Assignment 5

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22:04:29 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 --> 8(4x^3-3x^2-1)-6(4x^3+8x-2)= 32x^3-24x^2-8-24x^3-48x+12= 8x^3-24x^2+4-48x

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22:05:34 ** 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 --> My answer is correct

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22:24:07 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+2x^2-9+3x= -3x+2x^2-9

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22:28:34 ** 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 used FOIL because the book stated to use the FOIL method for problems 47-64 I copied the example in query using the Distributive Law and I do understand how to use this instead of FOIL

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22:29:42 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+x-x-1= x^2-1

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22:32:38 ** 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 --> I did not use Distributive Law and special product formulas have been confusing for me...Distributive Law is easier (for me) and I understand the solution shown using Distributive Law

The special products are very useful--those products appear frequently and it's worth the trouble to learn them. However if it's one or the other, the distributive law will always work.

Note that the special products are obtained by applying the distributive law.

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22:35:52 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 do not have a solution for #84 because I do not completely understand ""special product formula"" I do understand how to solve equations with special product formula such as (x-1)(x+1) or (x-7)(x+7) However, I do NOT understand how to solve equations with s.p.f. such as (2x + 3y)^2 or (x-5)^2

I'll show you the former:

The special product for the square of a binomial is (a + b)^ 2 = a^2 + 2 a b + b^2.

(2x + 3y)^2 is of the form (a + b)^2 with a = 2x and b = 3y.

Substituting a = 2x and b = 3y into (a + b)^ 2 = a^2 + 2 a b + b^2 we get

(2x + 3y)^2 = (2x)^2 + 2 * (2x) * (3y) + (3y)^2. Simplifying we have

(2x + 3y)^2 = 4 x^2 + 12 x y + 9 y^2.

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22:40:36 ** 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 --> I am going to have to work a few more equations that involve special product formulas.

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22:49:29 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 --> Hmmmm....According to my book, R.4.90 states: Explain why the degree ofthe sumof two polynomials of different degrees equals the larger of their degrees. :-) A polynomial in two variables is the sum of one or more monomials in two variables. The degree of a polynomial in two variables is the highest degree of all the monomials with nonzero coefficients.

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22:49:57 ** 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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22:52:12 Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE --> I was confident and doing well on the assignment until I reached the end of using special product formulas. Like I stated earlier, I do not understand fully how to work these equations. FOIL vs. Distributive Law...in my opinion, FOIL may be quicker, but Distributive Law is easier.

Hopefully my note will help.

FOIL doesn't work for longer expressions, such as (2x + 3 y - 4) ( 3x + 5 y - 2), but the distributive law does.

FOIL has its uses when it comes to factoring trinomials, but given the choice I would not introduce it before a student was grounded in the distributive law.

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Good work overall. Be sure to see my notes, and let me know if you have questions.