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20:27:43 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 --> has common factor of 9
9(4x^2 - 1) (4x^2 -1) is a perfect sq. so we follow the special formula =9(4x-1)^2 = 9 ( 2x-1)(2x-1).................................................
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20:29:15 ** 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. **......!!!!!!!!...................................
RESPONSE --> hmmm, I thought to always check for a common monomial factor first.
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20:35:14 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 --> this isn't number 28 from the book. # 28 is to factor x^2 + 10x + 25.
x^2 + 10x + 1 is prime because there are no factors of 1 that add up to 10..................................................
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20:36:26 ** 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. **......!!!!!!!!...................................
RESPONSE --> ok
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20:38:16 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 --> using the rule for factoring the sum of two cubes I get:
(x+5)(x^2 - 5x +25).................................................
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20:38:33 ** 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). **......!!!!!!!!...................................
RESPONSE --> right
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20:39:33 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 --> the factors of 16 that add up to -17 are -1 and -16
= (x-16)(x-1).................................................
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20:39:57 ** 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). **......!!!!!!!!...................................
RESPONSE --> right
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20:41:56 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^2 -3x) + (2x - 2) factor out common monomial 3x(x-1) + 2 (x-1) = (3x -1 )(x-1).................................................
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20:42:11 ** 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). **......!!!!!!!!...................................
RESPONSE --> right
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20:45:41 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 --> factor trinomial by multiplying a*c = 3*8=24 Find factors of 24 whose sum is -10. Substitute these numbers for middle term 10x and solve by grouping.
(3x^2 - 6x) - (4x +8) = 3x(x - 2) -4(x-2) = (x-2)(3x-4).................................................
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20:46:26 ** 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). **......!!!!!!!!...................................
RESPONSE --> I did mine the other way but I understand this method of trial and error as well.
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20:48:18 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 --> This is prime. Write in standard form: -x^2 +6x + 14
factor out negative one and change signs: -1(x^2 - 6x - 14) There are no two rational factors of -14 whose sum is -6 so the problem is prime..................................................
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20:48:57 ** 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. **......!!!!!!!!...................................
RESPONSE --> I am glad.
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