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Factor each of the given polynomial equations completely and find all of the solutions. 2x3 – 3x2 – 3x + 2 = 0 5x* – 28x3 + 84x - 45 = 0 x6 - 3x5 + 2x* – 6x3 – 8x2 + 24x = 0 x5 – 14x* + 74x3 – 184x² + 224x – 128 = 0 SHOW ALL OF YOUR WORK AND THE COMPLETE FACTORIZATIONS OF EACH POLYNOMIAL. (Attach extra pages)

Factor each of the given polynomial equations completely and find all of the solutions. 2x3 – 3x2 – 3x + 2 = 0 5x* – 28x3 + 84x - 45 = 0 x6 - 3x5 + 2x* – 6x3 – 8x2 + 24x = 0 x5 – 14x* + 74x3 – 184x² + 224x – 128 = 0 SHOW ALL OF YOUR WORK AND THE COMPLETE FACTORIZATIONS OF EACH POLYNOMIAL. (Attach extra pages)

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter9: Quadratic Functions And Equations
Section9.6: Solving Quadratic Equations By Using The Quadratic Formula
Problem 59PFA
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A little math riddle:
A little math riddle. Look inside your mind and you will find a cubic equation with three solutions is well defined. The first one is A not an integer it will be, the second a positive integer C, and the integer E is the least of all three. The second equation is a quartic with four solutions all real, the first is a rational number F, the second an integer I. The third and fourth are irrational, they are L and the opposite of L as a stated in a theorem that you know so well. Next, do not be perplexed by an equation with six zeros where one is complex, a pure imaginary number M. We should not forget the conjugate N, purely imaginary as well, is also a solution this I am sure you can tell. Now we can see the third is an integer O and the fourth a greater integer P. The fifth root is R of irrational type, and the opposite of R is the sixth solution that I hope you all like. It happened so fast, we have arrived at the last equation and it must be, a quintic polynomial with two complex conjugate zeros S and T, you must determine in which order they each shall be. Finally, this fifth degree equation has only one real solution Y, an integer zero of multiplicity three. Factor each of the given polynomial equations completely and find all of the solutions. 2x3 – 3x2 – 3x + 2 = 0 5x* – 28x3 + 84x - 45 = 0 x6 – 3x5 + 2x* – 6x3 – 8x2 + 24x = 0 x5 – 14x* + 74x3 – 184x2 + 224x – 128 = 0 SHOW ALL OF YOUR WORK AND THE COMPLETE FACTORIZATIONS OF EACH POLYNOMIAL. (Attach extra pages) Use the results to find the corresponding values for each letter and then solve the puzzle below. 1-i 0 V2 2i V2 -1 4 -2i 2i 5 V3 1+i 3 0 V3
Transcribed Image Text:A little math riddle. Look inside your mind and you will find a cubic equation with three solutions is well defined. The first one is A not an integer it will be, the second a positive integer C, and the integer E is the least of all three. The second equation is a quartic with four solutions all real, the first is a rational number F, the second an integer I. The third and fourth are irrational, they are L and the opposite of L as a stated in a theorem that you know so well. Next, do not be perplexed by an equation with six zeros where one is complex, a pure imaginary number M. We should not forget the conjugate N, purely imaginary as well, is also a solution this I am sure you can tell. Now we can see the third is an integer O and the fourth a greater integer P. The fifth root is R of irrational type, and the opposite of R is the sixth solution that I hope you all like. It happened so fast, we have arrived at the last equation and it must be, a quintic polynomial with two complex conjugate zeros S and T, you must determine in which order they each shall be. Finally, this fifth degree equation has only one real solution Y, an integer zero of multiplicity three. Factor each of the given polynomial equations completely and find all of the solutions. 2x3 – 3x2 – 3x + 2 = 0 5x* – 28x3 + 84x - 45 = 0 x6 – 3x5 + 2x* – 6x3 – 8x2 + 24x = 0 x5 – 14x* + 74x3 – 184x2 + 224x – 128 = 0 SHOW ALL OF YOUR WORK AND THE COMPLETE FACTORIZATIONS OF EACH POLYNOMIAL. (Attach extra pages) Use the results to find the corresponding values for each letter and then solve the puzzle below. 1-i 0 V2 2i V2 -1 4 -2i 2i 5 V3 1+i 3 0 V3
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