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course

I keep getting a script out of range error and can not access the querry program I seem to be able to try again later and normaly get it to work. Any ideas?

Anything can happen in Windows, but what you describe is not uncommon occurrence and could be unique to the setup of your specific machine. At what precise point were you in the running of the program when this occurred?

Åvàòä¯ãñ‰û•°ŒSƒâÂgŽfæºassignment #005

005. `query 4

College Algebra

09-15-2008

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11:26:38

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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11:27:43

** 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 -->

ok

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11:29:08

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

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RESPONSE -->

-6x+2x^2-9+3x

2x^2-3x-9

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11:29:44

** 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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11:30:18

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^2-1

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11:30:29

** 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

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11:33:13

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 -->

4x^2+6xy+6xy+9y^2

4x^2+12xy+9y^2

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11:33:23

** 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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11:34:06

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 -->

I dont understand these that well

If you were to multiply (x^3 + 6 x^2 - 7) ( x^2 + 3 x - 4) what would be the degrees of the two polynomials, and what would be the degree of their product?

Why is this so?

How can you generalize this to any two polynomials of given degree? How do the degrees of the two polynomials determine the degree of the product?

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11:34:38

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

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RESPONSE -->

Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.ok

Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.

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You have good answers on most of these questions.

&#Let me know if you have questions. &#