1. 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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

2. 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: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

3. 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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

4. 4 Standard 25 Evaluate: h(2)= x + 4x -2x + 3 3 2 h(2) = ( ) + 4( ) – 2( ) + 3 3 2 2 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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

5. 5 Standard 25 Evaluate: h(3)= x + 3x -6x + 1 3 2 h(3) = ( ) + 3( ) – 6( ) + 1 3 2 3 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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

6. 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 + 24 3 2 ; (y-2) y +4y-12 2 1 2 (y-2)(y+6) PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

7. 7 Standards 4, 25 Using synthetic division, The Remainder Theorem and the factor theorem: 3 1 1 -17 15 1 3 4 -5-5 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 + 15 3 2 y + 4y -5=0 2 1 2 y – 1=0 y + 5 =0 +1 y = 1 -5 y = -5 (1,0) (-5,0) zeros: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

8. 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 ) x x-x (2+ i ) + = = x -2x -x i -2x +x i + 4 - i 2 2 = x – 4x + 4 –(-1) 2 F O I L = x - 4x + 5 2 2 x - x -7x +15 3 2 x x 3 - 4x 2 + 5x - 3x 2 -12x +15 +3 3x 2 -12x +15 - The three zeros are: +(2- i ) x – (2- i ) =0 x= 2- i +(2+ i ) x – (2+ i ) =0 x= 2+ i x+3=0 -3 x= -3 i 2 = -1 recall: Using long division to find the other factor: x+3 f(x) =x - x -7x +15 3 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

9. 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 ) x x-x (5+ i ) + = = x -5x -x i -5x +x i + 25 - i 2 2 = x –10x +25 –(-1) 2 = x- 10x + 26 2 2 x -12x +46x -52 3 2 x x 3 -10x 2 +26x - -2x 2 +20x -52 -2 2x 2 +20x -52 - The three zeros are: +(5- i ) x – (5- i ) =0 x= 5- i +(5+ i ) x – (5+ i ) =0 x= 5+ i x-2=0 +2 x= 2 i 2 = -1 recall: Using long division to find the other factor: x-2 f(x) = x - 12x + 46x -52 3 2 F O I L PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

10. 10 Standard 25 r(x) = 5x + 4x -3x +6x -7x + 3 5 3 4 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 + 3 5 3 4 2 +5 +4 -3 +6 -7 +3 no 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( ) + 3 5 3 4 2 -x r(-x) = -5x + 4x +3x +6x +7x + 3 5 3 4 2 Now finding r(-x) and counting the number of changes in signs for the coefficients: yes no -5 +4 +3 +6 +7 +3 There is only one change, thus the number of negative real zeros is only 1. State the number of positive and negative real zeros in PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

11. 11 Standard 25 r(x) = 5x + 4x -3x +6x -7x + 3 5 3 4 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 10 12 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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

12. 12 Standard 25 r(x) = 8x - 3x +2x -8x -2x + 4 5 3 4 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 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( ) + 4 5 3 4 2 -x r(-x) = -8x - 3x -2x - 8x +2x + 4 5 3 4 2 Now finding r(-x) and counting the number of changes in signs for the coefficients: no yes no -8 -3 -2 -8 +2 +4 There 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 + 4 5 3 4 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

13. 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 10 12 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 + 4 5 3 4 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

14. 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 x-3=0 x – 3 x – 2 x+1 =0 xx (-3) x x -2 -3 (-2) + = x+1 F O I L = x- 5x + 6 2 x+1 f(x) x- 5x + 6 2 x+1x+1 X +6-5x +6x -5x 2 x 2 x 3 +6 + x x 3 -4x 2 = + 6 + x x 3 -4x 2 x= 2 -2 x-2=0 x= -1 +1 x+1=0 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

15. 15 Standard 3, 25 Write the polynomial function of least degree with integral coefficients whose zeros include 5, 2-2 i x= 5 x = 2-2 i x = 2+2 i This implies that we have zeros at the following values of x: -5 x-5=0 -(2-2 i ) x -(2-2 i ) =0 -(2+2 i ) x -(2+2 i ) =0 x – (2-2 i ) x – (2+2 i ) x-5 =0 (2+2 i ) (2-2 i ) x x-x (2+2 i ) + = x-5 = x -2x -2x i -2x +2x i + 4 -4 i 2 2 x –4x +4 –4(-1) 2 = x-5 F O I L = x- 4x + 8 2 x-5 f(x) x- 4x + 8 2 x-5x-5 X -40+20x +8x -4x 2 -5x 2 x 3 -40 + 28x x 3 -9x 2 = -40 + 28x x 3 -9x 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved