The student solutions shown below highlight the most commonly used approaches and also some that feature nice use of algebraic polynomial formulas.

Transcription

1 Print Assign Submit Solution and Commentary Online Resources Scoring Rubric [pdf] Teacher Packet [pdf] Strategy 11: Get Unstuck Strategy Examples [pdf] Polynomial Power [Problem #5272] Comments and Sample Solutions The key to solving this problem was taking the x + 1/x and treating it as a binomial. It could then be raised to a power, or multiplied by itself, to increase the powers of x. But whether raising to powers or multiplying, don't forget that there are middle terms to be considered! The student solutions shown below highlight the most commonly used approaches and also some that feature nice use of algebraic polynomial formulas. For the first question, almost everyone either squared the (x + 1/x) or multiplied it by itself. Some students, including Hari Vasireddy and Melodie Yu, used the formula that (a + b) 2 = a 2 + 2ab + b 2 to help with the squaring process. For the second question, the two most common methods were to either multiply three (x + 1/x) binomials together or to take the result of the first question and multiply that by another (x + 1/x) since it already represents two of them multiplied together. Melodie again used a neat formula for the sum of cubes, which says that a 3 + b 3 = (a + b)(a 2 - ab + b 2. By using x for a and 1/x for b she was able to quickly find x 3 + 1/x 3. In both problems, while the terms on the ends gave you the pieces you wanted, the terms in the middle cleaned up nicely and were able to be replaced with numeric values. It was pretty neat the way that worked! Another handy formula for the second problem shows the result of cubing a binomial: (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3. The coefficients of this expression, , are linked to Pascal's Triangle, and that's something that a number of people picked up on for the Extra question. Edel Auh and Rebecca Dun both talk about using Pascal's Triangle in their work, and Rebecca does a great job of explaining how it works in this context. Edel makes the excellent comment that in order to use it to find something like x 9 + 1/x 9 you need to know the values of at least some of the earlier expressions in the progression. Other people calculated several steps of the progression and then noticed a pattern. They saw that to find the value of a given step, it turned out to be five times the value from two steps before minus the value from one step before. That would certainly let you calculate the value of higher stages quickly as you could skip all the algebra and just use the results. But again, it requires knowing earlier values to reach the one you are trying to calculate. Can anyone find a way to go directly to the answer you want without having to know earlier values? Congratulations to Edel, Fatma, Liu, Haripriya, Rebecca, Melodie, and everyone else who solved this challenging problem. Nice work! - Riz From: Xingcheng L, age 13, Rancho San Joaquin Middle School x 2 + 1/x 2 would be equal to 23. x 3 + 1/x 3 is equal to 110. To find out what x 2 + 1/x 2 is, I multiplied the first equation, which was x + 1/x = 5, by itself. (x + 1/x)(x + 1/x) = 5*5 Page 1 of 10

2 Using the foil method, I quickly simplified and solved the equation to find what x 2 + 1/x 2 is. (x + 1/x)(x + 1/x) = 5*5 x 2 + x/x + x/x+ 1/x 2 = 25 x /x 2 = 25 x 2 + 1/x 2 = 23 To find out what x 3 + 1/x 3 is, I multiplied the first equation and the second one together. (x 2 + 1/x 2 )(x+1/x) = 23*5 x 3 + x + 1/x + 1/x 3 = 115 Then, I saw that there was an x + 1/x in the middle of the equation, and this could be replaced with 5. Then, you solve the equation. x /x 3 = 115 x 3 + 1/x 3 = 110 Reflection This POW was pretty easy, because all I had to do was multiply some equations together. At first, I thought that you were supposed to solve the problem by finding what x is, but after a while, I got the feeling that that would be hard, so I decided to just multiply the equations together. However, the extra was a little challenging because there was no real answer. Also, I only came to my pattern by Page 2 of 10

3 chance. Extra The pattern that I found was that x n + 1/x n = 5(x n-1 + 1/x n-1 ) - (x n-2 + 1/x n-2. However, this does not apply to the first and second equations. To find the pattern, I first found out what all the expressions up to x 7 + 1/x 7. x+1/x = 5 x 2 + 1/x 2 = 23 x 3 + 1/x 3 = 110 x 4 + 1/x 4 = 527 x 5 + 1/x 5 = 2525 x 6 + 1/x 6 = x 7 + 1/x 7 = Then, I got frustrated because I did not find a pattern, so I just did something random. I wanted to find how I could change into I just picked a random number, which turned out to be 5, and multiplied by 5. I got as the product. I then subtracted 2525 from 60490, and surprisingly, I got I tried this pattern with all of the other equations, and they all worked except for the first two. If you want to find the nth number in the sequence, then you multiply the previous number by 5 and then subtract the next previous one. This can also be said as x n + 1/x n = 5(x n-1 + 1/x n- 1 ) (x n-2 + 1/x n-2 ) Page 3 of 10

4 From: Haripriya V, age 13, Rancho San Joaquin Middle School x squared + 1/x squared is 23. x cubed + 1/x cubed is 110. EXTRA: (see explanation) In order to find the answer to x 2 + 1/x 2 I used the formula (a + b) 2 = a 2 + b 2 + 2ab. a = x b = 1/x (a + b) 2 = a 2 + b 2 + 2ab (x + 1/x) 2 = x 2 + 1/x (x + 1/x) = 5, so I replace that part of the equation with 5. (5) 2 = x 2 + 1/x = x 2 + 1/x 2 x squared + 1/x squared is 23. I calculated (x 2 + 1/x 2 )(x + 1/x). x 2 + 1/x 2 = 23 x + 1/x = 5 I multiply both values using the Distributive Property. (x 2 + 1/x 2 )(x + 1/x) x 3 + x + 1/x + 1/x 3 x 3 + 1/x 3 + (x + 1/x) = 23 times 5 x 3 + 1/x = 23 times 5 x 3 + 1/x 3 = 23 times 5-5 = EXTRA: After performing trial and error with exponents following the given pattern, I (with the help of my mother) was able to come up with a common equation that would work for the problems, excluding x + 1/x and x 2 + 1/x 2. x n + 1/x n = (x n-1 + 1/x n-1 )(x + 1/x) - (x n-2 + 1/x n-2 ) nth Term = (n-1)th Term x 5 - (n-2)th Term. Page 4 of 10

5 n=1 x + 1/x = 5 = 5 n=2 x 2 + 1/x 2 = 23 = 23 n=3 x 3 + 1/x 3 = 23x5-5 = 110 n=4 x 4 + 1/x 4 = 110x5-23 = 527 n=5 x 5 + 1/x 5 = 527x5-110 = 2,525 n=6 x 6 + 1/x 6 = 2525x5-527 = 12,098 n=7 x 7 + 1/x 7 = 12098x = 57,965 n=8 x 8 + 1/x 8 = 57965x = 277,727 n=9 x 9 + 1/x 9 = x =1,330,670 REFLECTION: It was challenging to figure out the future patterns, and required persistence. I enjoyed it, however, when I found out the method to finding the answer of any of that kind of problem. My mother helped me figure out the equation that would work for the extra. It was helpful when found out I could use the values from previous equations to figure out the answers for new ones. Being able to "split" large equations and group them using a parenthesis was also a plus. From: Rebecca D, age 13, Rancho San Joaquin Middle School The values for x^2 + 1/x^2 is 23 and x^3 + 1/x^3 is 110. Extra Credit: I saw that Pascal's Triangle showed the coffecients of the equations and that to find x^9 + 1/x^9, you only needed to find what the odd power equations were equal to which he First I squared x + 1/x = 5 to see if it equals x^2 + 1/x^2. (x + 1/x)^2 = (5)^2 x^ /x^2 = 25 Once I saw this, I subracted 2 from 25 in order to find what x^2 + 1/x^2 is equal to. x^2 + 1/x^2 = 23 Next, I cubed x + 1/x = 5 to see if it is close to x^3 + 1/x^3. (x + 1/x)^3 = (5)^3 x^3 + 3x + 3/x + 1/x^3 = 125 At first, this looked confusinng, but I saw this and substituted to find what x^3 + 1/x^3 is equal to. x^3 + 3(x + 1/x) + 1/x^3 = 125 x^3 + 3(5) + 1/x^3 = 125 x^ /x^3 = 125 x^3 + 1/x^3 = 110 So that's how I found x^2 + 1/x^2 = 23 and x^3 + 1/x^3 = 110. Page 5 of 10

6 Extra Credit: The first pattern I noticed was that this problem seemed to connect to Pascal's Triangle because it felt awfully familar to the POW, Sierpinski Stages, and what my teacher, Mr. Compton, was saying about this triangle pattern connecting to the coefficients of some equations. So then I looked up a picture of Pascal's Triangle and saw, just like Mr. Compton explained, that the triangle showed the cofficients of the equations. So then I tried x^5 + 1/x^5 and (x + 1/x)^5 to see if really does work. (x + 1/x)^5 = (5)^5 x^5 + 5x^3 + 10x + 10(1/x) + 5(1/x^3) + 1/x^5 = 3,125 x^5 + 5(x^3 + 1/x^3) + 10(x + 1/x) + 1/x^5 = 3,125 x^5 + 5(110) + 10(5) + 1/x^5 = 3,125 x^ /x^5 = 3,125 x^5 + 1/x^5 = 2,525 When I did this, I realized I could just find the values of the odd power equations to find x^9 + 1/x^9, so then I found the value of x^7 + 1/x^7 with (x + 1/x)^7. (x + 1/x)^7 = (5)^7 x^7 + 7x^5 + 21x^3 + 35x + 35(1/x) + 21(1/x^3) + 7(1/x^5) + 1/x^7 = 78,125 x^7 + 7(x^5 + 1/x^5) + 21(x^3 + 1/x^3) + 35(x + 1/x) + 1/x^7 = 78,125 x^7 + 7(2,525) + 21(110) + 35(5) + 1/x^7 = 78,125 x^7 + 17, , /x^7 = 78,125 x^7 + 1/x^7 = 57,965 Last but not least, I found the value of x^9 + 1/x^9 with (x + 1/x) ^9. (x + 1/x)^9 = (5)^9 x^9 + 9x^7 + 36x^5 + 84x^ x + 126(1/x) + 84(1/x^3) + 36(1/x^5) + 9(1/x^7) + 1/x^9 = 1,953,125 x^9 + 9(x^7 + 1/x^7) + 36(x^5 + 1/x^5) + 84(x^3 + 1/x^3) + 126(x + 1/x) + 1/x^9 = 1,953,125 x^9 + 9(57,965) + 36(2,525) + 84(110) + 126(5) + 1/x^9 = 1,953,125 x^ , , , /x^9 = 1,953,125 x^9 + 1/x^9 = 1,330,670 So that's the patterns and methods I used to make it easier to find the value of x^9 + 1/x^9, which is 1,330,670. Reflection: I would like to thank a lot of people for helping me on POW. First, I would like to thank Khue for showing me that finding x isn't necessary. Also, I would like to thank Jim for telling me that he used a equation (he didn't tell me what it was until after i solved it) because it was what lead me to the idea of Pascal's Triangle. Next, I would like to thank Mr. Compton for giving away all those hints about the POW, like how to solve x^2 + 1/x^2. Last, I would like to thank my mom for showing me an example of the process of how the triangle and these equations are linked (x + 1/x are the same coefficients as row 2 in the triangle). I found these answers reasonable because for x^2 + 1/x^2 and x^3 + 1/x^3, I knew they had to be close to 25 and 125. Also, for the Page 6 of 10

7 extra, I knew the value of x^7 + 1/x^7 was going to be a big number, x^9 + 1/x^9 had to be bigger which it is. The basic process of this POW is raise x + 1/x = 5 to whatever the power they ask and then find all the values inbetween and then subtract all those values from 5 to the something power and that will be the value. To check my answer for values, I looked over all my work and doublechecked to make sure I got the right answer. I also asked other kids at school if they got the same answer and they did. I found this POW very related to another POW, Sierpinski Stages, becuase they found have to do with Pascal's Triangle and some mysterious equation that really hard to find. From: Edel A, age 12, Rancho San Joaquin Middle School x^2 + 1/x^2 is equal to 23 and x^3 + 1/x^3 is equal to 110. Extra: One way is by using the formula x^n + 1/x^n = 5(x^(n-1) + 1/x^ (n-1)) - (x^(n-2) + 1/x^(n-2)). Using Pascal's Triangle is another way to solve higher value expressions. I decided to solve x^2 + 1/x^2 first. I realized that I could find (x^2 + 1/x^2) by squaring (x + 1/x) and then simplifying. x + 1/x = 5 (x + 1/x)^2 = 5^2 (x + 1/x)(x + 1/x) = 25 x^ /x^2 = 25 x^2 + 1/x^2 = 23 Therefore, I reached the conclusion that (x^2 + 1/x^2) is equal to 23. Next, I solved for (x^3 + 1/x^3). I realized that I could cube (x + 1/x) find out what (x^3 + 1/x^3) is equal to. (x + 1/x)(x + 1/x)(x + 1/x) = 5^3 (x^ /x^2)(x + 1/x) = 125 (x^3 + 2x + 1/x) + (x + 2/x + 1/x^3) = 125 (x^3 + 3x + 3/x + 1/x^3 = 125 (x^3 + 1/x^3) + (3x + 3/x) = 125 (x^3 + 1/x^3) + 3(x + 1/x) = 125 (x^3 + 1/x^3) + 3(5) = 125 (x^3 + 1/x^3) + 15 = 125 (x^3 + 1/x^3) = 110 Therefore, (x^3 + 1/x^3) is equal to 110. Extra: A method to finding out a higher value expression is by using the formula: x^n + 1/x^n = 5(x^n-1 + 1/x^n-1) - (x^n-2 + 1/x^n-2) For example: Solve: x^9 + 1/x^9 Need to know (x^(n-1) + 1/x^(n-1)) and (x^(n-2) + 1/x^(n-2)) which Page 7 of 10

8 is (x^8 + 1/x^8) and (x^7 + 1/x^7). 5(x^8 + 1/x^8) - (x^7 + 1/7) = (x^9 + 1/x^9) 5(277727) - (57965) = (x^9 + 1/x^9) = x^9 + 1/x^9 The other method that I found is related to Pascal's Triangle. I noticed that when I expand (x + 1/x)^n, the coefficients of the expanded equation were equal to the first half of the numbers in the nth row of Pascal's Triangle. They are coefficients to expressions in parantheses. The first expression is (x^n + 1/x^n). For each consecutive expression, the n decreases by two. This complete expression would be equal to 5^n since (x + 1/x) is equal to 5. After writing this new equation, I just needed to fill in the values and multiply to get my answer. Example: Solving for (x^9 + 1/x^9) 9th row of Pascal's Triangle (from left to middle): (x^9 + 1/x^9) + 9(x^7 + 1/x^7) + 36(x^5 + 1/x^5) + 84(x^3 + 1/x^3) + 126(x + 1/x) = 5^9 (x^9 + 1/x^9) + 9(57965) + 36(2525) + 84(110) + 126(5) = (x^9 + 1/x^9) + (521685) + (90900) + (9240) + (630) = (x^9 + 1/x^9) = (x^9 + 1/x^9) = The downside for both of these methods is that you need to know what other values of n in (x^n + 1/x^n) are equal to in order to use these methods. In reality, there is a lot more work, such as finding out (x^7 + 1/x^7), involved Reflection: This Problem of the Week was a bit difficult. I had help from adults on solving for (x^2 + 1/x^2), but I had to solve (x^3 + 1/x^3) on my own. At first, I thought that the equation I used above for solving (x^3 + 1/x^3) didn't work, but with the help of my tutor, I was able to find out that not only did it work, it gave me the correct answer. Then, for the Extra, I had help finding out the formula from my tutor and the method using Pascal's triangle from Mr.Compton. I didn't even think about looking at Pascal's Triangle, but once he explained it to me, I was able to understand how Pascal's Triangle was related to the problem. I was also able to simplify the method even more to discover an easier and quicker method. However when solving for (x^9 + 1/x^9), I got confused on what the number on the right side of the equation had to be, but my tutor helped me discover that it equaled to 5^9 and explain it. I would like to thank my tutor and my dad for helping me out and proofreading my explanation. I would also like to thank Mr. Compton and Brian S. for helping me out on the Extra. All in all, this Problem of the Week was a bit challenging, but I enjoyed it because it fits in with what we're learning in class right now. I think this Problem of the Week helped me learn more about polynomials and the way they work. I got to use FOIL, something I enjoy, substitution, and set up equations. Solving this was a lot of fun and I learned a lot. Page 8 of 10

9 From: Fatma., age 17, Gesamtschule Hattingen The value of the term x²+ 1/x² is 23. The value of the term x³+ 1/x³ is x²+1/x²=? I know that x+1/x=5 (x+1/x)²=5² x²+2*x*1/x+1/x²= 25 x²+2+1/x²= 25 x²+1/x²= 23 I square both sides I use the binomial formular I multiply the inner term I substact 2 on both sides 2. x³+1/x³=? (x+1/x)³=5³ I cube both sides (x+1/x)(x+1/x)(x+1/x)=125 I multiply the the fist two brackets (x²+2+1/x²)(x+1/x)=125 then i multiply it with the third x³+1x²/x+2x+2/x+1x/x²+1/x³=125 bracked x³+3x+3/x+1/x³=125 x³+3(x+1/x)+1/x³=125 I substitute the bracket for five x³+15+1/x³= I substract 15 on both sides x³+1/x³=110 From: Melodie Y, age 13, Rancho San Joaquin Middle School Answer: The value of x^2 + 1/x^2 is equal to 23 and I got the value of 110 for x^3 + 1/x^3. Explanation: First I noticed that I could use the equation a^2 + 2ab + b^2 to find out the value of x^2 + 1/x^2. I used x for a and 1/x for b. I would take x + 1/x = 5 and square it so I would get x^ /x^2 = 25. Then I would simplify it down to x^2 + 1/x^2 = 23 and that's how I got my first answer. a^2 + 2ab + b^2 (x + 1/x = 5)^2 x^2 + 2(x)(1/x) + (1/x)^2 = 25 x^ /x^2 = 25 x^2 + 1/x^2 = 23 Next to find the value of x^3 + 1/x^3, I used the equation a^3 + b^3 which is equal to (a + b)(a^2 - ab + b^2). I used that equation and used x + 1/x to substitute in. Like the first equation, a = x and b = 1/x. a^3 + b^3 = (a + b)(a^2 - ab + b^2) x + 1/x (x + 1/x)(x^2 - (x)(1/x) + (1/x)^2) Next I simplified it but I noticed that I already knew what x + 1/x was equal to, so I substituted that in for 5. 5(x^ /x^2) 5(x^2 + 1/x^2-1) After I simplified it, I noticed another part of the equation that we had the answer to, x^2 + 1/x^2 which was 23 so I substituted that for 23. Then I simplified and got my answer. 5(23-1) 5(22) 110 Page 9 of 10

10 Extra: I found that I could use the equation a^3 + b^3 = (a + b)(a^2 - ab + b^2) to find the value of the expression x^9 + 1/x^9. You replace a and b with a number that can be simplified in a^9 + 1/b^9. In this case it would be x^3 + 1/x^3. Reflection: I did not get this problem at all. I knew I wasn't supposed to try to solve it because that was the info Mr. Compton gave me but my head still kept trying to solve it. Melody was patient with me and explained it. I was still confused when she explained it but she explained it a little more and I eventually got it. Sidra tried to help me with x^3 + 1/x^3 but it was the wrong answer Drexel University. All Rights Reserved. Page 10 of 10