CHAPTER 4.2
CHAPTER 4 COMPLEX NUMBERS
PART 2 –Complex Solutions of Equations
TRIGONOMETRY MATHEMATICS CONTENT STANDARDS:
 17.0 - Students are familiar with complex numbers. They can represent a
complex number in polar form and know how to multiply complex
numbers in their polar form.
OBJECTIVE(S):
 Students will learn the fundamental theorem of algebra and the linear
factorization theorem.
 Students will learn how to find the solutions of polynomial equations.
 Students will learn how to find the zeros of polynomial functions.
 Students will learn how to find a polynomial given it’s zeros.
The Number of Solutions of a Polynomial Equation
The Fundamental Theorem of Algebra implies that a polynomial equation of degree n has
precisely n solutions in the complex number system. These solutions can be real or
complex and may be repeated.
The Fundamental Theorem of Algebra
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If f x is a polynomial of degree n, where n > 0, then f has at least one zero in the
complex number system.
Note that finding zeros of a polynomial function f is equivalent to finding solutions to the
polynomial equation _______________.
Linear Factorization Theorem
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If f x is a polynomial of degree n where n > 0, f has precisely n linear factors:
bg b gb gb g
f x  an x  c1 x  c2 ... x  cn
where c1 , c2 ,..., cn are complex numbers.
EXAMPLE 1: Real Zeros of Polynomial Functions
a.) The first-degree polynomial f x  x  2 has exactly one solution: x =
bg
___________.
CHAPTER 4.2
b.) The second-degree equation
x 2  6x  9  0
Second-degree equation.
Factor.
has exactly two solutions: x = __________ and x = _____________. (This is
called a repeated solution.)
c.) The third-degree polynomial equation
x3  4x  0
Third-degree equation.
Factor.
Factor.
has exactly three zeros: ______________, _________________, and
_______________.
CHAPTER 4.2
d.) The fourth-degree equation
x4 1 =
0
Fourth-degree equation.
=
Factor.
=
Factor.
has exactly four zeros: ________________, _________________,
________________, and _______________________.
Every second-degree equation, ax 2  bx  c  0 , has precisely two solutions given by the
quadratic formula.
x=
The expression inside the radical, __________________, called the discriminant, and can
be used to determine whether the solutions are real, repeated, or complex.
1. If b 2  4ac  0 , the equation has two complex solutions.
2. If b 2  4ac  0 , the equation has one repeated real solution.
3. If b 2  4ac  0 , the equation has two distinct real solutions.
CHAPTER 4.2
EXAMPLE 2: Using the discriminant
Use the discriminant to find the number of real solutions of each equation.
a.) 4 x 2  20 x  25  0
For this equation, a = _____, b = ____, and c = _____. So, the discriminant is
b 2  4ac =
=
=
Because the discriminant is _________, there is ________ ___________________
__________ solution.
b.) 13 x 2  7 x  2  0
For this equation, a = _____, b = ____, and c = _____. So, the discriminant is
b 2  4ac =
=
=
Because the discriminant is _________, there are ________ ___________________
solutions.
c.) 5 x 2  8 x  0
For this equation, a = _____, b = ____, and c = _____. So, the discriminant is
b 2  4ac =
=
=
Because the discriminant is _________, there are ________ ___________________
__________ solutions.
CHAPTER 4.2
Finding Solutions of Polynomial Equations
EXAMPLE 3: Solving a Quadratic Equation
Solve x 2  2 x  2  0 . Write complex solutions in standard form.
Using a = ____, b = ____, and c = ____, you can apply the Quadratic Formula as follows.
x
 b  b 2  4ac
2a
Quadratic Formula.
Substitute ___ for a, ____ for b,
and _____ for c.
Simplify.
Simplify.
Write in standard form.
Notice that two complex solutions are conjugates. That is, they are of the form
______________. This is not a coincidence, as indicated by the following theorem.
DAY 1
Complex Solutions Occur in Conjugate Pairs
If a  bi, b  0 , is a solution of a polynomial equation with real coefficients, the conjugate
__________is also a solution of the equation.
Be sure you see that this result is true only if the polynomial has real coefficients. For
instance, the result applies to the equation x 2  1  0 , but not to the equation x i  0 .
CHAPTER 4.2
EXAMPLE 4: Solving a Polynomial Equation
Solve x 4  x 2  20  0 .
x 4  x 2  20  0
Write original equation.
Partially factor.
Factor completely.
Setting each factor equal to zero yields the solutions x = ________, x = ________, x =
_________, and x = _________.
Finding Zeros of Polynomial Functions
The problem of finding the zeros of a polynomial function is essentially the same
problem as finding the solutions of a polynomial equation. For instance, the zeros of the
polynomial function
f  x   3x 2  4 x  5
are simply the solutions of the polynomial equation
3x 2  4 x  5  0
CHAPTER 4.2
EXAMPLE 5: Finding the Zeros of a Polynomial Function
Find all the zeros of f x   x 4  3x 3  6 x 2  2 x  60 given that 1 3i is a zero of f.
Because complex zeros occur in conjugate pairs, you know that _____________ is also a
zero of f. This means that both
and
are factors of f. Multiplying these two factors produces
x  1 3i x  1 3i 
=
=
=
=
Using long division, you can divide _______________________ into f to obtain the
following.
x 2  2 x  10 x 4  3x 3  6 x 2  2 x  60
bg =
Therefore, you have: f x
=
and you can conclude that the zeros of f are x = ______________, x =
________________, x = _______________, and x = _________________.
CHAPTER 4.2
EXAMPLE 6: Finding a Polynomial with Given Zeros
Find a fourth-degree polynomial function with real coefficients that has -1, -1, and 3i as
zeros.
Because _____ is a zero and the polynomial is stated to have real coefficients, you know
that the conjugate ______ must also be a zero. So, from the Linear Factorization
Theorem, f x  can be written as
f x  = ______________________________________________
For simplicity, let a = 1 to obtain
f x  =
=
EXAMPLE 7: Finding a Polynomial with Given Zeros
Find a cubic polynomial function f with real coefficients that has 2 and 1-i as zeros, such
that f 1  3 .
Because 1 – i is a zero of f, so is _______________. So,
f x  =
=
=
=
=
To find the value of a, use the fact that f 1  3 and obtain
f 1
=
=
=
So, a = ____ and it follows that
f x  =
=
DAY 2