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)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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A little math riddle:

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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