course MTH 158 I am having trouble with factoring, I'm going to keep praticing this section. ||㈌lط䆏[Eassignment #006
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14:07:20 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 --> First, 36x^2 and 9 have 9 in common: 9 (4x^2 -1) next: 9 (2x + 1) (2x - 1) confidence assessment: 3
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14:07:29 09-17-2007 14:07:29 ** 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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14:14:31 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 --> First you find a number whose product is 1 and whose sum is 10. There is no such number, so it is prime confidence assessment: 2
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14:14:51 09-17-2007 14:14:51 ** 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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14:17:42 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 --> By factoring the difference of two cubes: =(x - 5) (x^2 + 5x + 10) confidence assessment: 1
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14:17:48 09-17-2007 14:17:48 ** 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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14:20:45 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 --> By finding a number whose product is 16 and sum is -17 we find -1 and -16. x^2 - 16x - 1x + 16 grouping: x^2 - 1x and -16x + 16 x(x-1) and -16(x - 1) (x - 1) (x - 16) confidence assessment: 2
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14:20:55 ** 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 self critique assessment: 3
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14:23:54 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 --> Factor by grouping: 3x (x - 1) + 2 (x - 1) (3x + 2) (x - 1) confidence assessment: 3
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14:24:05 ** 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 self critique assessment: 3
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14:29:29 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 --> By finding a number whose product is 24 (3 * 8) and whose sum is -10, which is -12 and 2 3x^2 - 12x +2x +8 grouping: 3x^2 + 2x and -12x + 8 x (3x + 2) -4 (3x + 2) (x-4) (3x+2) confidence assessment: 2
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14:29:46 09-17-2007 14:29:46 ** 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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14:33:38 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 --> First you find two numbers whose product is 14 and whose sum is 6. There are no such numbers, so it is prime confidence assessment: 2
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14:33:49 09-17-2007 14:33:49 ** 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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