1

POLYNOMIAL FUNCTIONS

PROBLEM 4

PROBLEM 1

Standard 3, 4, 6 and 25

PROBLEM 3

PROBLEM 2

PROBLEM 8

PROBLEM 5

PROBLEM 7

PROBLEM 6

PROBLEM 9 PROBLEM 10

END SHOW

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2

STANDARD 3:

Students are adept at operations on polynomials, including long division.

STANDARD 4:

Students factor polynomials representing the difference of squares, perfect square trinomials, and the sum and difference of two cubes.

STANDARD 6:

Students add, subtract, multiply, and divide complex numbers

STANDARD 25:

Students use properties from number systems to justify steps in combining and simplifying functions.

ALGEBRA II STANDARDS THIS LESSON AIMS:

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3

ESTÁNDAR 6:

Los estudiantes suman, restan, multiplican, y dividen numeros complejos.

ESTÁNDAR 3:

Los estudiantes hacen operaciones con polinomios incluyendo divisi[on larga.

ESTÁNDAR 4:

Los estudiantes factorizan diferencias de cuadrados, trinomios cuadrados perfectos, y la suma y diferencia de dos cubos.

ESTÁNDAR 25:

Los estudiantes usan propiedades de los sistemas numéricos para combinar y simplificar funciones.

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4

Standard 25

Evaluate:

h(2)= x + 4x -2x + 33 2

h(2) = ( ) + 4( ) – 2( ) + 33 22 2 2

= 8 + 4(4) – 4 + 3

= 8 + 16 – 4 + 3

= 23

Evaluate using synthetic division:

2 1 4 -2 3

1

2

6 10

12 20

23

h(2) = 23

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5

Standard 25

Evaluate:

h(3)= x + 3x -6x + 13 2

h(3) = ( ) + 3( ) – 6( ) + 13 23 3 3

= 27 + 3(9) – 18 + 1

= 27 + 27 – 18 + 1

= 37

Evaluate using synthetic division:

3 1 3 -6 1

1

3

6 12

18 36

37

h(3) = 37

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6

Standards 4, 25

Using synthetic division, The Remainder Theorem and the factor theorem:

2 1 2 -20 24

1

2

4 -12

8 -24

0

+4-12

Two numbers that multiplied be negative twelve = (+)(-) or (-)(+)

Two numbers that added be positive 4=|(+)|>|(-)|

(-1)(12) -1+12=11(-2)(6) -2+6=4

Given the following polynomial and one of its factor find the other two:h(y)= y + 2y -20y + 243 2 ; (y-2)

y +4y-1221

y +4y-122

(y-2)(y+6)

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7

Standards 4, 25

Using synthetic division, The Remainder Theorem and the factor theorem:

3 1 1 -17 15

1

3

4 -5

12 -15

0

+4-5

Two numbers that multiplied be negative twelve = (+)(-) or (-)(+)

Two numbers that added be positive 4=|(+)|>|(-)|

(-1)(5) -1+5=4

Find all the zeros of the following function if one zero is 3

h(y)= y + y -17y + 153 2

y + 4y -5=021

y + 4y -5=02

y – 1=0 y + 5 =0+1 +1

y = 1-5 -5

y = -5

(1,0) (-5,0)

zeros:

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Standards 3, 4, 6, 25

Find all zeros of if 2+i is one zero of f(x)

If we have a factor x – (2+i ) then we have the conjugate as well according to the complex conjugate theorem

x – (2-i )

f(x) = x – (2-i ) x – (2+i )because a third degree polynomial has 3 roots.

(2+i ) (2-i ) (2-i )xx-x

-x

(2+i )+=

= x -2x -xi -2x +xi + 4 - i

2

2

= x – 4x + 4 –(-1)2

F O I L

= x - 4x + 5

2

x - 4x + 5

2

x - x -7x +153 2

x

x3 - 4x2 + 5x-

3x2 -12x +15

+3

3x2 -12x +15-

The three zeros are:

+(2-i ) +(2-i )x – (2-i )=0

x= 2-i

+(2+i )+(2+i )x – (2+i )=0

x= 2+i

x+3=0-3 -3x= -3

i 2 = -1

recall:

Using long division to find the other factor:

x+3

f(x) =x - x -7x +153 2

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Standards 3, 4, 6, 25

Find all zeros of if 5+i is one zero of f(x)

If we have a factor x – (5+i ) then we have the conjugate as well according to the complex conjugate theorem

x – (5-i )

f(x) = x – (5-i ) x – (5+i )because a third degree polynomial has 3 roots.

(5+i ) (5-i ) (5-i )xx-x

-x

(5+i )+=

= x -5x -xi -5x +xi + 25 - i

2

2

= x –10x +25 –(-1)2

= x- 10x + 26

2

x -10x + 26

2

x -12x +46x -523 2

x

x3 -10x2 +26x-

-2x2 +20x-52

-2

2x2 +20x-52-

The three zeros are:

+(5-i ) +(5-i )x – (5-i )=0

x= 5-i

+(5+i )+(5+i )x – (5+i )=0

x= 5+i

x-2=0+2 +2x= 2

i 2 = -1

recall:

Using long division to find the other factor:

x-2

f(x) = x - 12x + 46x -523 2

F O I L

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10

Standard 25

r(x) = 5x + 4x -3x +6x -7x + 35 34 2

Using Descartes’ rule of signs, we count the number of changes in sign for the coefficients of r(x)

r(x) = 5x + 4x -3x +6x -7x + 35 34 2

+5 +4 -3 +6 -7 +3

no

yes

yes

yes

yes

There are four changes of sign, thus the number of positive real zeros is 4 or 2 or 0

r( ) = 5( ) + 4( ) -3( ) +6( ) -7( ) + 35 34 2-x -x -x -x -x -x

r(-x) = -5x + 4x +3x +6x +7x + 35 34 2

Now finding r(-x) and counting the number of changes in signs for the coefficients:

yes

no

no

no

no

-5 +4 +3 +6 +7 +3There is only one change, thus the number of negative real zeros is only 1.

State the number of positive and negative real zeros in

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11

Standard 25

r(x) = 5x + 4x -3x +6x -7x + 35 34 2

•The number of positive real zeros is 4 or 2 or 0

•The number of negative real zeros is only 1.

We concluded that:

State the number of positive and negative real zeros in

Number of Positive Real Zeros

Number of Negative Real Zeros

Number of Imaginary Zeros

4

2

0

1 0

1 2

1 4

4+1+0=5

2+1+2=5

0+1+4=5

r(x)

x

With the graph we verify that we have only 1 negative zero and 4 imaginary zeros.

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12

Standard 25

r(x) = 8x - 3x +2x -8x -2x + 45 34 2

Using Descartes’ rule of signs, we count the number of changes in sign for the coefficients of r(x)

+8 -3 +2 -8 -2 +4

yes

yes

yes

no

yes

There are four changes of sign, thus the number of positive real zeros is 4 or 2 or 0

r( ) = 8( ) - 3( ) +2( ) - 8( ) -2( ) + 45 34 2-x -x -x -x -x -x

r(-x) = -8x - 3x -2x - 8x +2x + 45 3 4 2

Now finding r(-x) and counting the number of changes in signs for the coefficients:

no

no

no

yes

no

-8 -3 -2 -8 +2 +4There is only one change, thus the number of negative real zeros is only 1.

State the number of positive and negative real zeros in

r(x) = 8x - 3x +2x -8x -2x + 45 34 2

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13

Standard 25

•The number of positive real zeros is 4 or 2 or 0

•The number of negative real zeros is only 1.

We concluded that:

State the number of positive and negative real zeros in

Number of Positive Real Zeros

Number of Negative Real Zeros

Number of Imaginary Zeros

4

2

0

1 0

1 2

1 4

4+1+0=5

2+1+2=5

0+1+4=5

r(x)

x

With the graph we verify that we have only 1 negative zero and 2 positive real zeros and 2 imaginary zeros.

r(x) = 8x - 3x +2x -8x -2x + 45 34 2

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14

Standard 3, 25

Write the polynomial function of least degree with integral coefficients whose zeros include 3, 2, -1.

x= 3

This implies that we have zeros at the following values of x:

-3 -3

x-3=0

x – 3 x – 2 x+1 =0

x x (-3)xx-2

-3

(-2)+= x+1F O I L

= x- 5x + 6

2

x+1

f(x)

x- 5x + 6

2

x+1X

+6-5x+6x-5x2

x2

x3

+6+ xx3 -4x2

= + 6 + xx3 -4x2

x= 2-2 -2

x-2=0

x= -1+1 +1

x+1=0

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15

Standard 3, 25

Write the polynomial function of least degree with integral coefficients whose zeros include 5, 2-2i

x= 5 x = 2-2i x = 2+2i

This implies that we have zeros at the following values of x:

-5 -5

x-5=0

-(2-2i) -(2-2i)

x -(2-2i) =0

-(2+2i) -(2+2i)

x -(2+2i) =0

x – (2-2i ) x – (2+2i ) x-5 =0

(2+2i ) (2-2i ) (2-2i )xx-x

-x

(2+2i )+= x-5

= x -2x -2xi -2x +2xi + 4 -4i

2

2 x –4x +4 –4(-1)

2 =

x-5

x-5

F O I L

= x- 4x + 8

2

x-5

f(x)

x- 4x + 8

2

x-5X

-40+20x+8x-4x2

-5x2

x3

-40+ 28xx3-9x2

= -40+ 28xx3-9x2

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