Download 2.6 Fundamental Theorem of Algebra

1.
Perform the operation and write the
result in standard form
(2  3i )(5i )
(2  3i )
2.
Use the quadratic formula to solve
the quadratic equation
9 x  6 x  37  0
2
 Assignment
◦p. 266
◦# 11, 12, 16, 24, 26


Test Corrections Tomorrow After School
Notebook Quiz Next Week
•How
to use the Fundamental Theorem of
Algebra to determine the number of zeros of a
polynomial function and find all zeros of
polynomial functions, including complex zeros
•How
to find conjugate pairs of complex zeros
•How
to find zeros of polynomials by factoring

If f(x) is a polynomial function of
degree n, where n > 0, then f has at
least one zero in the complex number
system.
◦ Does it have to cross
◦ the x-axis to be a zero?
f ( x)  x  1
g ( x)  x  1
2
h( x)  ( x  4)
3


The highest degree of a polynomial is how
many complex zeros we should have.
How many zeros for the following?
◦ x3 + 2x
3
◦ 3x2 + 2x4 + x + 5
4
◦ 7x4 - 2x5 + x6 – 4x
6

Solve x3 + 6x – 7 = 0
◦ Possible zeros:
 P = ± 7, ±1
 Q = ±1
 (p/q) = ± 7, ±1
◦ How do we find the possible rational
zeros?
 Factor Theorem – please test the four
possibilities

X=1
X=1
1
0
6
-7
1
1
1
1
7
7
0
f ( x)  x  x  7
2
Now, what do we do once we have a quadratic equation?
 b  b 2  4ac
x
2a
 (1)  (1)  4(1)(7)

2(1)
2
 1  1  28

2
 1   27

2
 1  3 3i

2
1 3 3
 
i
2
2
 Using
the factor theorem,
synthetic division, and the
quadratic formula, we have:
◦ x3 + 6 x – 7
◦ =
1 3 3
1 3 3
i )) ( x  ( 
i ))
( x  1) ( x  ( 
2
2
2
2
1.
2.
3.
4.
Use the Rational root test to find the
possible rational zeros
Use the factor theorem to test the possible
rational zeros
Use synthetic division until you find a
quadratic equation
Use the quadratic formula to find the
complex zeros
 f(x)
1.
= x4 - 3x3 + x – 3
Use the Rational root test to find the
possible rational zeros
◦ P = ±3, ±1
◦ Q =± 1
◦ (p/q) = ±3, ±1
1.
1.
Use the factor theorem to test the
possible rational zeros
(p/q) = ±3, ±1
 f(x)
= x4 - 3x3 + x – 3
 Which ones work?
X = -1, 3
1.
Use synthetic division until you find a quadratic
equation
X=1
1
1
-3
0
1
-3
-1
4
-4
3
-4
4
-3
0
x  4x  4x  3
3
2
1.
Use synthetic division until you find a quadratic
equation
X=3
1
1
-4
4
-3
3
-3
3
-1
1
0
x  x 1
2
1.
Use the quadratic formula to find the complex
2

b

b
 4ac
zeros
x
2a
x  x 1
2
1
3
1
3
(x  ( 
i ))( x  ( 
i ))
2 2
2 2
 (1)  (1) 2  4(1)(1)

2(1)
1 1 4
2
1  3

2
1  3i

2
1
3
 
i
2 2

 f(x)
= x4 - 3x3 + x – 3
◦ Highest degree tells us there
must be 4 zeros
1
3
1
3
(x  ( 
i ))( x  ( 
i ))( x  3)( x  1)
2 2
2 2

Find all the zeros of the function and write
the polynomials as a product of linear factors
x  3x  4 x  2
3
2

Assignment
 p.
266
 # 38 – 44 even, 51 – 55 odd


Test Corrections today after school
Notebook Quiz Next Week
a + bi is a zero of f,
a – bi is also a zero of f and
 whenever
vice versa.
 Example:
1
3

i
2 2

1
3

i
2 2
Note that in Example 1, the two complex zeros
were conjugates.
6i
 6i
 2  3i
 2  3i
4i
4i
 1  7i
 1  7i
(2  5i )
1.
Given
2.
Conjugates
3.
Factor notation
4.
Rearrange
5.
FOIL
6.
Combine like terms
( x  4 x  4  (5))
Simplify
( x  4 x  9)
7.
(2  5i )(2  5i )
( x  (2  5i )( x  (2  5i )
(( x  2)  5i )(( x  2)  5i )
(( x  4 x  4)  ( x  2) 5i  ( x  2) 5i  (5))
2
2
2

How do we write the three initial
zeros?
◦ f(x) = x(x – 3)(x – i)

What is the fourth zero? (remember
that complex numbers come in
conjugates)
◦ (x + i)
f(x) = x(x – 1)(x – i)(x + i)

= (x2 – x)(x2 – 1).

= x4 – x3 - x2 + x.

2
,1,3  2i
2 3
 ( x  )( x  1)( x  (3  2i )( x  (3 
1,5i,5i
2i )
 ( x  1)( x  5i )( x  5i )
3
2
 ( x  )( x  1)(( x  3)  2i )(( x  3)  2i )
3
1
2
 ( x 2  x  )( x 2  6 x  9  (2))
3
3
3
1
2 2
2
 ( x  x  )( x  6 x  11)
3
3
1
2
11
22
 ( x 4  x 3  x 2  6 x 3  2 x 2  4 x  11x 2  x  )
3
3
3
3
17
25 2 23
22
 ( x 4  x3 
x  x )
3
3
3
3
 ( x  1)( x  25)
2
 ( x  x  25 x  25)
2


How do we find the conjugate pair of a
complex number?
How do we multiply the factored form of
complex numbers?

Factor f(x) = x4 –12x2– 13
◦ a) factor without square roots
◦ ( 2 – 13)( 2 + 1)
x
x
◦ b) factor with square roots.
◦
2
x 

◦ c) factor completely
◦
x 


13 x  13 x  1


13 x  13 x  i x  i 
 Find all zeros of
12x2 + 4x – 13
f(x) = x4 – 4x3 +
◦ given that (2 + 3i) is a zero.
 Knowing
about conjugates, what
must be another zero?
◦ Since 2 + 3i is a zero, 2 – 3i is also a zero.
That means that x2 – 4x + 13 is a factor of
f(x).


Given x2 – 4x + 13 is a factor of
f(x) = x4 – 4x3 + 12x2 + 4x – 13
What type of division could we use to find the
other factors?
◦ Long Division
2
x 1
x 2  4x  13 x 4  4x 3  12x 2  4x  13

4
3
x  4x  13x
2
 x 2  4x  13
2
x  4x 13
0
2
x 1
x 2  4x  13 x 4  4x 3  12x 2  4x  13

x 4  4x 3  13x 2
 x 2  4x  13
x 2  4x 13
0
x  1  ( x  1)( x  1)
2
of f(x) = x4 – 4x3 + 12x2 +
4x – 13
 Zeros
 After
using complex conjugates,
long division and factoring
◦ All the zeros of f are
◦ –1, 1, 2 + 3i, 2 – 3i.