assign6

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

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11:17:24 R.5.22 (was R.6.18). What do you get when you factor 36 x^2 - 9 and how did you get your result?

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RESPONSE --> (6x-3)(6x+3)

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11:17:37 ** 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 squars. **

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RESPONSE --> ok

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11:25:01 R.5.28 (was R.6.24 What do you get when you factor x^2 + 10 x + 1 and how did you get your result?

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RESPONSE --> (x+5)(2x+1)

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11:28:59 ** 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 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, 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. **

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RESPONSE --> I saw there was not a product and sum. I knew the answer i put was incorrect but i couldn't figure it out, now i see why.

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11:38:17 R.5.34 (was R.6.30). What do you get when you factor x^3 + 125 and how did you get your result?

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RESPONSE --> (x^3+5^3)=(x-5)(x^2+5x+15x) 5^3=125

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11:39:53 ** x^3+125 is the sum of two cubes, with 125 = 5^3. We know that a^3 + b^3 = (a+b) ( a^2 - 2 a b + b^2). So we write x^3+5^3 = (x+5)(x^2-5x+25). **

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RESPONSE --> I realize that it should have been 5^2 for the last number which is 25, where i put 15

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11:51:46 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?

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RESPONSE --> (x-16)(x-1) 16, 1 are factors of 16 and the sum of the two is equal to 17 but inorder to be negative 17, must put negative in the answer

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11:51:57 ** 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 we might have 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. The only possibility that gives us a + b = -17 is a = -1, b = -16. So we conclude that x^2 - 17 x + 16 = (x-16)(x-1). **

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RESPONSE --> ok

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11:56:26 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?

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RESPONSE --> (3x^2-3x)+(2x-2) 3x(x-1)+2(x-1) (3x+2)(x-1)

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11:56:32 ** 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). **

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RESPONSE --> ok

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12:04:11 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?

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RESPONSE --> (x+2)(x-12) factors are 2,12 of 24 which is the product of 8 and 3

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12:05:22 ** 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). **

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RESPONSE --> i dont understand how you get this. I thought you had to multiply 3 and 8 then factor that amt out

The factors have to multiply out to give you 3 x^2 - 10 x + 8.

(x + 2) ( x - 12) gives you x^2 - 10 x + 24, which is not the same thing as 3 x^2 - 10 x + 8. Your factors do get the -10x, but not the 3 x^2 or the +8

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12:10:09 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?

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RESPONSE --> It's prime, due to the fact that you could factor out a 2 but then you would have 2x and that's not right

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12:10:45 ** This expression factors, but not into binomtials 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 won't find the factors in the usual manner. The quadratic formula tells us that there are factors ( -6 + sqrt(6^2 - 4 * 14 * (-1) )) / (2 * -1) and ( -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. **

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RESPONSE --> ok

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See my note on the one problem where you had a question and let me know if you don't understand why your result doesn't work. You were using a process that works when the coefficient of x^2 is 1, but not in other cases. In other cases you have to try every possible pair of binomials.

assign6

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

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The factors have to multiply out to give you 3 x^2 - 10 x + 8.

(x + 2) ( x - 12) gives you x^2 - 10 x + 24, which is not the same thing as 3 x^2 - 10 x + 8. Your factors do get the -10x, but not the 3 x^2 or the +8

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See my note on the one problem where you had a question and let me know if you don't understand why your result doesn't work. You were using a process that works when the coefficient of x^2 is 1, but not in other cases. In other cases you have to try every possible pair of binomials.