R7 Assignment

course Mth 158

07/08 9:47 pm

If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution:

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

007. `* 7

* 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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Your solution:

X^2+4x+4/x^4-16=(x+2)(x+2)/(x-2)(x+2)(x^2+4)=x+2/(x-2)(x^2+4)

confidence rating #$&* 2

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Given Solution:

* * ** 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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Self-critique (if necessary):

I didn’t use the brackets.

The denominator should be grouped, as should the numerator:

(x+2) / ( (x-2)(x^2+4) ).

You can do that with parentheses. I use brackets to emphasize the grouping.

In any case I understood what you meant.

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Self-critique rating #$&* Fair

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Question:

* 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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Your solution:

{(x-2)/(4x)/[x^2-4x+4/(12x)]

(X-2)*12x/4x(x^2-4x+4=(x-2)*12x/4x(x-2)(x-2)=3/x-2

confidence rating #$&* 1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

[ ( 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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Self-critique (if necessary):

I worked on this one a very long time and am still very unsure about what I am doing, I understand better than I did, but I am still practicing.

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Self-critique rating #$&* Shaky OK

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Question:

* 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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Your solution:

(2x-5)/(3x+2)+(x+4)/3x+2

2(x-5)/3(x+2)+(x+4)/x(x+2)=2(x-5)+(x+4)/3(x+2)

confidence rating #$&* 1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

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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Self-critique (if necessary):

I really overthought this one when I see the solution it is so simple.

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Self-critique rating #$&*Fair

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Question:

* 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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Your solution:

(x-1)/x^3+x/(x^2+1)=x-1/x^4+x/(x^2+1)????????????

confidence rating #$&* -1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

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)]

STUDENT QUESTION

I got x^3 + x – x^2 – 1 + x^4 / (x^3)(x^2 + 1). Where did I go wrong?

INSTRUCTOR RESPONSE

As written your solution x^3 + x – x^2 – 1 + x^4 / (x^3)(x^2 + 1) indicates that you first divide x^4 by x^3, then multiply the result by x^2 + 1. You finally add this result to x^3 + x - x^2 - 1. I do not believe this is what you intended.

I believe every step in your solution was correct as you wrote it on paper, and that what you intended to say was correct.

However to put it correctly into typewriter notation, you need to group the denominator throughout, and in the last step you need to group your numerator.

Using brackets to emphasize the grouping of numerator and denominator:

[ x^3 + x – x^2 – 1 + x^4 ] / [(x^3)(x^2 + 1)]

We would then put the numerator into order of decreasing exponents, so arrive at an easily-recognized standard form:

[ x^4 + x^3 - x^2 + x - 1 ] / [(x^3)(x^2 + 1)]

STUDENT QUESTION

Why doesn’t the x^3 cancel out???

x^4+x^3-x^2+x-1/x^3(x^2+1)

In the first place you need to correct your signs of grouping. You numerator and denominator need to be grouped:

(x^4+x^3-x^2+x-1) / ( x^3(x^2+1) ).

The expression as you give it would is displayed according to the order of operations as follows:

whereas the expression (x^4+x^3-x^2+x-1) / ( x^3(x^2+1) ) is displayed as

,
which is what you intend.

Now to answer your question:

x^3 is a factor of the denominator, but it's not a factor of the numerator. For example x^3 won't divide into -1.
[ x^4 + x^3 - x^2 + x - 1 ] / [(x^3)(x^2 + 1)] means
x^4 / [(x^3)(x^2 + 1)] + x^3 / [(x^3)(x^2 + 1)] - x^2 / [(x^3)(x^2 + 1)] + x / [(x^3)(x^2 + 1)] - 1 / [(x^3)(x^2 + 1)],
by the distributive law. This would permit some simplification:
x^3 would divide out of the first term x^4 / [(x^3)(x^2 + 1)], leaving x / (x^2 + 1)
x^3 would also divide out of the second term x^3 / [(x^3)(x^2 + 1)], leaving 1 / (x^2 + 1).
However x^3 would not divide out of the remaining three terms.

STUDENT QUESTION

I had some trouble with this problem. I looked at your answer and I just could not figure out how you come up with the
single x in the step x^3+x-x^2-1+x^4 / (x^3)x^2+1
Starting with
(x-1)(x^2+1)/[ (x^3)(x^2+1)] + x^4/ [(x^3)(x^2+1)]
we multiply (x-1)(x^2+1), which is the numerator of the first fraction, to get x(x^2+1) - 1(x^2+1) = x^3 + x - x^2 - 1, which we write in the order x^3 - x^2 + x - 1.
Thus our expression becomes
(x^3 - x^2 + x - 1) [ (x^3)(x^2+1)] + x^4/ [(x^3)(x^2+1)] .
We continue by adding the numerators of the two fractions, which now have a common denominator, as indicated in the given solution.

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Self-critique (if necessary):
I got lost with this one and was glad to read the questions and answers I think I have a better idea of what I am doing now.

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Self-critique rating #$&*??

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Question:

* 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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Your solution:

x-3,x^3+3x,x^3-9x
x-3,x(x^2+3), x(x^2-9)
x-3, x(x^2+3), x(x-3)(x+3)
the LCM is x(x-3)(x+3)(x^2+3)

confidence rating #$&* 2
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

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:

• LCM = x(x-3)(x+3)(x^2+3)
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Self-critique (if necessary):

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Self-critique rating #$&* OK

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Question:

* R.7.64 (was R.7.60). Show how you found and simplified the difference 3x / (x-1) - (x - 4) / (x^2 - 2x + 1).

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Your solution:

3x/(x-1)-(x-4)/(x^2-2x+1)=3x/(x-1)-x-4/(x-1)(x-1)
3x(x-1)/(x-1)(x-1)-(x-4)(x-1)/(x-1)(x-1)=3x^2-3x-x-x+4/(x-1)^2=3x^2-4x+4)/(x-1)^2
confidence rating #$&* 2
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:
* * ** Starting with

3x / (x-1) - (x-4) / (x^2 - 2x +1)

we factor the denominator of the second term to obtain (x - 1)^2. Since the first denominator (x - 1) is already a factor of the second, our common denominator is (x - 1)^2.

To express the given expression in terms of the common denominator we then 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.
STUDENT QUESTION:

I don’t see where the 3x(x-1) comes in. When I tried working it on paper, I started with
3x / (x -1) – (x-4) / (x^2 - 2x + 1) and reduced to
3x / (x-1) – (x-4) / (x-1)^2
But then I tried to solve it by
(3x) – (x-4) / x-1 which isn’t right
INSTRUCTOR RESPONSE:

3x / (x-1) – (x-4) / (x-1)^2 consists of two fractions with different denominators. You can't do the subtraction until you have a common denominator.
The denominator of the first is (x - 1); the denominator of the second is (x - 1)^2. If you multiply the first denominator by (x - 1) you get the second denominator.
So to express the first fraction with the same denominator as the second you multiply its numerator and denominator both by ( x - 1 ). You get
(3x / (x - 1) ) * (x - 1) / (x - 1) = 3x ( x - 1) / (x - 1)^2.
So the subtraction becomes
3x ( x - 1) / ( x - 1) ^ 2 - (x - 4) / (x - 1)^2.
The denominator of the combined fraction is (x - 1)^2, while the numerator is (3 x * (x - 1) - (x - 4) ); this simplifies to 3 x^2 - 3 x - x + 4, whic gives us
(3x^2-3x-x-4) / (x-1)^2 = (3x^2 - 4x - 4) / (x-1)^2.

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Self-critique (if necessary):

In the solution the +4 became a -4 is that a typo or am I missing a step.

-(x-4) = -x - (-4) = -x + 4.

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Self-critique rating #$&* Fair

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 grab onto 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).

&#This looks good. See my notes. Let me know if you have any questions. &#

#$&*