1 c ollege a lgebra polynomial and rational functions (chapter3) l:16 1 university of palestine...

1

College Algebra

polynomial and Rational Functions (Chapter3)

L:16

1

University of PalestineIT-College

2

Objectives (Part I)

•Divide one polynomial by another using long division.

• Divide one polynomial by another using Synthetic Division

3

Divide Polynomial Polynomial Using Long Division

Step 1: Set up the long division. The divisor (what you are dividing by) goes on the outside of the box. The dividend (what you are dividing into) goes on the inside of the box.

Step 2: Divide 1st term of dividend by first term of divisor to get first term of the quotient.

The quotient (answer) is written above the division box. Make sure that you line up the first term of the quotient with the term of the dividend that has the same degree.

Step 3: Take the term found in step 2 and multiply it times the divisor.Make sure that you line up all terms of this step with the term of the dividend that has the same degree.

4

Divide Polynomial Polynomial Using Long Division

Step 4: Subtract this from the line above.Make sure that you subtract EVERY term found in step 3, not just the first one.

Step 5: Repeat until done.You keep going until the degree of the "new" dividend is less than the degree of the divisor.

Step 6: Write out the answer.Your answer is the quotient that you ended up with on the top of the division box. If you have a remainder, write it over the divisor in your final answer.

5

Example 1: Divide using long division

Step 1:

Step 2:

Step 3:

6

Step 4:

Step 5:

Step 6:

7

Example 2: Divide using long division

8

Synthetic Division is a process whereby the quotient and remainder can be determined when a polynomial function f is divided by g(x) = x - c.

Synthetic Division is a shorter version of polynomial long division where only the coefficients of each term in the dividend, divisor, and quotient are written.

Synthetic Division and the Remainder and Factor Theorems

Remainder Theorem

• If a polynomial f(x) is divided by x k, the remainder is f(k).

This means that we can apply synthetic division and the last number on the right, which is the remainder, will tell us what the functional value of c is

9

Example: Use synthetic division to find the quotient and remainder when

f x x x x

g x x

( )

( ) .

3 2 10

3

4 3 2 is divided by

x x 3 3

3 3 -1 2 0 -10

The coefficients of the divisor and dividend are written as follows:

10

3 3 -1 2 0 -10

3

11

3 3 -1 2 0 -10

- 9

3 -10

12

3 3 -1 2 0 -10

- 9 30

3 -10 32

13

3 3 -1 2 0 -10

- 9 30 - 96

3 -10 32 - 96

14

3 3 -1 2 0 -10

- 9 30 - 96 288

3 -10 32 - 96 278

Quotient:

Remainder: 278

3 10 32 963 2x x x

3

2789632103

3

1023 23234

xxxx

x

xxx

15

CautionNote: To avoid errors, use 0 as coefficient for any missing terms, including a missing constant, when setting up the division.

Example : Divide using synthetic division:

16

Division Algorithm

Let f(x) and g(x) be polynomials with g(x) of lower degree than f(x) and g(x) of degree of one or more. There exists unique polynomials q(x) and r(x) such that

f(x) = g(x) q(x) + r(x)

where either r(x) = 0 or the degree of r(x) is less than the degree of g(x).

17

Example

Use synthetic division to divide

2x3 3x2 11x + 7 by x 3.

Solution: Since x 3 in the form x k use this and the coefficients of the polynomial to obtain

3 2 3 11 7

18

Example continued

6 9 6

2 3 2 1 Remainder

Since the divisor x k has degree 1, the degree of the quotient will always be one less than the degree of the polynomial to be divided. Thus

3 2 3 11 7

Quotient

22

32 3 11 7

32

1

33 2

xx x

x x

xx

19

Applying the Remainder Theorem

Example: Let f(x) = 3x4 7x3 4x + 5. Use the remainder theorem to find f(2).

Solution: Use synthetic division with k = 2.

6 2 4 16

3 1 2 8 11 Remainder

By this result, f(2) = 11

2 3 7 0 4 5

If the polynomial f(x) is divided by x - c, then the reminder is f (c)

20

Factor Theorem If f(x( is a polynomial AND

1 (f(c) = 0, then x - c is a factor of f(x). 2 (x - c is a factor of f(x), then f(c) = 0

Keep in mind that the division algorithm is

dividend = divisor(quotient)+ reminder

21

Example :Use synthetic division to divide by x -2 .Use the result to find all zeros of f.

Note how the remainder is 0. This means that (x - 2)is a factor of

We need to finish this problem by setting this equal to zero and solving it:

23

Example : Solve the equation

given that 3/2 is a zero (or root) of it

The solution or zeros of this function are x =3/2 , -1 ,and 1.

24

Testing Potential Zeros

• A zero of a potential function f is a number k such that f(k) = 0. The zeros are the x-intercepts of the graph of the function.

• The remainder theorem gives a quick way to decide if a number k is a zero of a polynomial function defined by f(x). Use synthetic division to find f(k); if the remainder is 0, then f(k) = 0 and k is a zero of f(x). A zero of f(x) is called a root or solution of the equation f(x) = 0.

25

Example

Decide whether the given number k is a zero of f(x).

• f(x) = 2x3 2x2 34x 30 k = 1Solution: Since the remainder is 0, f(1) = 0, and 1 is a zero of

the polynomial function defined by

f(x) = 2x3 2x2 34x 30

• f(x) = 2x3 + 4x2 x + 5 k = 2

Solution: Since the remainder is 7, not 0, 2 is not a zero of f(x) = 2x3 + 4x2 x + 5 . In fact, f(2 ) = 7.

26

)Part II(Zeros of Polynomial Functions

27

Objectives (Part II)

• Use the Remainder and Factor Theorems

•Use the Rational Zeros Theorem

•Use Descartes’ Rule of Signs

28

Factor Theorem

By the remainder theorem, if f(k) = 0, then the remainder when f(x) is divided by x k is 0. This means that x k is a factor of f(x). Conversely, if x k is a factor of f(x), then f(k) must equal 0. This is summarized in the following factor theorem.

Factor Theorem

The polynomial x k is a factor of the polynomial f(x) if and only if f(k) = 0.

29

Example

Determine whether x 3 is a factor of f(x) for f(x) = x3 7x2 + 11x + 3.

Solution: By the factor theorem, x 3 will be the factor of f(x) only if f(3) = 0. Use synthetic division and the remainder theorem to decide.

3 12 3 1 4 1 0

3 1 7 11 3

f(x) = (x 3) (x2 4x 1).

Because the remainder is 0, x 3 is a factor. Additionally, we can determine from the coefficients in the bottom row that the other factor is x2 4x 1,

30

Rational Zeros Theorem

• The rational zeros theorem gives a method to determine all possible candidates for rational zeros of a polynomial function with integers.

If ,where

are integer coefficients and the reduced fraction is a rational zero

then P is a factor of the constant term ao , and q is a factor of the leading coefficient an We can use this theorem to help us find all of the POSSIBLE rational zeros or roots of a polynomial function.

31

Step 1: List all of the factors of the constant.

In the Rational Zero Theorem, p represents factors of the constant term. Make sure that you include both the positive and negative factors.

Step 2: List all of the factors of the leading coefficient.

In the Rational Zero Theorem, q represents factors of the leading coefficient. Make sure that you include both the positive and negative factors.

Step 3: List all the POSSIBLE rational zeros or roots.

This list comes from taking all the factors of the constant (p) and writing them over all the factors of the leading coefficient (q), to get a list of . Make sure that you get ever possible combination of these factors,

written as .

32

Example 1: Use the Rational Zero Theorem to list all the possible rational zeros for

Step 1: List all of the factors of the constant.(P)

The factors of the constant term 12 are

.

Step 2: List all of the factors of the leading coefficient.(q)

The factors of the leading coefficient -1 are .


Writing the possible factors as we get:

33

Example 2: List all of the possible zeros, use synthetic division to test the possible zeros, find an actual zero, and use the actual zero to find all the zeros of

List all of the possible zeros:

The factors of the constant term 100 are

The factors of the leading coefficient 1 are


Use synthetic division to test the possible zeros and find an actual zero: Recall that if you apply synthetic division and the remainder is 0, then c is a zero or root of the polynomial function.

34

X2 = is NOT a zero X2- = is NOT a zero

X4- = is a zero or root of our polynomial function

35

Use the actual zero to find all the zeros:

Since, x = -4 is a zero, that means (x+4) is a factor of our polynomial function.

The zeros of this function are x =-4, -5 and 5 .

36

Example 3: List all of the possible zeros, use synthetic division to test the possible zeros, find an actual zero, and use the actual zero to find all the zeros of

Step 1: List all of the factors of the constant.(P)

The factors of the constant term -16 are

Step 2: List all of the factors of the leading coefficient.(q)

The factors of the leading coefficient 1 are .



37

X=-1 is NOT a zero X=1 is a zero or root of our polynomial function

Use synthetic division to test the possible zeros and find an actual zero: Recall that if you apply synthetic division and the remainder is 0, then c is a zero or root of the polynomial function.


Since, x = 1 is a zero, that means (x-1) is a factor of our polynomial function.

38

The zeros of this function are x =1,4, -2 and 2 .

39

Using the Rational Zeros Theorem

Example: Do each of the following for the polynomial function defined by

f(x) 6x3 5x2 7x + 4.

• List all possible rational zeros.

• Find all rational zeros and factors f(x) into linear factors.

40

Using the Rational Zeros Theorem continued

Solution: • For a rational number to be a zero, p must be a factor of a0 =

4 and q must be a factor of a4 = 6. Thus, p can be 1, 2, 3, 4, and q can be 1, 2, 3, or 6. The possible rational zero , are

Use the remainder theorem to show that 1 is a zero.

6 11 4 6 11 4 0

p

q

p

q1 1 1 2 4

, , , , 1, , 2, 4.6 3 2 3 3

1 6 5 7 4

41

Using the Rational Zeros Theorem continued

The new quotient for the polynomial is 6x2 11x + 4. This factors to(3x 4)(2x - 1). Setting 3x 4 = 0 and 2x 1= 0

yields zeros and

Thus the rational zeros are and the linear

factors of f(x) are x + 1, 2x 1, and 3x 4 . Therefore,

f(x) = (x + 1)(2x 1) (3x 4) = 6x3 5x2 7x + 4.

1.

21 4

1, , and ,2 3

4

3

42

Descartes's Rule of Signs

Let be a polynomial where are real coefficients.

•The number of POSITIVE REAL ZEROS of f is either equal to the number of sign changes of successive terms of f(x(or is less than that number by an even number (until 1 or 0 is reached. (

•The number of NEGATIVE REAL ZEROS of f is either equal to the number of sign changes of successive terms of f (-x) or is less than that number by an even integer (until 1 or 0 is reached) •This can help narrow down your possibilities when you do go on to find the zeros.

43

Example : Find the possible number of positive and negative real zeros of using Descartes’s Rule of Signs

Possible number of positive real zeros:

There are 4 sign changes between successive terms, which means that is the highest possible number of positive real zeros.

To find the other possible number of positive real zeros from these sign changes, you start with the number of changes, which in this case is 4, and then go down by even integers from that number until you get to 1 or 0.

Since we have 4 sign changes with f (x), then there is a possibility of 4 or 4 - 2 = 2 or 4 - 4 = 0 positive real zeros.

44

Possible number of negative real zeros:

Note how there are no sign changes between successive terms. This means there are no negative real zeros.

45

Example :Find the possible number of positive and negative real zeros of using Descartes’s Rule of Signs.


There is only 1 sign change between successive terms, which means that is the highest possible number of positive real zeros. To find the other possible number of positive real zeros from these sign changes, you start with the number of changes, which in this case is 1, and then go down by even integers from that number until you get to 1 or 0.

there is exactly 1 positive real zero

46


There are 2 sign changes between successive terms, which means that is the highest possible number of negative real zeros.

To find the other possible number of negative real zeros from these sign changes, you start with the number of changes, which in this case is 2, and then go down by even integers from that number until you get to 1 or 0.

Since we have 2 sign changes with f(-x)), then there is a possibility of 2 or 2 - 2 = 0 negative real zeros.

47

Example 7: List all of the possible zeros, use Descartes’s Rule of Signs to possibly narrow it down, use synthetic division to test the possible zeros and find an actual zero, and use the actual zero to find all the zeros of

List all of the possible zeros:

The factors of the constant term -2 are

The factors of the leading coefficient 3 are


Before we try any of these, let’s apply Descartes’s Rule of Signs to see if it can help narrow down our search for a rational zero

48


Since we have 3 sign changes with f(x), then there is a possibility of 3 or 3 - 2 = 1 positive real zeros.


Note how there are no sign changes between successive terms. This means there are no negative real zeros.

Since we are counting the number of possible real zeros, 0 is the lowest number that we can have. This will help us narrow things down in the next step.

49

Use synthetic division to test the possible zeros and find an actual zero:

there are NO negative rational zeros

X=1 1 is NOT a zero X=2 is a zero or root of our polynomial function


50

The zeros of this function are x=2 , , and .

51

Number of Zeros

• Fundamentals of Algebra Every function defined by a polynomial of degree 1

or more has at least one complex zero.

• Number of Zeros Theorem A function defined by a polynomial of degree n has at

most n distinct zeros– Example: f(x) = x3 + 3x2 + 3x + 1 = (x + 1)3 is of degree 3,

but has only one zero, 1. Actually, the zero 1 occurs three times, since there are three factors of x + 1; this zero is called a zero of multiplicity 3.

52

Creating a Polynomial Function when Given Zeros

Step 1: Use the given zeros and the Linear Factorization Theorem to write out all of the factors of the polynomial function.

Keep in mind that if you are given a non-real complex zero, that its conjugate is also a zero. Also keep in mind that the degree tells you how many linear factors over the complex numbers (possibly real and not necessarily distinct) that you will have. The factors are written in the following way: if c is a zero than (x – c( is a factor of the polynomial function.

Step 2: Multiply all of the factors found in Step 1.

53

Example

• Find a function f defined by a polynomial of degree 3 that satisfies the given conditions. Zeros of 2, 3, and 4; f(1) = 3

Solution: These three zeros give x (2) = x + 2, x 3, x 4 as factors of f(x).

Since f(x) is to be of degree 3, these are the only possible factors by the number of zeros theorem. Therefore, f(x) has the form f(x) = a(x + 2)(x 3)(x 4) for some real number a.

54

Example continued

To find a, use the fact that f(1) = 3.

f(1) = a(1 + 2)(1 3)(1 4) = 3

a(3)(2)(3) = 3

18a = 3

a =

Thus, f(x) = (x + 2)(x 3)(x 4),

or

f(x) =

1

61

6

3 21 5 14

6 6 3x x x

55

Example 5: Find an nth degree polynomial function where n =2 - 3i , 2 + 3i , and 4 are zeros; f(3) = -20.

56

This problem gave another condition, f (3)= - 20 This will help us find in this problem.

Putting it all together we get

57

f x x x x( ) 2 3 23 123 2Example: Discuss the real zeros of

There are at most three zeros, since the function is a polynomial of degree three.

Using Descartes’ Rule of Signs, f(x) has one sign change. So, there is one positive real zero.

f x x x x( ) ( ) ( ) ( ) 2 3 23 123 2

2 3 23 123 2x x x

Using Descartes’ Rule of Signs, f (-x) has two sign changes. So, there are two or zero negative real zeros.

58

Ex: Find the real zeros of f x x x x x x( ) 5 4 3 29 20 12

Factor f over the reals.

First, determine the nature of the zeros.

Since the polynomial is degree 5, there are at most five zeros.

Using Descartes’ Rule of Signs, there are three or one positive real zero(s).

f x x x x x x( ) 5 4 3 29 20 12

Using Descartes’ Rule of Signs again, there are two or no negative real zeros.

59

Now, list all possible rational zeros p/q by factoring the first and last coefficients of the function.

f x x x x x x( ) 5 4 3 29 20 12

p: , , 3, 4, 6, 12 1 2

q: 1pq

: 1, 2, 3, 4, 6, 12

Now, begin testing each potential zero using synthetic division. If a potential zero k is in fact a zero, then x - k divides into f (remainder will be zero) and is a factor of f.

60

f x x x x x x( ) 5 4 3 29 20 12

3 1 1 - 9 -1 20 -12

- 3 6 9 - 24 12

1 - 2 - 3 8 - 4 0

Test k = -3

Thus, -3 is a zero of f and x + 3 is a factor of f.

f x x x x x x( ) ( )( ) 3 2 3 8 44 3 2

Test k = -2 2 1 - 2 - 3 8 - 4

- 2 8 -10 4

1 - 4 5 - 2 0

Thus, -2 is a zero of f and x + 2 is a factor of f.f x x x x x x( ) ( )( )( ) 3 2 4 5 23 2

61

We know from Descartes’ Rule of Signs that there are no more negative real zeros.

Test k = 1 1 1 - 4 5 - 2

1 - 3 2

1 - 3 2 0

Thus, 1 is a zero of f and x - 1 is a factor of f.

f x x x x x x( ) ( )( )( )( ) 3 2 1 3 22

f x x x x x x( ) ( )( )( ) 3 2 4 5 23 2

f x x x x x x( ) ( )( )( )( )( ) 3 2 1 1 2

f x x x x x( ) ( )( )( ) ( ) 3 2 1 22

1 c ollege a lgebra polynomial and rational functions (chapter3) l:16 1 university of palestine...

Documents