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02.18.2019: One Lesson of Math - Solving Rational Equations and Factoring Four-Term Polynomials by U


Today's soundtrack is Tristania: Widow's Weeds, a gothic operatic metal album.


I've been thinking about making some changes to my daily routine. I've realized that I'm spending a lot of time sitting at my desk working on improving myself intellectually, but I haven't been spending as much time in other areas that I have opportunities for growth, which are:

  • Physical health

  • Social engagement

  • Time outdoors, and

  • Time spent with family.


Here are the actions I've taken so far:


  • Time with family:

  • I recently found a program called Forest that helps keep you focused by locking your phone, and limiting your web browser to certain websites. It's been helpful to me already; I've spent less time reading nonsense on Reddit, and my little forest is coming along nicely. It's helped me to better appreciate time spent with my family, and I'm feeling more engaged.

  • Physical health:

  • For a while now, I've been thinking about doing this; yesterday, I finally got started: Starting at 1, I'm going to do a cumulative increase of one push-up per day. So, yesterday I did one push-up. Today, I did two. Tomorrow, I'll do three. To help me keep track, I installed a counter widget on my phone that lets me track how many days have elapsed since a certain date. I set the start date to the 16th, so that it will always show me how many days have passed since, which will tell me how many push-ups I need to do that day. I wonder how far I'll get!

  • I installed SleepTown, another app by the same people who made Forest. The app is designed to provide incentive to stay off your phone betweeen certain hours - for me, between 11pm and 6am.

 

Now then, on to the math!


So far in this unit, I've learned how to add, subtract, multiply, and divide rational expressions. Today, I'll be learning how to find the value of a variable by solving an equation with rational coefficients or expressions.


Let's go through the steps.

 

How to solve a rational equation


  1. Find all non-permissible values

  2. Find a common denominator

  3. First, factor if possible

  4. Multiply each numerator by the common denominator, then clear the fractions

  5. Since we are originally dividing each numerator by its denominator, this lets us work with whole numbers instead

  6. Solve for x

  7. If quadratic, factor

  8. If not quadratic, solve for x algebraically

  9. Confirm that the solution(s) are permissible

  10. If a solution is not permissible, we call it "an extraneous root" (Pearson's Pre-Calculus 11, p. 579) and say that there is no solution.

  11. Using substitution, verify your answer

 

While doing the assignment, I came across a concept that's brand new to me: If I have a four-term polynomial that I need to factor, and factoring by grouping isn't working, I need to use the Rational Root Theorem to find possible roots, then test them using synthetic division. Now, even though my textbook asked me a question requiring the knowledge of these concepts, they did not mention the concepts at all, so it took me a fair bit of digging to figure this out. Here's how it works:


The Rational Root Theorem states that x = ± P/Q, with P representing the factors of our ending constant, and Q representing the factors of our "leading integer coefficient" (Cole's World of Mathematics), and x representing all possible roots.


  1. The equation I was working with was 2x³ - 2x² - 10x - 6 = 0.

  2. I started by factoring out the common factor of 2, giving me 2(x³ - x² - 5x - 3) = 0.

  3. I identified the factors of my ending constant, 3: 1 and 3.

  4. I identified the factors of my leading integer coefficient, 1: 1.

  5. Plugging 1 and 3 into x = ± P/Q gave me ±1 and ±3.


Now that I knew my possible factors, I needed to check them using synthetic division.


Synthetic Division Steps

  1. I started by listing my information:

  2. Coefficients from the polynomial: 1, -1, -5, -3.

  3. Possible roots: 1, -1, 3, -3.

  4. Next, I started testing my roots, looking for a remainder of 0:

  5. Multiply the possible root by the first coefficient:

  6. 11 = 1

  7. Add the second coefficient to the product:

  8. 1 + -1 = 0

  9. Multiply the possible root by the sum:

  10. 0 1 = 0

  11. Add the third coefficient to the product:

  12. 0 + -5 = -5

  13. Multiply the possible root by the sum:

  14. -51 = -5

  15. Add the fourth coefficient to the product:

  16. -5 + -3 = -2

You'll notice that my result was not zero, meaning that 1 was not one of my roots. After performing those same steps on my other three possible roots, I was able to confirm that the two roots are 3 and -1. Once knew my roots, I needed to divide my polynomial by the known roots:


I started by dividing x³ - x² - 5x - 3 by (x - 3) and got the result x² + 2x + 1. I divided that result by (x + 1) and got the result (x + 1).

  • So 2(x³ - x² - 5x - 3) can be factored as 2(x-3)(x² +2x + 1), and 2(x-3)(x² +2x + 1) can be factored as 2(x-3)(x+1)(x+1), which can be factored as 2(x-3)(x+1)².


Going back to the beginning of this equation, we can tell that our non-permissible values of x are 0 and -1. So we know that we can rule out (x+1) as an answer. That brings us to 2(x-3). Solved algebraically, we can say that x = 3.

 

That's it for today. What a wild ride!

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