1 polynomial functions problem 4 problem 1 standard 3, 4, 6 and 25 problem 3 problem 2 problem 8...
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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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