course Mth 158 ߣYߑiĊassignment #005
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12:45:49 Query R.4.36 (was R.5.30). What is the single polynomial that is equal to 8 ( 4 x^3 - 3 x^2 - 1 ) - 6 ( 4 x^3 + 8 x - 2 )?
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RESPONSE --> 32x^3 - 24x ^2 - 8 - 24x^3 - 48x + 12= 8x^3 - 24x^2 - 48x + 4
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12:50:58 ** ERRONEOUS STUDENT SOLUTION: To make this problem into a single polynomial, you can group like terms together. (8-6)+ (4x^3-4x^3) + (-3x^2) + (8x) + (-1+2). Then solve from what you just grouped...2 (-3x^2+8x+1). INSTRUCTOR CORRECTION: 8 is multiplied by the first polynomial and 6 by the second. You can't isolate them like that. Starting with 8 ( 4 x^3 - 3 x^2 - 1 ) - 6 ( 4 x^3 + 8 x - 2 ) use the Distributive Law to get 32 x^3 - 24 x^2 - 8 - 24 x^3 - 48 x + 12. Then add like terms to get 8x^3 - 24x^2 - 48x + 4 **
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RESPONSE --> That was my solution. I don't understand what the u is in the equation.
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12:52:29 Query R.4.60 (was R.5.54). What is the product (-2x - 3) ( 3 - x)?
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RESPONSE --> -6x^2 - 4xy - 9xy - 6y^2= -6x^2 - 13xy - 6y^2
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12:54:05 ** Many students like to use FOIL but it's much better to use the Distributive Law, which will later be applied to longer and more complicated expressions where FOIL does not help a bit. Starting with (-2x - 3) ( 3 - x) apply the Distributive Law to get -2x ( 3 - x) - 3 ( 3 - x). Then apply the Distributive Law again to get -2x(3) - 2x(-x) - 3 * 3 - 3 ( -x) and simiplify to get -6x + 2 x^2 - 9 + 3x. Add like terms to get 2 x^2 - 3 x - 9. **
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RESPONSE --> I'm sorry. I looked at problem 63. I had the right answer on 60.
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12:55:35 Query R.4.66 (was R.5.60). What is the product (x - 1) ( x + 1) and how did you obtain your result using a special product formula?
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RESPONSE --> I used the difference of 2 squares. (x^2 - a^2 x^2-1
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12:56:11 ** Starting with (x-1)(x+1) use the Distributive Law once to get x ( x + 1) - 1 ( x+1) then use the Distributive Law again to get x*x + x * 1 - 1 * x - 1 * 1. Simplify to get x^2 +- x - x + - 1. Add like terms to get x^2 - 1. **
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RESPONSE --> ok, but the difference of 2 squares was easier.
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13:00:45 Query R.4.84 (was R.5.78). What is (2x + 3y)^2 and how did you obtain your result using a special product formula?
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RESPONSE --> I used the special formula for the squares of binomials x^2 + 2ax + a^2 4x^2 + 12xy + 9y^2
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13:01:09 ** The Special Product is (a + b)^2 = a^2 + 2 a b + b^2. Letting a = 2x and b = 3y we get (2x)^2 + 2 * (2x) * (3y) + (3y)^2, which we expand to get 4 x^2 + 12 x y + 9 y^2. **
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RESPONSE --> ok
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13:07:18 Query R.4.105 \ 90 (was R.5.102). Explain why the degree of the product of two polynomials equals the sum of their degrees.
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RESPONSE --> Because of the law of exponents. When multiplying two polynomials, you add the exponents.
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13:08:27 ** STUDENT ANSWER AND INSTRUCTOR COMMENTS: The degree of the product of two polynomials equals the sum of their degrees because you use the law of exponenents and the ditributive property. INSTRUCOTR COMMENTS: Not bad. A more detailed explanation: The Distributive Law ensures that you will be multiplying the highest-power term in the first polynomial by the highest-power term in the second. Since the degree of each polynomial is the highest power present, and since the product of two powers gives you an exponent equal to the sum of those powers, the highest power in the product will be the sum of the degrees of the two polynomials. Since the highest power present in the product is the degree of the product, the degree of the product is the sum of the degrees of the polynomials. **
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RESPONSE --> Very detailed answer. Thanks.
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13:09:04 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE -->
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RłCYֻӅѦ assignment #006 006. `query 6 College Algebra 09-04-2008
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13:12:31 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 --> (6x - 3)(6x + 3) I knew that the square root of 36 was 6 and the square root of 9 was 3. Because I has a -, one of my signs had to be -.
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13:12:50 ** 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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13:13:46 R.5.32 \ 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 -->
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ȗCԀw assignment #006 006. `query 6 College Algebra 09-04-2008 Ŗi{nJ}v㞩 assignment #006 006. `query 6 College Algebra 09-04-2008 x͎njxvkF assignment #006 006. `query 6 College Algebra 09-04-2008
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13:20:18 R.5.32 \ 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 --> For problem 32, I had 25x^2 + 10x + 1 My answer is (5x + 1)^2. zI recognized it as factors of a perfect square.
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13:20:42 ** 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 --> My book doesn't have this problem.
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13:23:09 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 --> I used the formula for the sum of two cubes. (x-5)(x^2 - 5x + 25)
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13:23:23 ** 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 --> ok
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13:24:41 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 --> I found the factors of 16 that would equal -17. (x-1)(x-16)
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13:25:22 ** 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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13:28:12 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 --> I first grouped like so: (3x^2 - 3x) + (2x - 2) Then I factored each grouping. 3x(x-1) + 2(x - 1) = (x-1)(3x +2)
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13:28:24 ** 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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13:31:08 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 --> I multiplies 3 and 8 and found the factors that would give me -10. I then regrouped and factored. 3x^2 - 6x - 4x - +8 3x(x-2) - 4(x - 2) (x-2)(3x - 4)
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13:31:54 ** 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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13:32:44 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 --> There are no factors of 14 that will equal 6, so this is prime.
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13:33:33 ** 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 --> Thank goodness!
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