query 4

course mth 158

this was rough and hard to understand on this lesson

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‹R{Ƀassignme Uw~Ł College Algebra 01-29-2006

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10:04:09 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^3-23x^2-8 24x^3+48x-12 8x^3+15x-20

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10:04:31 ** 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 --> ok

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10:04:57 Query R.4.60 (was R.5.54). What is the product (-2x - 3) ( 3 - x)?

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RESPONSE --> 2x

You need to show more detail than that in your solution so I have an idea of what you are thinking. With more information I can try to help you address any errors in your procedures.

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10:05:04 ** 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 -->

In a good self-critique you need identify the specific things you do and do not understand in the given solution, and either demonstrate your understanding or ask specific questions about what you dont understand.

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10:05:15 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 --> 1

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10:05:21 ** 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 -->

You need to document your solution and self-critique when it doesn't agree with the given solution.

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10:05:51 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 --> 6xy^2

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10:05:58 ** 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 --> ok

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10:17:34 Query R.4.90 (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 --> cause its the sum of the product

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10:17:38 ** 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 --> ok

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ɎX⫁ɷ\ assignment #006 Uw~Ł College Algebra 01-29-2006

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10:22:17 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 --> (9)4x^2-1 you factor out 9

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10:23:01 ** 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 --> their both factoring

In fact, neither your solution nor the given solution is completely factored.

It is best to factor out the 9 in the first step, as you did. You get

9(4x^2-1).

Then 4 x^2 - 1 factors into (2x - 1)(2x + 1). So the final solution should read

9(2x-1)(2x+1).

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10:24:03 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 --> it cant be

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10:24:56 ** 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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10:25:56 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 --> x63=125cubed root so it is 5

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10:26:13 ** 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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10:27:11 ** 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). **

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RESPONSE --> x=16 x=1

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10:28:07 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 --> u cant

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10:28:15 ** 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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10:29:15 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 --> not possible just 7 and 2 thats all

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10:29:29 ** 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 --> ok

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򵐘҆d|h|Ǜ assignment #007 Uw~Ł College Algebra 01-29-2006

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10:31:41 Query R.7.10 (was R.7.6). Show how you reduced (x^2 + 4 x + 4) / (x^4 - 16) to lowest terms.

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RESPONSE --> x+2+x+=====1/4

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10:31:48 ** We factor the denominator to get first (x^2-4)(x^2+4), then (x-2)(x+2)(x^2+4). The numerator factors as (x+2)^2. So the fraction is (x+2)(x+2)/[(x-2)(x+2)(x^2+4)], which reduces to (x+2)/[(x-2)(x^2+4)]. **

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RESPONSE --> ok

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10:32:20 Query R.7.28 (was R.7.24). Show how you simplified[ ( x - 2) / (4x) ] / [ (x^2 - 4 x + 4) / (12 x) ].

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RESPONSE --> x-2 x=2 x-2 x=4

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10:32:23 Query R.7.28 (was R.7.24). Show how you simplified[ ( x - 2) / (4x) ] / [ (x^2 - 4 x + 4) / (12 x) ].

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RESPONSE --> ok

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10:32:36 ** [ ( x - 2) / (4x) ] / [ (x^2 - 4 x + 4) / (12 x) ] = (x-2) * / 4x * 12 x / (x^2 - 4x + 4) = (x-2) * 12 x / [ 4x ( x^2 - 4x + 4) ] = 12 x (x-2) / [4x ( x-2) ( x-2) ] = 3/(x - 2) **

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RESPONSE --> ok

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10:37:57 Query R.7.40 (was R.7.36). Show how you found and simplified the sum (2x - 5) / (3x + 2) + ( x + 4) / (3x + 2).

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RESPONSE --> multiplt top and bottom by 3x+2 then foil and get 6x^2-11x-16/12x^+20x+8

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10:38:01 Query R.7.40 (was R.7.36). Show how you found and simplified the sum (2x - 5) / (3x + 2) + ( x + 4) / (3x + 2).

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RESPONSE --> ok

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10:38:12 ** We have two like terms so we write (2x-5)/(3x+2) + (x+4)/(3x+2) = [(2x-5)+(x+4)]/(3x+2). Simplifying the numerator we have (3x-1)/(3x+2). **

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RESPONSE --> ok

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10:38:34 Query R.7.52 (was R.7.48). Show how you found and simplified the expression(x - 1) / x^3 + x / (x^2 + 1).

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RESPONSE -->

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10:38:38 ** Starting with (x-1)/x^3 + x/(x^2+1) we multiply the first term by (x^2 + 1) / (x^2 + 1) and the second by x^3 / x^3 to get a common denominator: [(x-1)/(x^3) * (x^2+1)/(x^2+1)]+[(x)/(x^2+1) * (x^3)/(x^3)], which simplifies to (x-1)(x^2+1)/[ (x^3)(x^2+1)] + x^4/ [(x^3)(x^2+1)]. Since the denominator is common to both we combine numerators: (x^3+x-x^2-1+x^4) / ) / [ (x^3)(x^2+1)] . We finally simplify to get (x^4 +x^3 - x^2+x-1) / ) / [ (x^3)(x^2+1)] **

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RESPONSE --> ok

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10:38:55 Query R.7.58 (was R.7.54). How did you find the LCM of x - 3, x^3 + 3x and x^3 - 9x, and what is your result?

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RESPONSE --> 3

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10:39:00 ** x-3, x^3+3x and x^3-9x factor into x-3, x(x^2+3) and x(x^2-9) then into (x-3) , x(x^2+3) , x(x-3)(x+3). The factors x-3, x, x^2 + 3 and x + 3 'cover' all the factors of the three polynomials, and all are needed to do so. The LCM is therefore: x(x-3)(x+3)(x^2+3) **

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RESPONSE --> ok

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10:39:06 Query R.7.64 (was R.7.60). Show how you found and simplified the difference3x / (x-1) - (x - 4) / (x^2 - 2x + 1).

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RESPONSE --> ok

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10:39:10 ** Starting with 3x / (x-1) - (x-4) / (x^2 - 2x +1) we factor the denominator of the second term to obtain (x - 1)^2. To get a common denominator we multiply the first expression by (x-1) / (x-1) to get 3x(x-1)/(x-1)^2 - (x-4)/(x-1)^2, which gives us (3x^2-3x-x-4) / (x-1)^2 = (3x^2 - 4x - 4) / (x-1)^2. DRV**

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RESPONSE --> ok

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10:39:20 QUESTION FROM STUDENT: On the practice test I'm having problems with problem #5 I don't know where to start or how to set it up. I'm probably missing something simple and will probably feel stupid by seeing the solution. Could you help with this problem. A retailer is offering 35% off the purchase price of any pair of shoes during its annual charity sale. The sale price of the shoes pictured in the advertisement is $44.85. Find the original price of the shoes by solving the equation p-.35p = 44.85 for p. INSTRUCTOR RESPONSE: It's very easy to get ahold of the wrong idea on a problem and then have trouble shaking it, or to just fail to look at it the right way. Nothing stupid about it, just human nature. See if the following makes sense. If not let me know. p - .35 p = 44.85. Since p - .35 p = 1 p - .35 p = (1 - .35) p = .65 p we have .65 p = 44.85. Multiplying both sides by 1/.65 we get p = 44.85 / .65 = etc. (you can do the division on your calculator); you'll get something near $67).

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RESPONSE --> ok

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»Sՠߓa assignment #008 Uw~Ł College Algebra 01-29-2006

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10:41:49 Extra question. What is the simplified form of sqrt( 4 ( x+4)^2 ) and how did you get this result?

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RESPONSE --> (2x+2)^2 4x+4

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10:42:02 ** sqrt(a b) = sqrt(a) * sqrt(b) and sqrt(x^2) = | x | (e.g., sqrt( 5^2 ) = sqrt(25) = 5; sqrt( (-5)^2 ) = sqrt(25) = 5. In the former case x = 5 so the result is x but in the latter x = -5 and the result is | x | ). Using these ideas we get sqrt( 4 ( x+4)^2 ) = sqrt(4) * sqrt( (x+4)^2 ) = 2 * | x+4 | **

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RESPONSE --> ok

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10:42:15 Extra Question: What is the simplified form of (24)^(1/3) and how did you get this result?

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RESPONSE --> 8

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10:42:28 ** (24)^(1/3) = (8 * 3)^(1/3) = 8^(1/3) * 3^(1/3) = 2 * 3^(1/3) **

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RESPONSE --> ok

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10:43:03 Extra Question: What is the simplified form of (x^2 y)^(1/3) * (125 x^3)^(1/3) / ( 8 x^3 y^4)^(1/3) and how did you get this result?

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RESPONSE --> ok

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10:43:10 ** (x^2y)^(1/3) * (125x^3)^(1/3)/ ( 8 x^3y^4)^(1/3) (x^(2/3)y^(1/3)* (5x)/[ 8^(1/3) * xy(y^(1/3)] (x^(2/3)(5x) / ( 2 xy) 5( x^(5/3)) / ( 2 xy) 5x(x^(2/3)) / ( 2 xy) 5 ( x^(2/3) ) / (2 y) **

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RESPONSE --> ok

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10:43:58 Extra Question: What is the simplified form of 2 sqrt(12) - 3 sqrt(27) and how did you get this result?

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RESPONSE --> 6-3Sqr t 27

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10:44:03 ** 2* sqrt(12) - 3*sqrt(27) can be written as 2* sqrt (4*3) - 3 * sqrt (9*3) by factoring out the maximum possible perfect square in each square root. This simplifies to 2* sqrt (4) sqrt(3) - 3 * sqrt (9) sqrt(3) = 2*2 sqrt 3 - 3*3 * sqrt 3 = } 4*sqrt3 - 9 * sqrt3 = -5sqrt3. **

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RESPONSE --> ok

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10:44:12 Extra Question: What is the simplified form of (2 sqrt(6) + 3) ( 3 sqrt(6)) and how did you get this result?

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RESPONSE --> 3

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10:44:18 ** (2*sqrt(6) +3)(3*sqrt(6)) expands by the Distributive Law to give (2*sqrt(6) * 3sqrt(6) + 3*3sqrt(6)), which we rewrite as (2*3)(sqrt6*sqrt6) + 9 sqrt(6) = (6*6) + 9sqrt(6) = 36 +9sqrt(6). **

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RESPONSE --> ok

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10:44:33 Query R.8.42. What do you get when you rationalize the denominator of 3 / sqrt(2) and what steps did you follow to get this result?

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RESPONSE --> 3

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10:44:38 ** Starting with 3/sqrt(2) we multiply numerator and denominator by sqrt(2) to get (2*sqrt(2))/(sqrt(2)*sqrt(2)) = (3 sqrt(2) ) /2.

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RESPONSE --> ok

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10:44:58 Query R.8.46. What do you get when you rationalize the denominator of sqrt(3) / (sqrt(7) - sqrt(2) ) and what steps did you follow to get this result?

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RESPONSE --> 3 sqrt 5

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10:45:02 ** Starting with sqrt(3)/(sqrt(7)-sqrt2) multiply both numerator and denominator by sqrt(7) + 2 to get (sqrt(3)* (sqrt(7) + 2))/ (sqrt(7) - 2)(sqrt(7) + 2). Since (a-b)(a+b) = a^2 - b^2 the denominator is (sqrt(7)+2 ) ( sqrt(7) - 2 ) = sqrt(7)^2 - 2^2 = 7 - 4 = 3 so we have sqrt(3) (sqrt(7) + 2) / 3.

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RESPONSE --> ok

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10:45:49 Extra Question: What steps did you follow to simplify (-8)^(-5/3) and what is your result?

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RESPONSE --> -3/4

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10:45:54 ** (-8)^(-5/3) = [ (-8)^(1/3) ] ^-5. Since -8^(1/3) is -2 we get [-2]^-5 = 1 / (-2)^5 = -1/32. **

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RESPONSE --> ok

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10:46:06 query R.8.64. What steps did you follow to simplify (8/27)^(-2/3) and what is your result?

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RESPONSE --> 4/9

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10:46:11 ** Starting with (8/27)^(-2/3) we can write as (8^(-2/3)/27^(-2/3)). Writing with positive exponents this becomes (27^(2/3)/8^(2/3)) 27^(2/3) = [ 27^(1/3) ] ^2 = 3^2 = 9 and 8^(2/3) = [ 8^(1/3) ] ^2 = 2^2 = 4 so the result is (27^(2/3)/8^(2/3)) = 9/4. **

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RESPONSE --> ok

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10:47:13 Extra Question: What steps did you follow to simplify 6^(5/4) / 6^(1/4) and what is your result?

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RESPONSE --> 7/1.3

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10:47:17 ** Use the laws of exponents (mostly x^a / x^b = x^(a-b) as follows: 6^(5/4) / 6^(1/4) = 6^(5/4 - 1/4) = 6^1 = 6. **

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RESPONSE --> ok

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10:47:48 Extra Question: What steps did you follow to simplify (x^3)^(1/6) and what is your result, assuming that x is positive and expressing your result with only positive exponents?

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RESPONSE --> x^6

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10:47:51 ** Express radicals as exponents and use the laws of exponents. (x^3)^(1/6) = x^(3 * 1/6) = x^(1/2). **

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RESPONSE --> ok

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10:48:01 Extra Question: What steps did you follow to simplify (x^(1/2) / y^2) ^ 4 * (y^(1/3) / x^(-2/3) ) ^ 3 and what is your result, assuming that x is positive and expressing your result with only positive exponents?

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RESPONSE --> xy^3/4

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10:48:05 ** (x^(1/2) / y^2) ^ 4 * (y^(1/3) / x^(-2/3) ) ^ 3 = x^(1/2 * 4) / y^(2* 4) * y^(1/3 * 3) / x^(-2/3 * 3)= x^2 / y^8 * y / x^(-2) = x^2 * x^2 / y^7 = x^4 / y^7. **

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RESPONSE --> ok

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10:48:27 query R.8.96. Factor 8 x^(1/3) - 4 x^(-2/3), x <> 0.

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RESPONSE --> 4x

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10:48:31 ** To factor 8x^(1/3)- 4x^(-2/3) we first need to write the expression without negative exponents. To accomplish this we multiply through by x^(2/3) / x^(2/3), obtaining (8 x^(1/3 + 2/3) - 4x^(-2/3 + 2/3) / x^(2/3) = (8 x - 4) / x^(2/3). We then factor 2 out of the numerator to obtain 4 ( 2x - 1) / x^(2/3). Other correct forms include: ( 4x^(1/3) ) ( 2 - ( 1/x) ) 8 x^(1/3) - 4 / x^(2/3). **

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RESPONSE --> ok

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10:48:54 Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE --> im lost on this I dont get it I guess ill study harder

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"

You did not get most of the questions in this assignment, and should therefore have included a detailed self-critique after seeing the given solutions.

I don't critique your answers until I've seen your self-critique, and that's for two reasons. In the first place the self-critique process is very important to your learning process, and in the second place I cannot help you if I dont know what you do and do not understand. The first step in the process of critique is yours, and it's important that you take that step unless you are very sure that you completely understand the given solution (which I believe often you do).

I suggest that you copy this document into Notepad, insert self-critiques after the given solution on those problems where your solution differed from the given solution, and submit the resulting document so I can help you address anything you might not be clear on.