Download section 4.5

Chapter 4
More Nonlinear
Functions and
Equations
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4.5
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The Fundamental
Theorem of Algebra
Apply the fundamental theorem of algebra
Factor polynomials having complex zeros
Solve polynomial equations having complex
solutions
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Fundamental Theorem of Algebra
A polynomial f(x) of degree n, with n ≥ 1, has at least
one complex zero.
The fundamental theorem of algebra guarantees
that every polynomial has a complete
factorization, if we are allowed to use complex
numbers.
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Number of Zeros Theorem
A polynomial of degree n has at most n distinct
zeros.
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Example: Classifying zeros
All zeros for the given polynomial are distinct. Use
the figure to determine
graphically the number
of real zeros and the
number of imaginary
zeros.
f(x) = 3x3 – 3x2 – 3x – 5
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Example: Classifying zeros
Solution
Real zeros correspond to x-intercepts; imaginary
zeros do not.
Graph has one xintercept so there is one
real zero.
Since f is degree 3 and
all zeros are distinct,
there are two imaginary
numbers.
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Example: Constructing a
polynomial with prescribed zeros
Represent a polynomial f(x) of degree 4 with
leading coefficient 2 and zeros of 3, 5, i and i
in complete factored form and expanded form.
Solution
Let an = 2, c1 = 3, c2 = 5, c3 = i, and c4 = i.
f(x) = 2(x + 3)(x  5)(x  i)(x + i)
Expanded: 2(x + 3)(x  5)(x  i)(x + i)
= 2(x + 3)(x  5)(x2 + 1)
= 2(x + 3)(x3  5x2 + x  5)
= 2(x4  2x3  14x2  2x  15)
= 2x4  4x3 – 28x2 – 4x – 30
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Conjugate Zeros Theorem
If a polynomial f(x) has only real coefficients
and if a + bi is a zero of f(x), then the
conjugate a – bi is also a zero of f(x).
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Example: Finding imaginary zeros
of a polynomial
Find the zeros of f(x) = x4 + x3 + 2x2 + x + 1 given
one zero is i.
Solution
By the conjugate zeros theorem it follows that i
must be a zero of f(x).
(x + i) and (x  i) are factors
(x + i)(x  i) = x2 + 1, using long division we can find
another quadratic factor of f(x).
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Example: Finding imaginary zeros
of a polynomial
Long division
x2  x  1
x 2  0x  1 x 4  x 3  2x 2  x  1
x 4  0x 3  x 2
x x x
3
2
x  0x  x
3
2
x  0x  1
2
x 2  0x  1
So,
0
x4 + x3 + 2x2 + x + 1 = (x2 + 1)(x2 + x + 1)
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Example: Finding imaginary zeros
of a polynomial
Use the quadratic formula to find the zeros of
x2 + x + 1.
b  b 2  4ac
x
2a
x
  
1 12  4 1 1

21
1
3
x i
2
2
The four zeros are:
1
3
i and   i
2
2
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Polynomial Equations with Complex
Solutions
We can solve polynomial equations by writing them
in their complete factored form and then use the fact
that the solutions to the equation are the zeros of
f(x).
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Example: Solving a polynomial
equation
Solve x3 = 3x2  7x + 21.
Solution
Rewrite the equation as f(x) = 0, where
f(x) = x3  3x2 + 7x  21.
We can use factoring by grouping or graphing to
find one real zero of f(x). Let’s use graphing in
this example.
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Example: Solving a polynomial
equation
The graph shows a zero
at 3. So, x  3 is a
factor.
Now use synthetic division by 3.
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Example: Solving a polynomial
equation
1 –3
3
1
0
x3  3x2 + 7x  21 = (x  3)(x2 + 7)
3
Synthetic division
x30
or
x3
or
x3
or
7 –21
0 21
7
0
x 70
2
x 2  7
x  i 7
The solutions are 3 and x  i 7.
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