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5.7 Apply the Fundamental Theorem of Algebra
Goal  Classify the zeros of polynomial functions.
Your Notes
VOCABULARY
Repeated Solution
For the equation f(x) = 0, k is a repeated solution if and only if the factor (x  k) has a
degree greater than 1 when f is factored completely.
THE FUNDAMENTAL THEOREM OF ALGEBRA
Theorem: If f(x) is a polynomial of degree n where n __ 0, then the equation f(x) = 0
has at least __one__ solution in the set of complex numbers.
Corollary: If f(x) is a polynomial of degree n, then the equation f(x) = 0 has exactly _n_
solutions provided each solution repeated twice is counted as _2_ solutions, each solution
repeated three times is counted as _3_ solutions and so on.
Example 1
Find the number of solutions or zeros
Find the number of solutions or zeros for each equation or function.
a. Because x3  3x2 + 9x  27 = 0 is a __third__ degree polynomial equation, it has
__three__ solutions.
b. Because f(x) = x4 + 6x3  32x is a __fourth__ degree polynomial function, it has
__four__ zeros.
Checkpoint Complete the following exercise.
1. State the number of zeros of f(x) = x3  2x2  9x + 18.
three zeros
Your Notes
Example 2
Find the zeros of a polynomial function
Find all zeros of f(x) = x5  5x4  9x3  5x2  8x + 12.
Solution
1. Find the rational zeros of f. Because f is a fifth-degree function, it has __five__ zeros.
The possible rational zeros are __±1, ±2, ±3, ±4, ±6, and ±12__ . Using synthetic
division, you can determine that _2_ is a zero repeated twice and _1_ is also a zero.
2. Write f(x) in factored form. Dividing f by its known factors gives a quotient of
__x2  2x + 3__ . So, f(x) = __(x  2)2 (x + 1) (x2  2x + 3)__
3. Find the complex zeros of f. Use the quadratic formula to factor the trinomial into
linear factors.
f(x) = ___(x  2)2 (x + 1)[x  (1 + i 2 )][x  (1  i 2 )]___
The zeros of f are __1, 2, 2, 1 + i 2 , and 1  i 2
__
Checkpoint Find all zeros of the polynomial function.
2. f(x) = x4  7x3 + 13x2 + x  20
1, 4, 2 + i, 2  i
COMPLEX CONJUGATES THEOREM
If f is a polynomial function with __real__ coefficients, and __a + bi__ is an imaginary
zero of f, then __a  bi__ also a zero of f.
IRRATIONAL CONJUGATES THEOREM
Suppose f is a polynomial function with __rational__ coefficients, and a and b are
b
rational numbers such that b is irrational. If __a + b __ is a zero of f, then __a  __
is also a zero of f.
Your Notes
Example 3
Use zeros to write a polynomial function
Write a polynomial function f of least degree that has real coefficients, a leading
coefficient of 1, and 2 and 3 + i as zeros.
Because the coefficients are real and 3 + i is a zero, __3  i__ must also be a zero. Use
the three zeros and the factor theorem to write f(x) as a product of three factors.
f(x)
=
( _x + 2_ )[x  ( _3 + i_ )][x  ( _3  i_ )]
Factored form
=
( _x + 2_ )[ _(x- 3)  i_ ][ _(x  3) + i_ ]
Regroup terms.
=
(x+ 2)[(x  3)2  i2]
Multiply.
=
=
(x+ 2)[x2  6x+ 9  (1)]
(x+ 2)(x2  6x+ 10)
Expand, use i2 = 1.
Simplify.
=
x3  6x2 + 10x+ 2x2  12x + 20
Multiply.
=
x3  4x2  2x + 20
Combine like terms.
You can check this
result by evaluating
f(x) at each of its
three zeros.
Checkpoint Complete the following exercise.
3. Write a polynomial of least degree that has rational coefficients, a leading
coefficient of 1, and 4 and 1 + 6 as zeros.
f(x) = x3  6x2 + 3x + 20
DESCARTES' RULE OF SIGNS
Let f(x) = anxn + an 1xn  1 + …….+ a2x2 + a1x + a0 be a polynomial function with real
coefficients.
 The number of __positive__ real zeros of f is equal to the number of changes in sign of
the coefficients of _f(x)_ or is less than this by an _even_ number.
 The number of __negative__ real zeros of f is equal to the number of changes in sign
of the coefficients of __f(x)__ or is less than this by an __even__ number.
Your Notes
Example 4
Use Descartes'rule of signs
Determine the possible numbers of positive real zeros, negative real zeros, and
imaginary zeros for
f(x) = 2x5  7x4 + 12x3 + 2x2 + 4x + 6.
Solution
f(x) = 2x5  7x4 + 12x3 + 2x2 + 4x + 6
The coefficients in f(x) have _2_ sign changes, so f has __2 or 0__ positive real zero(s).
f(x) = 2(x)5  7(x)4 + 12(x)3 + 2(x)2 + 4(x) + 6
f(x) = _2x5  7x4  12x3 + 2x2  4x + 6_
The coefficients in f(x) have _3_ sign changes, so f has _3 or 1_ negative real zero(s).
Positive real
zeros
Negative real
zeros
Imaginary
zeros
Total
zeros
_2_
_3_
_0_
_5_
_2_
_1_
_2_
_5_
_0_
_3_
_2_
_5_
_0_
_1_
_4_
_5_
Checkpoint Complete the following exercise.
4.
Determine the possible numbers of positive real zeros, negative real zeros,
and imaginary zeros for f(x) = 3x5  4x4 + x3 + 6x2 + 7x  8.
Positive Negative
real zeros real zeros
Imaginary
zeros
Total
zeros
3
2
0
5
3
0
2
5
1
2
2
5
1
0
4
5
Homework
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