Fundamental Theorem of Algebra

Every polynomia l equat ion 0

with complex coefficients has a t least

one root .

P x

Theorem

2

3 2

Every polynomia l of degree can be

expressed as product of linea r factors.

3 2 degree: 2

= 2 1

6 11 6 degree: 3

= 1 2 3

n

n

x x

x x

x x x

x x x

Theorem

1 21 2

1 2

1 2

Every polynomia l equa t ion 0

of degree has a t most dist inct roots.

In genera l, a polynomia l equa t ion can be

writ ten as

0

, ,..., and a re the dist inct roots and

...

mk k k

m

m

P x

n n

P x a x r x r x r

r r r

k k

m

k n

1 21 2

if

1, is a simple root .

2, is a double root .

, is a root of mult iplicity .

mk k k

m

i i

i i

i i

P x a x r x r x r

k r

k r

k m r m

Example 1.6.12

2 4

Determine the roots of

1 3 5 0

dist inct roots: 1, 3, 5

1 is a simple root .

3 is a double root .

5 is a root of mult iplicity 4.

P x x x x

Theorem

A polynomia l equat ion 0 of degree

has exact ly roots, a root of mult iplicity

being counted as roots.

P x

n n

k k

Example 1.6.13

2

Form an equa t ion which has

1 as a double root

2 and 4 as simple roots

and no others.

1 2 4 0x x x

Theorem

Let 0 be a polynomia l equa t ion with

rea l coefficients. Then if 0 has a

complex root , then it s conjuga te is a lso a

root of the equa t ion.

P x

P x

a bi a bi

Example 1.6.14

Form an equa t ion with in tegra l

coefficients and with lowest possible

degree having 1 2 and 2 3 as

roots.

12 3 2 3 0

2

i

x x i x i

2 2

2

2

12 3 2 3 0

2

2 1 2 3 2 3 0

2 1 2 3 2 3 0

2 1 2 3 0

2 1 4 4 9 1 0

2 1 4 13 0

x x i x i

x x i x i

x x i x i

x x i

x x x

x x x

Theorem

The roots of 0 a re precisely

the addit ive inverses of the roots of

0.

2 is a root of 0

2 is a root of 0

P x

P x

P x

P x

Example 1.6.15

5 3 2

5 3 2

5 3 2

5 3 2

Obta in an equa t ion whose roots a re

the nega t ives of the roots of

2 3 4 2 0

2 3 4 2 0

2 3 4 2 0

2 3 4 2 0

x x x x

P x x x x x

P x x x x x

x x x x

Variation of Signs

descending powers

varia t ion of sign

If the terms of a re a rranged in

of , we say tha t

a occurs when two

successive terms have different signs.

P x

x

Example 1.6.16

5 4 2

5 3 2

Determine the number of va r ia t ion of

signs for each polynomia l.

1. 2 3 4

var ia t ion of signs: 3

2. 2 3 4 2

var ia t ion of signs: 4

x x x x

x x x x

Descartes Rule of Signs

The of the

polynomia l equa t ion 0 with

rea l coefficients is

number of posit ive roots

number of var ia t

equa l to the

in

or less than tha t by

ion of signs

an even number.

P x

P x

Descartes Rule of Signs

The of 0

is

number of negat ive roots

number of posit ive roots the of 0.

P x

P x

Example 1.6.17

7 4 3

7 4 3

Determine the possible number of posit ive,

nega t ive, and complex roots of

2 4 2 5 0

2 4 2 5 0

posit ive roots: 2 or 0

nega t ive roots: 3 or 1

complex roots: 6,4 or 2

P x x x x x

P x x x x x

Rational Root Theorem

2

0 1 2

0

Consider

... 0, 0

with in tegra l coefficients.

If is a root , where and a re

rela t ively pr ime in tegers, then is a

factor of and is a factor of .

n

n n

n

a a x a x a x a

pp q

q

p

a q a

Example 1.6.18

3 2

3 2

3 2

Solve 2 3 7 3 0

: 1, 3 : 1, 2

1 3: 1, 3, ,

2 2

2 3 7 3 0

2 3 7 3 0

posit ive roots: 3 or 1 nega t ive roots: 0

x x x

p q

p

q

x x x

x x x

3 2

3 2

2 3 7 3 0

1 3: 1,3, ,

2 2

2 3 7 3

1 2 3 7 3

2 1 6

2 1 6

1 is not a root

3

x x x

p

q

x x x

3 2

2

2 3 7 3

12 3 7 3

2

1 1 3

2 2 6

1 1 is a root and is a factor .

2 2

12 2 6 0

2

0

x x x

x

x x x

2

2

2

12 2 6 0

2

12 3 0

2

10 3 0

2

1 1 11

2 2

1 1 11 1 11, ,

2 2 2

x x x

x x x

x x x

ix x

i iS S

Example 1.6.19

4 3 2

4 3 2

4 3 2

Solve 8 14 13 6 0

: 1, 2, 3, 6 : 1

: 1, 2, 3, 6

8 14 13 6 0

8 14 13 6 0

posit ive roots: 0

nega t ive roots: 4 or 2 or 0

x x x x

p q

pq

x x x x

x x x x

4 3 2

4 3 2

3 2

8 14 13 6 0

: 1, 2, 3, 6

8 14 13 6

1 1 8 14 13 6

1 7 7 6

1 7 7 6 0

1 is a root , 1 is a factor .

1 7 7 6 0

x x x x

pq

x x x x

x

x x x x

3 2

3 2

2

1 7 7 6 0

: 1, 2, 3, 6

7 7 6

6 1 7 7 6

6 6 6

1 1 1 0

6 is a root , 6 is a factor

1 6 1 0

x x x x

pq

x x x

x

x x x x

2

2

1 6 1 0

1 0 6 0 1 0

1 6 1 1 1

1 1 4 1 1

2

1 3 1 3

2 2

1 3 1 31, 6, ,

2 2

x x x x

x x x x

x x a b c

x

i

i iS S

Equations in Linear or Quadratic Forms

Some equa t ions can be converted to

linear or quadra t ic equa t ions.

equa t ions involving ra t iona l expression s

equa t ions involving radica ls

equa t ions in quadra t ic form

Example 1.6.20

EQUATIONS & INEQUALITIES

Solution: The LCD of the RE is

Multiplying both sides by the LCD:

2

1 2 7

2 1 2x x x x

2 1 .x x

21 2 7

2 1 2 12 1 2

x x x xx x x x

Checking the results shows that the LCD 0 for .

Therefore, the solution set is .

2 x

2 SS

1 2 2 7 x x 2 2 x x

21 2 7

2 1 2 12 1 2

x x x xx x x x

Example 1.6.21

Solution:

The LCD of the fractions is .

Multiplying both sides by the LCD yields:

2

3 2 3

3 2 5 6x x x x

3 2 x x

3 2 2 3 3 x x3 6 2 6 3

3

x x

x

Checking the results shows that the LCD = 0 for . Thus, 3 is NOT a solution, hence, there is NO SOLUTION to the equation.

Therefore, the solution set is .

3x

SS

Example 1.6.22

3 3 1x x

2 2

3 3 1x x

square both sides of the

equat ion not the equat ion

23 9 6 1x x x

29 7 2 0x x 9 2 1 0x x

9 2 0 1 0x x

2

19

x x

3 1 3x x

Checking:

2:

9x

1 :x

22 8

3 19 3

2 23

9 3

2 is an extraneous solut ion

9

2

1 3 1 3 3 1 3

1S S

Example 1.6.23

2 5 1 2x x

2

2 5 1 4x x

22 5 2 2 3 5 1 4x x x x

23 4 2 2 3 5 4x x x

23 2 2 3 5x x x

2 29 4 2 3 5x x x 23 2 2 3 5x x x

2 29 8 12 20x x x

2 12 20 0x x

10 2 0x x

10 0 2 0

10 2

x x

x x

Checking:

10 :x 2 10 5 10 1 8 4

10 is an ext raneous root .

2 :x 2 2 5 2 1 4 2

2 is an ext raneous solut ion

SS

Example 1.6.24

4 21 2x x 2

2 22 1 0x x

2

2 1 0x 2 1 0x

2 1x

x i

,SS i i

Example 1.6.25

22 2

6 1 5 1 6 0x x

2

Let 1ux

26 5 6 0u u

2 3 3 2 0u u

3

2u

2

3u

2 31

2x

4, 6

5S S

3

2u

2

3u

2 2

13x

2 4 3x x

5 4x

4

5x

3 6 2x x

6x