zeros cannot be determined from the graph.

Complex Zeros Come

in Conjugate Pairs

Complex Zeros Come in Conjugate Pairs

As you may have noticed from the examples

so far, the complex zeros of polynomials with

real coefficients come in pairs.

• Whenever a + bi is a zero, its complex conjugate a – bi is also a zero.

Conjugate Zeros Theorem

If the polynomial P has real coefficients,

and if the complex number z is a zero of P,

then its complex conjugate is also a zero

of P.

z

Conjugate Zeros Theorem—Proof

Let

P(x) = anxn + an-1xn-1 + . . . + a1x + a0

where each coefficient is real.

• Suppose that P(z) = 0.

• We must prove that . 0P z

Conjugate Zeros Theorem—Proof

We use the facts that:

• The complex conjugate of a sum of two complex numbers is the sum of the conjugates.

• The conjugate of a product is the product of the conjugates.

Conjugate Zeros Theorem—Proof

• This shows that is also a zero of P(x), which proves the theorem.

1

1 1 0

11 1 0

11 1 0

11 1 0

0 0

n n

n n

n nn n

n nn n

n nn n

P z a z a z a z a

a z a z a z a

a z a z a z a

a z a z a z a

P z

z

E.g. 6—A Polynomial with a Specified Complex Zero

Find a polynomial P(x) of degree 3

that has integer coefficients and zeros

½ and 3 – i.

• Since 3 – i is a zero, then so is 3 + i by the Conjugate Zeros Theorem.

E.g. 6—A Polynomial with a Specified Complex Zero

That means P(x) has the form

12

12

2 212

212

3 2132

(Diff. of Squares Formula)

3 3

3 3

3

6 10

13 5

P x a x x i x i

a x x i x i

a x x i

a x x x

a x x x

E.g. 6—A Polynomial with a Specified Complex Zero

To make all coefficients integers,

we set a = 2 and get:

P(x) = 2x3 – 13x2 + 26x – 10

• Any other polynomial that satisfies the given requirements must be an integer multiple of this one.

E.g. 7—Counting Real and Imaginary Zeros

Without actually factoring, determine how

many positive real zeros, negative real zeros,

and imaginary zeros this polynomial could

have:

P(x) = x4 + 6x3 – 12x2 – 14x – 24

E.g. 7—Counting Real and Imaginary Zeros

There is one change of sign.• So, by Descartes’ Rule of Signs, P has

one positive real zero.

Also, P(–x) = x4 – 6x3 – 12x2 + 14x – 24

has three changes of sign.• So, there are either three or one negative real

zero(s).

E.g. 7—Counting Real and Imaginary Zeros

So, P has a total of either four or two

real zeros.

• Since P is of degree 4, it has four zeros in all, which gives the following possibilities.

Positive Real Negative Real Imaginary

1 3 0

1 1 2

Linear and Quadratic Factors

Linear and Quadratic Factors

We have seen that a polynomial factors

completely into linear factors if we use

complex numbers.

If we don’t use complex numbers,

a polynomial with real coefficients can always

be factored into linear and quadratic factors.

Linear and Quadratic Factors

A quadratic polynomial with no real

zeros is called irreducible over the real

numbers.

• Such a polynomial cannot be factored without using complex numbers.

Linear and Quadratic Factors Theorem

Every polynomial with real coefficients

can be factored into a product of linear

and irreducible quadratic factors with

real coefficients.

Linear and Quadratic Factors Theorem—Proof

We first observe that, if c = a + bi

is a complex number, then

• The last expression is a quadratic with real coefficients.

2 2

2 2 22

x c x c x a bi x a bi

x a bi x a bi

x a bi

x ax a b

Linear and Quadratic Factors Theorem—Proof

If P is a polynomial with real coefficients, by

the Complete Factorization Theorem,

P(x) = a(x – c1)(x – c2) ··· (x – cn)

• The complex roots occur in conjugate pairs.• So, we can multiply the factors corresponding to each

such pair to get a quadratic factor with real coefficients.• This results in P being factored into linear and

irreducible quadratic factors.

E.g. 8—Factoring into Linear and Quadratic Factors

Let P(x) = x4 + 2x2 – 8.

(a) Factor P into linear and irreducible

quadratic factors with real coefficients.

(b) Factor P completely into linear factors

with complex coefficients

E.g. 8—Linear & Quadratic Factors

• The factor x2 + 4 is irreducible since it has only the imaginary zeros ±2i.

4 2

2 2

2

2 8

2 4

2 2 4

P x x x

x x

x x x

Example (a)

E.g. 8—Linear & Quadratic Factors

To get the complete factorization, we

factor the remaining quadratic factor.

22 2 4

2 2 2 2

P x x x x

x x x i x i

Example (b)