course mth 158 assignment #005005. `query 4
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00:30:43 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 --> 7x^3 - 24x^2 - 44
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00:33:50 ** 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 8úx^3 - 24úx^2 - 48úx + 4 **
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RESPONSE --> Got the equation right to this point: 32 x^3 - 24 x^2 - 8 - 24 x^3 - 48 x + 12 Obviously need to go back to second grade because I can't subtract 32-24 in my head. I also left off the 48x in the answer. I have got to stop working at midnight. :-) Sorry about that. I do have the process down though.
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00:35:14 Query R.4.60 (was R.5.54). What is the product (-2x - 3) ( 3 - x)?
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RESPONSE --> (-2x - 3) ( 3 - x) = -6x + 2x2 - 9 + 3x = 2x^2 - 3x - 9
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00:35:36 ** 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 --> ok
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00:37:36 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 --> (x - 1) ( x + 1) = x^2 - 1 The special product formula was (x-a)(x+a) = x^2 + a^2
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00:38:11 ** 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 --> seems much easier to just use the special product formula than to distribute out all the factors
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00:49:18 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 --> (2x + 3y)^2 = 2x^2 + 2*3y*2x + 3y^2 = 2x^2 + 12xy + 3y^2 The special formula was (x+a)^2 = x^2 + 2ax + a^2
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00:50:58 ** 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 --> didn't square 3, only squared y
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00:53:19 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 --> The degree of the product of two polynomials equals the sum of their degrees because of the formula: (a^n)(a^m) = a^n+m This theory can be tested with the values of: a = 2 n = 3 m = 4 (2^3)(2^4) = (8)(16) = 128 2^3+4 = 2^7 = 128
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00:54:41 ** 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 --> should have been a bit more descriptive, rather than formulaic
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00:56:50 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> surprised that Algebra II from the early 90's is slowly coming back to me. It's all making sense. I think my biggest downfall is careless errors. If I can remember all of the formulas, I'll be doing OK.
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