Real Zeros of Polynomial Functions

Section 2.3

Objectives

• Use long division to divide polynomials by other polynomials.

• Use synthetic division to divide polynomials by binomials of the form (x-k).

• Use the Remainder and Factor Theorems.

• Use the Rational Zero Test to determine possible rational zeros of polynomial functions.

• Use Descarte’s Rule of Signs and the Upper and Lower Bound Rules to find zeros of polynomials.

Dividing Polynomials

Dividing a polynomial by a polynomial uses a “long division” technique that is similar to the process known as long division in dividing two numbers, which is reviewed on the next slide.

Dividing Polynomials

1 6 8 43 29 5 258 37 6 344 32

Divide

43 into 72.

Multiply

1 times 43.

Subtract

43 from 72.

Bring down

5.

Divide

43 into 295.

Multiply

6 times 43.

Subtract

258 from 295.

Bring down

6.

Divide

43 into 376.

Multiply

8 times 43.

Subtract

344 from 376.

Nothing to

bring down

.

We then write our result as 168 32 43 .

Dividing Polynomials

As you can see from the previous example, there is a pattern in the long division technique.

Divide Multiply Subtract Bring down Then repeat these steps until you can’t bring down or divide any longer.

We will incorporate this same repeated technique with dividing polynomials.

Division of Polynomials

Dividing Polynomials

Long division of polynomials is similar to long division of whole numbers. When you divide two polynomials you can check the answer using the following: dividend = (quotient • divisor) + remainder The result is written in the form: dividend  divisor  quotient + remainder divisor

Long Division of Polynomials

• Procedure: 1.

Arrange the terms of both polynomials in descending order of the exponents of the variable.

2.

3.

4.

If either the dividend or the divisor has missing terms, insert these terms with coefficients of 0.

Divide Check, by multiplying the divisor times the quotient and adding the remainder.

• If the remainder is zero, then the divisor is a factor of the dividend.

Example

: Divide

x

2 + 3

x

– 2 by

x

– 1 and check the answer.

x

 1

x

+ 2

x

2

x

2  +

x

3

x

 2 2

x

– 2 2

x

+ 2 – 4 1.

2.

x x

2 

x

2

x x

(

x

 1 ) 

x

2 

x



x

3.

(

x

2  3

x

)  (

x

2 

x

)  2

x

4.

5.

x

2 (

x

2

x

 2

x x

 1 )  2

x

 2  2 remainder – 4 Answer:

x

+ 2 +

x

 1 Check: (

x

+ 2) (

x

+ 1) + ( – 4) 6.

( 2

x

 2 )  ( 2

x

 2 )   4 =

x

2 + 3

x

– 2 correct quotient divisor remainder dividend

Example

: Divide 4

x +

2

x

3 – 1 by 2

x

– 2 and check the answer.

x

2 +

x

+ 3 2

x

 2 2

x

2

x

3 3  0 – 2

x

2

x

2  4

x

 1 Write the terms of the dividend in descending order.

Since there is no

x

2 dividend, add 0

x

2 term in the as a placeholder.

2

x

2 2

x

2 + 4

x

– 2

x

6

x

– 1 6

x

– 6 5 1.

2

x

3 

x

2 2.

x

2 ( 2

x

 2 )  2

x

3  2

x

2 2

x

3.

2

x

3  ( 2

x

3  2

x

2 )  2

x

2 4.

2

x

2 2

x



x

5.

x

( 2

x

 2 )  2

x

2  2

x

Answer:

x

2 +

x

+ 3 5  2

x

 2 Check: (

x

2 +

x

+ 3)(2

x

= 4

x +

2

x

3 – 1 – 2) + 5 6.

( 2

x

2  4

x

)  ( 2

x

2  2

x

)  6

x

7.

6

x

 3 8.

3 ( 2

x

 2 )  6

x

 6 2

x

9.

( 6

x

 1 )  ( 6

x

 6 )  5  remainder

Example

: Divide

x

2 – 5

x

+ 6 by

x

– 2.

x

– 3

x

 2

x

2  5

x

2 – 2

x x

 6 – 3

x

+ 6 – 3

x

+ 6 0 Answer:

x

– 3 with no remainder.

Check: (

x

– 2)(

x

– 3) =

x

2 – 5

x

+ 6

Example

: Divide

x

3 + 3

x

2 – 2

x

+ 2 by

x

+ 3 and check the answer.

x

2 + 0

x

– 2

x

 Note: the first subtraction 3 eliminated two terms from the dividend.

x

3

x

3  3

x

2 + 3

x

2  2

x

 2 0

x

2 – 2

x

+ 2 – 2

x

– 6 Therefore, the quotient skips a term.

8 Answer:

x

2 – 2 + 8

x

 3 Check: (

x

+ 3)(

x

2 – 2) + 8 =

x

3 + 3

x

2 – 2

x

+ 2

Your Turn:

x 2  3x 2 x 1 x  2 + 2  x 2 + x 2x + 2 0

CHECK:

(x + 1)(x + 2) = x 2 + 2x + x + 2 = x 2 + 3x + 2

Divide

Your Turn:

x 3 4  - x 2 0x 3 + 2x  x 2 - 2  by .

 3 x 1 -x 3 + x 2 2x 2 + 0x -2x + 1 x 3  x 2  3 3

Polynomial Long Division:

Example: 2x 4 x 2 + 3x 3 + 5x – 1 – 2x + 2

2x 2

x

2 

2

x



2 2

x

4 

3

x

3 

0

x

2 

5

x



1

2x 4 = 2x 2 x 2

x

2 

2

x



2

2x 2 +7x +10

2

x

4 

3

x

3 

0

x

2 

5

x



1

7x 3 = x 2 7x

7x 3 -( 7x 3 - 4x 2 +5x - 14x 2 +14x 10x 2 - 9x ) -1 11x - 21

remainder

The solution is written:

2

x

2



7

x



10



x

2 11

x

 

2

x

21



2 • Quotient + Remainder over Divisor

Your Turn:

• y 4 + 2y y 2 2 – y + 5 – y + 1 • Answer: y 2 + y + 2 + y 2 3 – y + 1

(5

x

3 – 4

x

2 + 7

x

Your Turn:

 2) ÷ (

x

2 + 1). Solution:

x

2  0

x

 1 5

x

5

x

3   4 4

x

2  7

x

 2 5

x

3  0

x

2  5

x

Quotient is 5

x

– 4 with a remainder   4 4

x x

2 2   2 0

x x

  of 2

x

+ 2.

2

x

 2 4 2 (5

x

3 – 4

x

2 + 7

x

 2) ÷ (

x

2 + 1) = 5

x

 4  2

x

 2

x

2  1

Division Algorithm for Polynomials

Let

f

(

x

) and

d

(

x

) be two polynomials, with degree of

d

(

x

) greater than zero and less than the degree of

f

(

x

). Then there exists unique polynomials

q

(

x

) and

r

(

x

) such that

f

(

x

) =

d

(

x

) •

q

(

x

) +

r

(

x

) Dividend = Divisor • Quotient + Remainder where either

r

(

x

) = 0 or the degree of

r

(

x

) is less than the degree of

d

(

x

).

The polynomial

r

(

x

) is called the remainder.

Synthetic Division

• A Quick method of dividing polynomials by a divisor of the form (x-k).

• Procedure: 1. Arrange the terms of the dividend in descending order and represent each term by it’s coefficient only, use zeros to represent the coefficients of any missing terms.

2. The divisor must be of the form (x-k) or rewritten in a form equivalent to (x-k).

3. Divide, by multiplying and adding.

4. Synthetic division yields the quotient and the remainder.

Synthetic Division

2x 3 +7x 2 +10x+15 / x+2

Synthetic Division

Use synthetic division to find the quotient: (x 3 – 3x 2 + x + 6) ÷ (x – 3) Step 1: Write the coefficients of the polynomial, and the k-value of the divisor on the left 3 1 -3 1 6

Synthetic Division

Use synthetic division to find the quotient: (x 3 – 3x 2 + x + 6) ÷ (x – 3) Step 2: Draw a line and write the first coefficient under the line.

3 1 -3 1 1 6

Synthetic Division

Use synthetic division to find the quotient: (x 3 – 3x 2 + x + 6) ÷ (x – 3) Step 3: Multiply the k-value, 3, by the number below the line and write the product below the next coefficient.

3 1 -3 3 1 6 1

Synthetic Division

Use synthetic division to find the quotient: (x 3 – 3x 2 + x + 6) ÷ (x – 3) Step 4: Write the sum of -3 and 3 below the line.

3 1 1 -3 3 0 1 6

Synthetic Division

Use synthetic division to find the quotient: (x 3 – 3x 2 + x + 6) ÷ (x – 3) Repeat steps 3 and 4.

3 1 1 -3 3 0 1 0 1 6

Synthetic Division

Use synthetic division to find the quotient: (x 3 – 3x 2 + x + 6) ÷ (x – 3) Repeat steps 3 and 4.

3 1 1 -3 3 0 1 0 1 6 3 9

Synthetic Division

Use synthetic division to find the quotient: (x 3 – 3x 2 + x + 6) ÷ (x – 3) The remainder is 9 and the resulting numbers are the coefficients of the quotient.

3 1 1 -3 3 0 1 0 1 x 2 + 1 + 9 x – 3 6 3 9

Synthetic Division

(x 3 – 3x 2 + 4) ÷ (x – 2) Method 1 x 2 - x – 2 x – 2 x 3 - (x 3 – 3x – 2x 2 ) 2 + 0x + 4 -x 2 - (-x 2 + 0x + 2x) –2x + 4 - (–2x + 4) 0 2 Method 2 1 -3 0 4 1 2 -1 -2 -2 -4 0 (x 2 – x – 2)

Synthetic Division

• Synthetic division is derived from the recursive nature of the algorithm. (Recursive – specifies a beginning value and a way to calculate each successive value) • This method relies on the relationships between the coefficients of x in the product of the quotient and divisor.

Example

Use synthetic division to find the quotient and remainder.

  4

x

5 

x

4  6

x

3  2

x

2  50   (

x

 2) 2 –4 –4 1 –8 –7 6 –14 –8 2 –16 –14 0 –28 –28 50 –56 –6

Note

: We must write a 0 for the missing term.

The quotient is – 4

x

4 remainder is –6.

– 7

x

3 – 8

x

2 – 14

x

– 28 and the  4

x

4  7

x

3  8

x

2  14

x

 28 

x

6  2

Your Turn:

• Use synthetic division to divide 5x 3 +8x 2 -x+6 by x+2.

• 5x 2 -2x+3 • Use synthetic division to divide x 4 -10x 2 -2x+4 by x-3.

•

x

3  3

x

2

x x

11  3

Synthetic Division

• Divide 6x 3 +7x 2 +x+1 by 2x+3 • 2x+3 = 2(x+3/2) • (6x 3 +7x 2 +x+1)/(2x+3) = (1/2)[(6x 3 +7x 2 +x+1)/(x+3/2)] • 3

x

2

x

2

x

5



3

Synthetic Division

REMAINDER THEOREM

Let f(

x

)

= 2x

4

+ x

3

– 3x

2

– 5

What is

f

(2)?

f

(

2

) = 2(

2

) 4 + (

2

) 3 – 3(

2

) 2 – 5

f

(2) = 2(16) + 8 – 3(4) – 5

f

(2) = 32 + 8 – 12 – 5

f(2) = 23

This is the same as the remainder when divided by (x – 2):

2 2 1 –3 0 –5 2 4 5 10 7 14 14 28 23

f(2) = 23

The Remainder Theorem

There are two parts of the

remainder theorem

: 1.

If the polynomial f(x) is divided by (x – a), the remainder will be a number that is equal to f(a).

1. i.e.. If f(x) is divided by x – 4, f(4) will give the value of the remainder.

2.

Dividend = (quotient ∙ divisor) + remainder 1. Also can see this as f(x) = [q(x) ∙ (x – a)] + f(a).

2. The quotient is always a polynomial with one degree less than f(x).

– Synthetic division is helpful in solving these problems (this can also be called synthetic substitution).

– The quotient may also be called a depressed polynomial because it has one less degree than the original polynomial.

The Remainder Theorem

That it means: – Used to find the remainder.

– When dividing a polynomial by a divisor of the form (

x-k

), the Remainder Theorem gives us two ways of calculating the remainder.

1) Use synthetic division to divide by (

x-k

).

2) Evaluate the function at

f(k)

.

The Remainder Theorem

Given P(x) = 3x 3 – 4x 2 find the remainder.

+ 9x + 5 is divided by x – 6, Method 1 6 3 -4 9 5 18 84 558 3 14 93 563 Method 2 P(6) = 3(6) 3 – 4(6) 2 + 9(6) + 5 = 3(216) – 4(36) + 54 + 5 = 648 – 144 + 54 + 5 = 563

The Remainder Theorem

Use synthetic division and direct substitution to find f(4) when f(x) = x 4 – 6x 3 + 8x 2 + 5x + 13.

4 1 -6 8 5 13 4 -8 0 20 1 -2 0 5 33 OR f(4) = 4 4 – 6(4) 3 + 8(4) 2 + 5(4) + 13 =256 – 384 + 128 + 20 + 13

f(4) = 33

The Remainder Theorem

Give the factors of x 3 – 11x 2 + 36x – 36 if one factor is x – 6.

6 1 -11 36 -36 6 -30 36 1 -5 6 0 So, after we divide the polynomial by x – 6 we are left with x 2 – 5x + 6 which we can solve by factoring into (x – 3)(x - 2).

This means the factors are

(x – 6)

,

(x – 3)

, and

(x - 2)

.

This can also be written in the f(x) = quotient ∙ divisor + remainder.

This would look like f(x) = (x 2 – 5x + 6)(x – 6) + 0 or f(x)=(x-2)(x-3)(x-6)..

The Remainder Theorem

• It is usually easiest to evaluate a polynomial for a specific value of “x” by using synthetic division and the remainder theorem • Example: Given

f

   • Find:

f x

5  2

x

4 

x

3  5 5

f

 

 1995 1 1  2 5 3 1 15 16 0 80 80 0 400 400  5 2000 1995

Your Turn:

• Find the remainder when f(x) =4x 3 +10x 2 -3x-8 is divided by (x+1).

• Solution: f(-1)=1, the remainder is 1.

The Remainder Theorem Problems 1.

2.

3.

Use synthetic division to do (4x 3 – 9x 2 – 10x – 2) divided by (x – 3). Then write the answer in the form f(x) = [quotient ∙ divisor] + remainder.

Given f(x) = 4x 2 + 6x – 7, find f(-5) by synthetic division or direct substitution.

Find the factors of x 3 + 6x 2 5).

– x – 30 if one factors is (x + Answers: 1) (4x 2 + 3x – 1)(x – 3) – 5 2) f(-5) = 63 3) (x + 5), (x – 2), and (x + 3)

The Remainder Theorem Problem • • • • Show that x=1 is a solution of the equation 2x 3 -x 2 +8=7x+2, and find any further solutions.

The first part is simple substitution, but you must show the result.

2(1) 3 -(1) 2 +8=7(1)+2 9=9, so x=1 is a solution.

Now have to solve the cubic equation 2x 3 -x 2 -7x+6=0. We know x=1 is a solution, so (x-1) is a factor. Using synthetic division, 1 2 -1 -7 6 2 1 -6 2 1 -6 0 So the quadratic factor is 2x 2 +x-6. This can be factored to (2x-3)(x+2). Thus the remaining solutions are 3/2 and -2.

Review-The Remainder Theorem

The Remainder Theorem

If a polynomial

f

(

x

) is divided by

x

remainder is equal to

f

(

k

).

–

k

, the

THE FACTOR THEOREM

k

is Zero of a Polynomial Function if

P

(

k

) = 0

Example

Decide whether the given number is a zero of

P

.

(a) (b) 2  ; 2 ;

P

(

x

) 

x

3  4

x

2  9

x

 10

P

(

x

)  3 2

x

3 

x

2  3 2

x

Analytic Solution

(a) 2 (b)  2 1 1 3 2 3 2  4 2  2  9 4   1 3  4 5  10 10 0 3 2 8 19 2 0  19  19 The remainder is zero, so

x

= 2 is a zero of

P

.

The remainder is not zero, so

x

= –2 is not a zero of

P

.

The Factor Theorem • From the previous example, part (a), we have

P

(

x

)

x

 2 

x

2  2

x

 5 indicating that

x –



P

(

x

)  (

x

 2 )(

x

2 2 is a

factor

of

P

(

x

).

 2

x

 5 ), •

Factor Theorem

: • (x – a) is a factor of f(x) if and only if the remainder (or f(a)) is equal to zero.

• This is a good way to find the first factor of a polynomial.

The Factor Theorem

• When a polynomial division results in a zero remainder – The divisor is a factor –

f(x) = (x – k) q(x) + 0

• This would mean that

f(k) = 0

– That is … k is a zero of the function • It also means the quotient, q(x), is a factor of f(x).

Example using the Factor Theorem Determine whether the second polynomial is a factor of the first.

P

(

x

)  4

x

3  24

x

2  48

x

 32 ;

x

 2

Solution

Use synthetic division with

k

= –2.

 2 4 4  24 8 16  48 32 16  32 32 0 Since the remainder is 0,

x

+ 2 is a factor of

P

(

x

), w here

P

(

x

)   ( 4

x

3

x

  2 24 )( 4

x

2

x

2   48 16

x x

 32  16 ).

Example using the Factor Theorem Use Remainder Theorem (evaluate

f(1)

) to determine whether x – 1 is a factor of x 3 – x 2 – 5x – 3.

Let x 3 – x 2 – 5x – 3 = 0 f(1) = (1) 3 – (1) 2 – 5(1) - 3 f(1) = 1 – 1 – 5 - 3 f(1) = -8 Since f(1) does not equal zero, x – 1 is not a factor.

Factoring a polynomial using the Factor Theorem • Factor f(x) = 2x 3 + 11x 2 + 18x + 9 • Given f(-3)=0 • Since f(-3)=0 • x-(-3) or x+3 is a factor • So use synthetic division to find the others!!

Factoring a polynomial using the Factor Theorem cont.

-3 2 11 18 9 -6 -15 -9 2 5 3 0 So…. 2x 3 + 11x 2 + 18x + 9 factors to: (x + 3)(2x 2 + 5x + 3) Now keep factoring gives you: (x+3)(2x+3)(x+1)

Your Turn:

• Show that (x+3) is a factor of f(x)=x 3 -19x-30. Then find the remaining factors f(x).

• Solution: f(-3) = 0, since the remainder is 0, (x+3) is a factor.

Remaining factors are (x-5) and (x+2).

Example using the Factor Theorem The function f(x)=x 3 +ax 2 -2x+b has (x+1) as a factor, and leaves a remainder of -3 when divided by x-2. Find the values of a and b.

-1 1 a -2 b 2 1 a -2 b -1 1-a 1+a 2 4+2a 4+4a 1 a-1 -1-a 1+a+b 1 2+a 2+2a 4+4a+b 4+4a+b=-3 1+a+b=0 Solve the system: a+b=1 4a+b=-7 ∴ a=-2, b=1

Review-The Factor Theorem

The Factor Theorem

A polynomial

f

(

x

) has a factor (

x

only if

f

(

k

)=0.

–

k)

, if and

THE RATIONAL ZERO THEOREM

The Rational Zeros Theorem

The Rational Zeros Theorem

Let

P

(

x

) 

a n x n



a n

 1

x n

 1   

a

1

x



a

0 , where

a n

 0 define a polynomial function w ith integer coefficien ts.

, If

p

/

q

is a rational number wri tten in lowest ter ms, and if

p

/

q

is a zero of

P

, then

p

is a factor of the constant term

a

0 , and

q

is a factor of the leading coefficien t

a n

.

The Rational Zero Theorem

What it Means: 1. The Rational Zero Theorem allows us to identify possible rational zeros of the polynomial. We then check the possible zeros, using the Remainder and Factor Theorems, to determine if they are actual zeros.

 factors of constant term factors of the leading coefficient 2. Does not agree the polynomial has any rational zeros.

Example Rational Zeros Theorem

Find all possible rational zeros :

f

   6

x

2  10

x

 4 All factors of 4 All factors of 6 : :  1,   1,  2,  2,  4 3,  6 All possible ratios of p q :  1 1 ,  1 2 ,  1 3 ,  1 6 ,  2 1 ,  2 2 ,  2 3 ,  2 6 ,  4 ,  1 4 2 ,  4 3 ,  4 6  1 ,  1 2 ,  1 3 ,  1 6 ,  2 ,  1,  2 3 ,  1 3 ,  4 ,  2,  4 3 ,  2 3 Simplified Possibilit ies :  1 6 ,  1 3 ,  1 2 ,  2 3 ,  1 ,  4 3 ,  2 ,  4

Your Turn:

• Find all possible rational zeros of f(x) = 6x 3 -x 2 +9x+4.

• Solution: ±1, ±2, ±4, ±1/6, ±1/3, ±1/2, ±2/3, ±4/3

Finding Rational Zeros

• • • • •

Example: Find the rational zeros for

f(x) =

x 3 + x 2 – 10x + 8 p /q

represents

all possible zeros

the

constant

, and

q

where

p

are all factors of are all factors of the

leading coefficient

.

p

:

all

the factors of

8 q

:

all

factors of

1 + 1, + 2, + 4, + 8 + 1

Here are all of the possible zeros for the function:

p + 1, + 2, + 4, + 8 q + 1

Possible zeros are

+ 1, + 2, + 4, + 8

So which one do you pick? Pick any. Find one that is a zero using synthetic division...

• f(x) =

x 3 + x 2 – 10x + 8

possible zeros

+ 1, + 2, + 4, + 8

Let’s try

1 : 1 1 1 1 – 10 8 1 2 – 8 1 2 – 8 0

is a zero of the function The depressed polynomial is

x 2

Find the zeros of

x 2 + 2x

(by using the quadratic formula)…

(x + 4)(x x = – – – 8 2) = 0 4, x = 2

The zeros of f(x) are

1,

by factoring or

– 4 + 2x

, and

– 2 8

Finding Zeros of a Polynomial Function

• Use Rational Zeros Theorem to find all possible rational zeros.

• Use Synthetic Division and Remainder Theorem to try to find one rational zero (remainder will be zero).

• If “n” is a rational zero, factor original polynomial as (x – n)q(x).

• Test remaining possible rational zeros in q(x). If one is found, factor again as in previous step.

• Continue in this way until all rational zeros have been found.

• See if additional irrational or non-real complex zeros can be found by solving a quadratic equation.

Example

• Find all zeros of:

f

• Find all solutions to:

x

5  

x

5 2

x

 3 2 

x

3 2

x

 2 2 

x

3 2

x

  3

x

2   2 0 • Rational Zeros Theorem says the only possible rational zeros are:  1 and  2 • See if -1 is a zero:  1 1 0  2  2  3  2 • Conclusion: 1  1  1 1  1 1  1 1  2 2 0  1 is x 4 a zero, x 3 x 2 (x x  1) is 2  a factor and another factor is :

Example Continued

 possible rational zeros:  1 and   2 • Check to see if -1 is also a zero of this: • Conclusion:  1 1 1  1  1  2  1 2 1  1  1  2  2 2 0 1 is  x 3 a zero, 2x 2  (x x  1) is 2  a factor and another factor is :

Example Continued

  possible rational zeros:  1 and  2 • Check to see if -1 is also a zero of this: • Conclusion:  1 1 1  2  1  3 1 3 4  2  4  6 1 is NOT a zero, so try another possible zero : 1

Example Continued

• Check to see if 1 is a zero: 1 1 1  2 1  1 1  1 0  2 0  2 • Conclusion: 1 is NOT a zero, so try another possible zero : 2

Example Continued

• Check to see if 2 is a zero: 2 1 1  2 2 0 • Conclusion: 1 0 1  2 2 0 2 is  x 2 a zero,  1  (x  2) is a factor and another factor is :

Example Continued

• Summary of work done:

f

  

x

5 

f x

 2

x

1

 

3

x

  2 2 2

x





x

 2 3 

x

1   2 1 is a zero of multiplici ty two, 2 is a zero, and the other two zeros can be found by solving

x

2  1  0 :

x

2  1  0

x

2   1

x

 

i Distinct zeros : - , , i, - i

Your Turn:

• 1.

Find rational zeros of

f

(x)=x 3 +2x 2 -11x-12 List possible q=1 p=-12 p/q= ±1,± 2, ± 3, ± 4, ± 6, ±12 2.

Test: 1 1 2 -11 -12 -1 1 2 -11 -12 p/q=1 1 3 -8 p/q=-1 1 3 -8 -20 -1 -1 12 1 1 -12 0 3.

Since -1 is a zero: (x+1)(x 2 +x-12)=

f(

x) Factor: (x+1)(x-3)(x+4)=0 x=-1 x=3 x=-4

Your Turn: Find the zeros of

g(x) =

6x 3 + 4x 2 – 14x + 4

Factors of p:

+ 1, + 2, + 4

Factors of q

: + 1, + 2, + 3, + 6

• • • •

Possible zeros are

+ 1, + 2, + 3, + 4, + 4/3, + 1/2, + 1/3, + 1/6, + 2/3

We’ll try

1 1 6 4 –14 4 6 10 –4 6 10 –4 0 1

is a zero, and

6x 2

polynomial.

Factor

6x 2 + 10x + 10x – 4 – 4

is the depressed

… 2(3x 2 + 5x – 2) 2(3x + 1)(x – 2)

Find the zeros…

2(3x + 1)(x – 2) = 0 3x + 1 = 0 x – 2 = 0

x = –1/3 x = 2

The zeros are

1, –1/3,

and

2

Example :

• 2.

f(x

)=10x 4 -3x 3 -29x 2 +5x+12 1.

List: q=10 p=12 p/q= ± 1/1, ± 2/1, ± 3/1, ± 4/1, ± 6/1,±12/1, ± 3/2, ± 1/5, ± 2/5, ± 3/5, ± 6/5, ± 12/5, ± 1/10, ± 3/10, ± 12/10 Check possible zeros using synthetic division: Check: p/q= -3/2 -3/2 10 -3 -29 5 12 -15 27 3 -12 10 -18 -2 8 0 Yes it works * (x+3/2)(10x 3 -18x 2 -2x+8)* (x+3/2)(2)(5x 3 -9x 2 -x+4) -factor out GCF (2x+3)(5x 3 -9x 2 -x+4) -multiply 1 st factor by 2

Repeat finding zeros for:

•

g

(x)=5x 3 -9x 2 -x+4

1.

q=5 p=4 p/q: ±1, ±2, ±4, ±1/5, ±2/5, ±4/5 Check possible zeros of depressed equation: 4/5 5 -9 -1 4 p/q=4/5 4 -4 -4 5 -5 -5 0 (2x+3)(x-4/5)(5x 2 -5x-5)= (2x+3)(x-4/5)(5)(x 2 -x-1)= mult.2

nd factor by 5 (2x+3)(5x-4)(x 2 -x-1)= -now use quad for last *-3/2, 4/5, 1 ± 5 , 2

Your Turn:

• Find all zeros of f(x) = x 4 -6x 3 -7x 2 +6x-8 • Solution: 1, -1, 2, 4

Review-Theorems

The Remainder Theorem

If a polynomial

f

(

x

) is divided by

x

to

f

(

k

).

–

k

, the remainder is equal

The Factor Theorem

The polynomial

x

– only if

P

(

k

) = 0.

k

is a factor of the polynomial

P

(

x

) if and

The Rational Zeros Theorem

Let

P

(

x

) 

a n x n



a n

 1

x n

 1   

a

1

x



a

0 , where

a n

 0 define a polynomial function w ith integer coefficien ts.

, If

p

/

q

is a rational number wri tten in lowest ter ms, and if

p

/

q

is a zero of

P

, then

p

is a factor of the constant term

a

0 , and

q

is a factor of the leading coefficien t

a n

.

Other Tests For Zeros of Polynomials

• Descartes’s Rule of Signs – Used to help identify the number of real zeros of a polynomial.

• Upper and Lower Bound Rules – Used to give an upper or lower bound of real zeros of a polynomial, which can help eliminate possible real zeros.

Descartes’ Rule of Signs

Let

P P

( (

x x

) be a polynomial function with real coefficients and a nonzero constant term. The number of positive real zeros of ) is either: 1. The same as the number of variations of sign in

P

(

x

), or 2. Less than the number of variations of sign in

P

(

x

) by a positive even integer.

The number of negative real zeros of

P

(

x

) is either: 3. The same as the number of variations of sign in

P

( 

x

), or 4. Less than the number of variations of sign in

P

( 

x

) by a positive even integer.

A zero of multiplicity

m

must be counted

m

times.

Descartes’ Rule of Signs

• A

variation in sign

means that two consecutive (nonzero) coefficients have opposite signs.

• The polynomial x 3 -3x+2 has two variations in sign.

Example

: Use Descartes’s Rule of Signs to determine the possible number of positive and negative real zeros of

f

(

x

) = 2

x

4 – 17

x

3 + 35

x

2 + 9

x

– 45. The polynomial has

three

+ to – variations in sign.

+ to –

f

(

x

) = 2

x

4 – 17

x

3 + 35

x

2 + 9

x

– 45 – to +

f

(

x

) has either three positive real zeros or one positive real zero.

f

( –

x

) = 2( – =2

x

4

x

) 4 + 17 – 17(–

x

3 + 35

x

) 3

x

2 + 35( – – 9

x x

– 45 ) 2 + 9( –

x

) – 45 One change in sign

f

(

x

) has one negative real zero.

f

(

x

) = 2

x

4 – 17

x

3 + 35

x

2 + 9

x

– 45 = (

x

+ 1)(2

x

– 3)(

x

– 3)(

x

– 5) .

Example

• What does Descartes’ rule of signs tell us about the number of positive real zeros and the number of negative real zeros?

 3

x

4  6

x

3 

x

2  7

x

 2  3

x

4  6

x

3 

x

2  7

x

 2 There are two variations of sign, so there are either two or zero positive real zeros to the equation.

Example continued

P

(

x

) 3(

x

) 4

x

) 3

x

) 2  3

x

4  6

x

3 

x

2  7

x

 2 2 The number of negative real zeros is either two or zero.

What this tells us; Total Number of Zeros 4 Positive Negative Nonreal 2 2 0 2 0 2 0 2 2 0 0 4

Your Turn:

Example

Determine the possible number of positive real zeros and negative real zeros of

P

(

x

) =

x

4 – 6

x

3 + 8

x

2 + 2

x

– 1.

We first consider the possible number of positive zeros by observing that

P

(

x

) has three variations in signs.

+ x

4 – 6

x

3 + 8

x

2 + 2

x

– 1 1 2 3 Thus, by Descartes’ rule of signs,

f

has either 3 or 3 – 2 = 1 positive real zeros.

For negative zeros, consider the variations in signs for

P

( 

x

).

P

( 

x

) = ( =

x

 4

x

) 4 + 6 – 6(

x

3 

x

+ 8 )

x

3 2 + 8( – 2

x



x

) 2 – 1 + 2( 

x

)  1 Since there is only one variation in sign,

P

(

x

) has only one negative real root.

Total number of zeros 4 Positive: 3 1 Negative: 1 1 Nonreal: 0 2

Upper and Lower Bound Rules Let

P

(

x

) define a polynomial function of degree

n

 1 with real coefficients and with a positive leading coefficient. If

P

(

x

) is divided synthetically by

x

–

c

, and (a) if

c

> 0 and all numbers in the bottom row of the synthetic division are nonnegative, then

P

(

x

) has no zero greater than

c

; (b) if

c

< 0 and the numbers in the bottom row of the synthetic division alternate in sign (with 0 considered positive or negative, as needed), then

P

(

x

) has no zero less than

c

.

Example: Show that the real zeros of

P

(

x

) = 2

x

4 satisfy the following conditions.

(a) No real zero is greater than 3.

(b) No real zero is less than –1.

– 5

x

3 + 3

x

+ 1 Solution a)

c

> 0 6 3 9 36 2 1 3 12 37 All are nonegative.

No real zero greater than 3.

b)

c

< 0    6 7  7 4 2  7 7  4 5 The numbers alternate in sign.

No zero less than  1.

How to Use the Upper and Lower Bound Rules • Finding the real zeros of

f



x

3  2

x

2  3

x

 6 we check the possible zero 2 by synthetic division: 2 1 1 2 2 4  3 8 5  6 10 4 • Notice bottom row. What does this tell us about zeros?

  has no zero greater than 2 • Now we don’t need to check any possible zeros greater than 2.

How to Use the Upper and Lower Bound Rules • Using the same function:

f

  

x

3  2

x

2  3

x

 6 check the possible zero -3  3 1 1 2  3  1  3 3 0  6 0  6 • Notice bottom row. What does this tell us about zeros?

f

  has no zero less than  3 • So we don’t need to check any possible zeros less than -3.