real zeros of polynomial functions long division and synthetic division

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Real Zeros of Polynomial Functions Long Division and Synthetic Division () () () () fx dx qx rx Quotient divisor dividend

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Real Zeros of Polynomial Functions

Long Division and Synthetic Division

( ) ( ) ( ) ( )f x d x q x r x

Quotient

divisor dividend

Warm-up1-2 Factor and Simplify each expression.

2

12 241.

2

x

x x

2

2

12.

3 4

x

x x

12x

( 1)

( 4)

x

x

212 ( )

x

( 2)x x

( 1) ( 1)

x x

(( )4) 1 x x

Warm-up3-4 Factor each expression

completely.33. 27x

3 24. 8 x y

2( 3)( 3 9) x x x

2 2(2 )(4 2 ) x y x xy y

Warm-up5-6 Multiply (FOIL) each expression.

25. ( 5)( 6) x x

6. ( 8)( 8) x x x

3 2( 6 5 30) x x x

3 216 64 x x x

Warm-up

7. Evaluate( ) ( )

F ( ) 2 4,

f x h f x

or f x x findh

2, 0 h

Warm-up – Answer #7

2 4( ) ( )

F ( ) ,

f x h f x

or f x findxh

( ) ( ) [ 2( ) 4] ( 2 4)

2 2 4 2 4

2

2 0 ,

f x h f x

h hx h x

hh

h

h

h

x x

Multiplication EquationsCan be written as two or more division

equations.3 2 2 23 61 54 2 ( )( )x x x xx x

3

2

23 4 1

5 61 2

2x xx

x

x

x

3 22 5

32

2

4 26

1

xx x x

r

x

o

x

Think about this in terms of AreaIf I have a box lid with a length of x2+5x+6 and a width of x-2, what is the area of the box?2 5 6x x

2x 3 23 4 12

x x x

Area

3 2 2 23 61 54 2 ( )( )x x x xx x

LengtAr Wea thh id

Dividing by a Monomial

• Properties of exponents are used to divide a monomial by a monomial and a polynomial by a monomial.

4 2 2

2

24 12

12

x x y

x y

4 2 2

2 2

24 12

12 12

x x y

x y x y

22x y

Division by a Polynomial• A polynomial can be divided by a

divisor of the form x-r by using long division or a shortened form of long division called synthetic division.

• Long division is similar to long division of real numbers. This method works for dividing by any type of term.

• Synthetic division uses only the coefficients of the terms in the process. To use synthetic division your divisor must be linear!

Polynomials Consider the graph of f(x) = 6x3 - 19x2 +

16x - 4

•Notice that the graph appears to cross the x-axis at 2. •From the graph we know that x = 2 appears to be a zero meaning f(2) = 0.•If x = 2 is a zero then we also know f(x) has a factor of (x-2).•How many additional zeros does the graph have? •This means that there exists a 2nd degree polynomial q(x) such that

•To find q(x) without a calculator we will use long or synthetic division.

2( ) ( ) ( )f x x q x

What is the degree of the polynomial?

How many real zeros does the graph have?

Division Rules

Dividend equals divisor times quotient plus remainder.

( ) ( ) ( ) ( )f x d x q x r x

( ) ( )( )

( ) ( )

f x r xd x

q x q x

( )

( )

f x remainderQuotient

q x divisor

Quotient

divisor dividend

Long Division• Use a process similar to long

division of whole numbers to divide a polynomial by a polynomial.

• Leave space for any missing powers of x in the dividend.

• Write the remainder as you would with whole numbers. (Remainder over the divisor)

Example 1 – Long Division

1. Divide the “x” into 6x3

2. Multiply the 6x2 by the divisor (x-2)

3. Subtract4. Bring down the “16x”Repeat:5. Divide “x” into 7x2 6. Multiply (-7x) times (x-

2)7. Subtract8. Bring Down the 4Repeat…

3 22 6 19 16 4x x x x

3 26 19 16 4 divided by 2x x x x

6x2

3 26 12x x( )20 7x 16x

-7x

27 14x x ( )

0 2x 4( ) 2 4x

+2

0

Remainder is Zero

Example 2 – Long Division3 23 5 10 3

Divide 3 1

x x x

x

3105313 23 xxxx

2x

233 xx 26x x10

x2

xx 26 2 x12 3

4

x12 4

7

13

7

x

Subtract!!

Subtract!!

Subtract!!

remainder

divisor

The Remainder Theorem

If a polynomial f(x) is divided by x – k, the remainder is r = f(k).

82)( 2 xxxf

)3(f 8)3(2)3( 2 869

5

823 2 xxx

x

xx 32

8x

1

3x5

xx 32

3 x

Example 2 Concluded

Using the remainder theorem we know that f(-1/3) = - 7 and since the remainder is not 0, x = -1/3 is not a root of the function.

3 23 5 10 3

3 1

x x x

x

2 7 2 4

3 1x x

x

1

73

f

3 1 0

3 1

1

3

x

x

x

3 2( ) 3( ) 5(1 / 3 1 / 3 1 / 3) 10( 1 33)/f 3 23( 1 / 3) 5( 1 / 3) 10( 1 / 3) 3

3 1 5 1 10 13

1 27 1 9 1 3

1 5 10 3

9 9 3 17

Long Division Practice 1-3

31. Divide 1 by 1x x

13 22. 6 19 6 3x x x x

2 1x x

26 2x x

3 273. Find the quotient of .

3

x

x

2 3 9x x

Answers Practice 1-3

P1

3

3 2

2

2

2

1 -1

1

1

1

0

x x

x x

x

x x

x

x x

x

P2

3 2

3 2

2

2

2

3 6 -19 6

6 18

-

3

-2 6

2 6

6 2

0

x x x x

x x

x x

x x

x

x

x x

P3

3

3 2

2

2

2

3 27

3

3

3 9

9 27

9 27

3

9

0

x x

x x

x

x x

x

x

x x

Synthetic Division

• Shortened form or “short cut” of Long division

• Must have a linear divisor• Only uses coefficients not

variables• Zeros must be included to hold

the place of any power of x that is missing.

Synthetic DivisionThe pattern for synthetic division of a cubic polynomial is summarized as follows. (The pattern for higher-degree polynomials is similar.)1. Write the coefficients of the dividend in a upside-down division symbol.2. Take the zero of the divisor, and write it on the left.3. Carry down the first coefficient.4. Multiply the zero by this number. Write the product under the next coefficient.5. Add.6. Repeat as necessary

Example 3 - Synthetic Division

Divide x4 – 10x2 – 2x + 4 by x + 3

1 0 -10 -2 4-3

1

-3

-3

+9

-1

3

1

-3

1

3

4210 24

x

xxx3

1

x

13 23 xxx

RRemainder

CConstant

xx2x3

Multiply

Add

Coefficients→

Example 4 – Synthetic Division

2 7 -4 -27 -18+2

2

4

11

22

18

36

9

18

0

4 3 22 7 – 4 – 27 – 18Divide

2

x x x x

x

RCXX2X3

3 22 11 18 9x x x 4 3 22 7 – 4 – 27 – 18

2

x x x x

x

Example 4 - Synthetic Division

• Is (x-2) a factor of the function?

• Why or Why not?

• What is the value of f(2)?

• Where would you find this point on a graph?

Yes

3 2 4 3 2( 2)(2 11 18 9) 0 2 7 – 4 – 27 – 18x x x x x x x x

Definition of Division( ) ( ) ( ) ( )f x d x q x r x

f(2) = 0

On the x-axis at x = 2It is an x-intercept or root of the function.

Synthetic Division Practice 4-6

3 24 2 7 2 1 3. ( ) ( )x x x x

4 25 3 4 2 1 1. ( ) ( )x x x x

36 2 3 2. ( ) ( )x x x

2 42 1

3x x

x

3 23 3 1x x x

2 92 6

2x x

x

Answers Practice 4-6

P42 7 2 1

6 3 3

2 1 1 4

P5 P63

x2 x C R

3 0 4 2 1

3 3 1 1

3 3 1 1 0

1

x3 x2 x C R

1 0 2 3

2 4 12

1 2 6 9

2

x2 x C R

2 42 1

3x x

x

3 23 3 1x x x 2 9

2 62

x xx