#$&* course MTH 158 If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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Given Solution: 3 * * ** 36x^2-9 is the difference of two squares. We write this as (6x)^2-3^2 then get (6x-3)(6x+3), using the special formula difference of two squares. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I had to look up the second part of the answer but the first part I had. ------------------------------------------------ Self-critique Rating:2 ********************************************* Question: R.5.32 \ 28 (was R.6.24) What do you get when you factor 25 x^2 + 10 x + 1 and how did you get your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5x^2+2x+1 is what I get when I factor out 5 from each thing but I do not get why the equation answer has ### in it. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The coefficient of x^2 in the expression 25 x^2 + 10 x + 1 is 25. So when the expression is factored into the form (## x + ##) ( ## x + ##), where each ## could stand for any number, the coefficients of x (i.e., the numbers in front of the x's) have to multiply out to 25. 25 can only be expressed as a product of whole numbers are 25 * 1 or 5 # 5. So if the expression can be factored with whole numbers, it has to have one of the following forms: (25 x + ##) ( x + ##) or (5 x + ##) ( 5 x + ##). Since the constant number in our original expression is 1, the product of the numbers ## in either factorization must be 1. The only way two whole numbers can have a product of 1 is if they are both 1. So we have only two possible forms for the factored expression: (25 x + 1) ( x + 1) or (5 x + 1) ( 5 x + 1). If you multiply out both expressions using the distributive law, you will find that only the second one has the desired product. The solution given here is a sort of last-resort brute-force method, meant to be used when nothing else works. It's also useful for really understanding what's going on with this sort of factorization, but of course there are other ways of looking at this problem. With practice we naturally learn the various ways of thinking about factorization. THE REST OF THIS SOLUTION ADDRESSES THE DIFFERENT EXPRESSION x^2 + 10 x + 1, which is different that the expression of the given problem. * * ** STUDENT SOLUTION: x^2+10x+1 is prime because there are no integers whose product is 10 and sum is 1 INSTRUCTOR COMMENTS: The sum should be 10 and the product 1. I agree that there are no two integers with this property. Furthermore there are no two rational numbers with this property. So you would never be able to find the factors by inspection. However that doesn't mean that there aren't two irrational numbers with the property. For example 10 and 1/10 come close to the criteria, with product 1 and sum 10.1. The quadratic formula tells you in fact that the two numbers are ( -10 + sqrt( 10^4 - 4 * 1 * 1) ) / (2 * 1) and ( -10 - sqrt( 10^4 - 4 * 1 * 1) ) / (2 * 1) . Since 10^2 - 4 = 96 is positive, these are real numbers, both irrational. So the polynomial isn't prime. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I dont understand how you received the (5x+#) (x+#) in your answers. ------------------------------------------------ Self-critique Rating:2
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Given Solution: x^3+125 is the sum of two cubes, with 125 = 5^3. We know that a^3 + b^3 = (a+b) ( a^2 - a b + b^2). So we write x^3+5^3 = (x+5)(x^2-5x+25). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I got the pluses and minuses mixed up in the last factoring equation, other than that it was correct. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: R.5.46 (was R.6.42). What do you get when you factor x^2 - 17 x + 16 and how did you get your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The way that I figured out on how to do this is to find a number that pluses or minuses to get the -17x and times to get the 16 so it would equal (x-16)(x-1). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** x^2-17x+16 is of the form (x + a) ( x + b) = x^2 + (a + b) x + ab, with a+b = -17 and ab = 16. If ab = 16 then, if a and b happen to be integers, we have the following possibilities: a = 1, b = 16, or a = 2, b = 8, or a = -2, b = -8, or a = 4, b = 4, or a = -1, b = -16, or a = -4, b = -4. These are the only possible integer factors of 16. In order to get a + b = -17 we must have at least one negative factor. So the possibilities are reduced to a = -2, b = -8, or a = -1, b = -16, or a = -4, b = -4. The only of the these possibilities that gives us a + b = -17 is a = -1, b = -16. So we conclude that x^2 - 17 x + 16 = (x-16)(x-1). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: R.5.52 (was R.6.48). What do you get when you factor 3 x^2 - 3 x + 2 x - 2 and how did you get your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You can solve this factoring problem by grouping the first and the second with the third and fourth like; (3x^2-3x)(2x-2). Then you would factor out whats in common between both like; 3x(x-1) 2(x-1) which would leave you (3x+2)(x-1). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** This expression can be factored by grouping: 3x^2-3x+2x-2 = (3x^2-3x)+(2x-2) = 3x(x-1)+2(x-1) = (3x+2)(x-1). ** ADDITIONAL EXPLANATION: To see that (3x^2-3x)+(2x-2) = 3x(x-1)+2(x-1) apply the distributive law to each term in the second expression: 3x ( x - 1) = 3 x^2 - 3x, and 2 ( x - 1) = 2x - 2. To see that 3x(x-1)+2(x-1) = (3x+2)(x-1) apply the distributive law as follows: (3x + 2) ( x - 1) = 3x * (x - 1) + 2 * (x - 1). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I looked up in my old notes on how to factor when it comes to having 4 factors. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: R.5.64 (was R.6.60). What do you get when you factor 3 x^2 - 10 x + 8 and how did you get your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You first find out which set gives you this when you foil and I found out that (3x-4) (x-2) is the right answer in this situation. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** Possibilities are (3x - 8) ( x - 1), (3x - 1) ( x - 8), (3x - 2) ( x - 4), (3x - 4) ( x - 2). The possibility that gives us 3 x^2 - 10 x + 8 is (3x - 4) ( x - 2). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I had to look up how to do this but when I figured it out how to I easily finished this problem. ------------------------------------------------ Self-critique Rating:2 ********************************************* Question: R.5.82 (was R.6.78). What do you get when you factor 14 + 6 x - x^2 and how did you get your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I have also figured out that if you find two numbers that times to get the number that is in the place of 14 then you have your factor and I found out that (-x+7)(x-2) does the trick. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** This expression factors, but not into binomials with integer coefficients. We could list all the possibilities: (x + 7) ( -x + 2), (x + 2) ( -x + 7), (x + 14) ( -x + 1), (x + 1)(-x + 14) but none of these will give us the desired result. For future reference: You often cannot find the factors by factoring in the usual manner; however it is always possible to find the factors of a second-degree trinomial using the quadratic formula. The quadratic formula applied to this problem tells us that the factors are (x - z1) * (x - z2), were z1 and z2 are the solutions to the equation 14 + 6 x - x^2 0. Using the formula we find that z1 = ( -6 + sqrt(6^2 - 4 * 14 * (-1) )) / (2 * -1) and z2 = ( -6 - sqrt(6^2 - 4 * 14 * (-1) ) ) / (2 * -1) . Since sqrt(6^2 - 4 * 14 * (-1) ) = sqrt(36 + 56) = sqrt(92) is a real number these solutions are real numbers but again, as in a previous example, they aren't rational numbers and nobody could ever find them by inspection. This is not something you're expected to do at this point. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: R.5.82 (was R.6.78). What do you get when you factor 14 + 6 x - x^2 and how did you get your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I have also figured out that if you find two numbers that times to get the number that is in the place of 14 then you have your factor and I found out that (-x+7)(x-2) does the trick. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** This expression factors, but not into binomials with integer coefficients. We could list all the possibilities: (x + 7) ( -x + 2), (x + 2) ( -x + 7), (x + 14) ( -x + 1), (x + 1)(-x + 14) but none of these will give us the desired result. For future reference: You often cannot find the factors by factoring in the usual manner; however it is always possible to find the factors of a second-degree trinomial using the quadratic formula. The quadratic formula applied to this problem tells us that the factors are (x - z1) * (x - z2), were z1 and z2 are the solutions to the equation 14 + 6 x - x^2 0. Using the formula we find that z1 = ( -6 + sqrt(6^2 - 4 * 14 * (-1) )) / (2 * -1) and z2 = ( -6 - sqrt(6^2 - 4 * 14 * (-1) ) ) / (2 * -1) . Since sqrt(6^2 - 4 * 14 * (-1) ) = sqrt(36 + 56) = sqrt(92) is a real number these solutions are real numbers but again, as in a previous example, they aren't rational numbers and nobody could ever find them by inspection. This is not something you're expected to do at this point. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! ********************************************* Question: R.5.82 (was R.6.78). What do you get when you factor 14 + 6 x - x^2 and how did you get your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I have also figured out that if you find two numbers that times to get the number that is in the place of 14 then you have your factor and I found out that (-x+7)(x-2) does the trick. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** This expression factors, but not into binomials with integer coefficients. We could list all the possibilities: (x + 7) ( -x + 2), (x + 2) ( -x + 7), (x + 14) ( -x + 1), (x + 1)(-x + 14) but none of these will give us the desired result. For future reference: You often cannot find the factors by factoring in the usual manner; however it is always possible to find the factors of a second-degree trinomial using the quadratic formula. The quadratic formula applied to this problem tells us that the factors are (x - z1) * (x - z2), were z1 and z2 are the solutions to the equation 14 + 6 x - x^2 0. Using the formula we find that z1 = ( -6 + sqrt(6^2 - 4 * 14 * (-1) )) / (2 * -1) and z2 = ( -6 - sqrt(6^2 - 4 * 14 * (-1) ) ) / (2 * -1) . Since sqrt(6^2 - 4 * 14 * (-1) ) = sqrt(36 + 56) = sqrt(92) is a real number these solutions are real numbers but again, as in a previous example, they aren't rational numbers and nobody could ever find them by inspection. This is not something you're expected to do at this point. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!