course
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16:58:31 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 --> You get 4x^2-1 or 36x^2-9=4(x^2-1) You get this by finding the common monomial factor which is 4. You factor that out to get your answer.
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16:59:59 ** 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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17:00:56 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 --> there are no common factors so you get the same answer, its in simplest form?
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17:01:18 ** 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 --> ok
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17:02:46 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 --> Its not two squares so it stays the same too...125 isn't a sqrt
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17:02:56 ** 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 forgot about cubes
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17:05:55 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+1)(X-16) rules of factoring a second-degree polynomial. the factors are 1 and -16 because, to check this you add the 1 and -16 and it gives you -17.
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17:06:27 ** 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 --> I needed to put a subtract infront of the 16
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17:21:25 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 --> first add the like terms -3x+2x the factors will be (3x+2)(x-1)
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17:21:36 ** 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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17:23:50 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 --> ? in simpliest form?
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17:24:17 ** 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 --> Ok
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17:28:22 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 --> I don't know, i need to understand how to find the answers to these
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17:29:03 ** 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 --> So i need to use the quadratic formula.
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