5.7
Apply the Fundamental Theorem of Algebra
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.
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 )][x
- (1 - i )]___
The zeros of f are
__-1, 2, 2, 1 + i , and 1 - 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 rational numbers such that is irrational. If __a + __ is a zero of f, then __a - __
is also
a zero of f.
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. |
|
|
= |
|
Multiply. |
|
|
= |
(x+ 2)[x2
- 6x+ 9 - (-1)] |
Expand, use i2
= -1. |
|
|
= |
|
Simplify. |
|
|
= |
|
Multiply. |
|
|
= |
|
Combine like terms. |
You
can check this result by evaluating f(x) at each of
its three zeros.
Complete the following exercise.
1.
Write a polynomial of
least degree that has rational coefficients, a leading coefficient of 1, and 4
and 1 + 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.
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_ |
Complete the following
exercise.
2.
Determine the possible numbers of positive real
zeros, negative real zeros, and imaginary zeros for f(x) = 3x5
- 4x4 +
x3 + 6x2 + 7x - 8.
|
Positive real zeros |
Negative real zeros |
Imaginary zeros |
Total zeros |
|
3 |
2 |
0 |
5 |
|
3 |
0 |
2 |
5 |
|
1 |
2 |
2 |
5 |
|
1 |
0 |
4 |
5 |