You may submit revisions for Exam #2 to earn back points. The maximum number of points you may earn back is half of what you lost. For example, if your exam score is 60, then you lost 40 points. When you submit your revisions, you may earn back a maximum of 20 points.

In order to earn back points you must submit the following:

• your original exam
• a detailed explanation of why your solution was incorrect and how you will correct the problem
• your revised work

Due date is Tuesday, 10/31/2017.

On Tuesday 10/24/2017 Exam #2 was returned. On your exam was written:

• your WeBWorK grade, PLEASE NOTE, THIS SCORE WAS AUTOMATICALLY COMPUTED FOR ALL SETS EVEN OPEN ONES, so once you complete the open sets, your grade will change to reflect that. By the end of the semester all sets will be closed and WeBWorK will compute an accurate homework grade.
• your exam average, if you missed Exam #2, this will count as your dropped grade and your exam average will just be your score from Exam #1.
• your midsemester grade, calculated as 10% WeBWorK and 90% exam average. No quiz grades were added in here.
• your midsemester letter grade which was “P” pass, “BL” for borderline or “U” for unsatisfactory.

Note that November 10th is the last day you may drop the class and receive a “W” grade. If you are thinking about dropping the class, please see me to discuss this and check with the financial aid office if you are receiving financial aid.

It is also important to remember that your final grade is composed of your best 3 of 4 exams, your final exam, quizzes and WeBWorK. Please review the course syllabus for the breakdown.

My office hours are 10 am -12 pm on Tuesdays, room N602B. No appointment for those hours is necessary, so stop by with any questions or for help with course material. If that time does not work with your schedule, please contact me and we can set up an alternate meeting time.

Prof. Bonanome

Please check WeBWorK, new problem sets have been assigned for Polynomial Inequalities as well as Exponential and Logarithmic Functions!

Prof. Bonanome

Due to a conflict I will not be holding office hours today at 10-12. If you need to see me, I will be in my office from 3-4.

Prof. Bonanome

Session 10 and 11 on 10/5/2017 and 10/10 covered

• finding roots of polynomial functions
• using roots to factor polynomial functions
• the fundamental theorem of algebra
• complex roots
• rational functions
• graphing rational functions including how to locate vertical and horizontal asymptotes, removable discontinuities and intercepts
• WeBWork problems sets were extended and a new set was opened

Exam #2 will take place next Tuesday, 10/17/2017. The exam will begin promptly at 8 am. No extra time will be given for students arriving late. Please remember to bring your calculators. The exam will cover material from Sessions #6-#11. See here for a Sample Exam and here for Sample Exam solutions.

Is there a simple explanation why degree 5 polynomials (and up) are unsolvable?

“We can solve (get some kind of answer) equations like:

But why is there no formula for an equation like

I’m not sure if this has anything to do with the Galois theory, but is there a simple explanation as to why degree 5 polynomials (and up) are unsolvable?”

Posted on Mathematics Stack Exchange by Pacerier. This explanation comes from trb456:

“Solvable means solvable by radicals, and that means that, starting from the polynomial equation, you can only do 1) field arithmetic (+,−,×,÷), or 2) “extracting roots”; e.g. square roots, cube roots, etc. It is the case, by Abel-Ruffini first and then by Galois, that there is no general “formula” for solving polynomials above degree 4. Naively, that suggests that the formula gets “too complicated” at some point…

Galois found that the way to measure “too complicated” is by checking which roots of the polynomial can be “switched around”, or permuted, while maintaining certain equations of the roots. For example, if you are working over the rational numbers, then you can’t switch around any rational number without changing important relationships. That seems obvious. But what might seem strange is that for a polynomial like x^2−2, whose roots are sqrt(2) and -sqrt(2), you can switch these around and not hurt any other arithmetic!

The way to formalize what it means to “switch around” roots is thought group theory, and there is a group that corresponds to how the roots of a polynomial can be switched around called the Galois group. Finally, if this group is “too complicated” (i.e. too many ways to permute the roots), then that group and its corresponding polynomial are not solvable by radicals. In the case of 5th degree polynomials, if it were possible to “invert” the polynomial x^5−x−1 (i.e. solve it directly like we can x^5), I believe this is all that would we needed for all 5th degree polynomials to be solvable by radicals. So as you see, it’s just a “little bit” too complicated, and it gets worse as the degree increases.

I’m leaving out lots of details, but the other answers and links fill in those details. But I hope this gives you a flavor if what’s going on.”

Is there a simple explanation why degree 5 polynomials (and up) are unsolvable?,
https://math.stackexchange.com/users/452/trb456),
https://math.stackexchange.com/q/176591

In Session #9 we:

• reviewed graphs of basic polynomial functions such as f(x)=x^2, f(x)=x^3, f(x)=x^4 and their properties
• considered more complication graphs such as f(x)=x^4 – 6.5x^3 + 12.5x^2 -7x
• made observations about graphs such as “graphs of degree n have at most n roots and (n-1) extrema”
• pg. 117 in the text summarizes
• worked on WeBWorK homework sets

In Session 8 we:

• reviewed the division algorithm for numbers
• adapted this to dividing polynomial expressions
• gave a formal definition for a polynomial function
• Exam #1 was returned, see solutions here.