Transcript The Fundamental Theorem

Fundamental Theorem
The Fundamental
Theorem of Algebra
The Fundamental Theorem

Quadratic Formula Examples

1. Solve: x2 – 2x + 1 = 0
x =
=
–b ±
 b2 – 4ac
2a
+2 ±
 (–2)2 – 4(1)(1)
2(1)
2 ± 0
=
2
=1
Solution set: { 1 }
Question: Is there an easier way ?
9/2/2013
Fundamental Theorem
2
The Fundamental Theorem

Quadratic Formula Examples
x =

–b ±
 b2 – 4ac
2a
2. Solve: x2 – 4x – 5 = 0
x =
+4 ±
 (–4)2 – 4(1)(–5)
2(1)
4 ±  36
=
2
=
–1
5
Solution set: { –1, 5 }
Question: Is there an easier way ?
9/2/2013
Fundamental Theorem
3
The Fundamental Theorem

Quadratic Formula Examples
x =

–b ±
 b2 – 4ac
2a
3. Solve: x2 – 2x + 5 = 0
x =
+2 ±
 (–2)2 – 4(1)(5)
2(1)
2 ±  –16
=
2
= 1 ± 2i
Solution set:
{ 1 – 2i , 1 + 2i }
Question: Is there an easier way ?
9/2/2013
Fundamental Theorem
4
The Fundamental Theorem

Complex Solutions

Complex solutions always occur in
conjugate pairs
WHY ?

Depends on discriminant of the
quadratic formula :
x =
–b ±
 b2 – 4ac
2a
Note ± radical
9/2/2013
Fundamental Theorem
5
The Fundamental Theorem

Complex Solutions

Quadratic formula :
x =
b2 – 4ac
9/2/2013
–b ±
 b2 – 4ac
2a
Discriminant
b2 – 4ac = 0
b2 – 4ac > 0
One real solution
Two real solutions
b2 – 4ac < 0
Two complex conjugate
solutions
Fundamental Theorem
6
The Fundamental Theorem

The Fundamental Theorem of Algebra
A polynomial of degree n ≥ 1 has
at least one complex zero

Consequence:
Every polynomial has a complete
factorization
WHY ?
 For polynomial f(x) there exists a zero k1
such that
9/2/2013
f(x) = (x – k1)Q1(x)
Fundamental Theorem
7
The Fundamental Theorem
For polynomial f(x) there exists a zero k1
such that
f(x) = (x – k1)Q1(x)
where deg Q1(x) = deg f(x) – 1
Apply the Fundamental Theorem to Q1(x)
Q1(x) = (x – k2)Q2(x)
so
f(x) = (x – k1)(x – k2)Q2(x)
where deg Q2(x) = deg f(x) – 2
9/2/2013
Fundamental Theorem
8
The Fundamental Theorem
f(x) = (x – k1)(x – k2)Q2(x)
where deg Q2(x) = deg f(x) – 2
For deg f(x) = n , apply n times :
f(x) = (x – k1)(x – k2) • • • (x – kn)Cn
where Cn is a nonzero constant
f(x) is now completely factored !
9/2/2013
Fundamental Theorem
9
The Fundamental Theorem

The Number of Zeros Theorem
A polynomial of degree n has at most
n distinct zeros

From the Complete Factorization:
f(x) = (x – k1)(x – k2)...(x – kn)Cn

If all ki are distinct there are n distinct zeros
... otherwise one or more zeros are
repeated zeros
9/2/2013
Fundamental Theorem
10
The Fundamental Theorem

The Number of Zeros Theorem

Examples

1. f(x) = 5x(x + 3)(x – 1)(x – 4)
How many distinct zeros ?

9/2/2013
2. f(x) = 4x2(x – 2)3(x – 7)
How many distinct zeros ?
Total zeros ?
Fundamental Theorem
11
The Fundamental Theorem

9/2/2013
The Conjugate Zeros Theorem

If f(x) is a polynomial with only real
coefficients then its complex zeros
occur in conjugate pairs

Examples
3
 1. f(x) = x + x
= x(x2 + 1)
= x(x + i)(x – i) Zeros ?
Fundamental Theorem
12
The Fundamental Theorem

9/2/2013
The Conjugate Zeros Theorem

If f(x) is a polynomial with only real
coefficients then its complex zeros
occur in conjugate pairs

Examples
4
2
 2. f(x) = x + 5x – 36
Let u = x2
= u2 + 5u – 36
= (u – 4)(u + 9)
= (x2 – 4)(x2 + 9)
= (x + 2)(x – 2)(x + 3i)(x – 3i) Zeros ?
Fundamental Theorem
13
The Fundamental Theorem





Solving Equations Using Zeros



Factor polynomial
Set completely factored form equal to 0
Apply zero product property
Examples
4
2
 1. f(x) = x + 5x – 36
= (x + 2)(x – 2)(x + 3i)(x – 3i) = 0
Thus x + 2 = 0 OR x – 2 = 0
OR x + 3i = 0 OR x – 3i = 0
Solution set: { –2 , 2 , –3i , 3i }
9/2/2013
Fundamental Theorem
14
The Fundamental Theorem





9/2/2013
Solving Equations Using Zeros



Factor polynomial
Set completely factored form equal to 0
Apply zero product property
Examples
3
 2. f(x) = x + x
= x (x + i)(x – i) = 0
Thus
x = 0 OR x + i = 0 OR x – i = 0
Solution set: { 0, –i, i }
Fundamental Theorem
15
Think about it !
9/2/2013
Fundamental Theorem
16