Assignment 7

course mth 158

6/28

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

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

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

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

First you factor the numerator by factoring a trinomial, then you break down the denomenator by finding

the square. Then you reduce.

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):

OK

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Self-critique rating #$&*OK

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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 - 4 x + 4) / (12 x) ]

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

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

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

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

[ 12x * (x - 2)] / [4x * (x -2) (x -2)]

(3x) / (x-2)

First you factor, then combine numerators and denomenators, then divide 12x by 4x, then reduce.

confidence rating #$&*3

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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):

OK

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Self-critique rating #$&*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)

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

(3x - 1) + (3x + 2)

- 1 / 2

Since the denomenators are the same, you simply add the numerators, then reduce.

confidence rating #$&*3

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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):

??? Could you not further reduce by eliminating the 3x????

3x is not a factor of the numerator or the denominator, so you couldn't do that.

If you pick just about any nonzero value for x and substitute it into the expression (3x - 1) / (3x + 2), you won't get 1/2.

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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^2 + 1)]/ [(x^3) * (x^2 +1)]} + [(x) * (x^3)] / [(x^3)(x^2+1)]

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

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

confidence rating #$&*2

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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):

OK

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Self-critique rating #$&*OK

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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

x^3 + 3x = x(x^2 + 3)

x^3 + 9x = x(x^2 + 9) = x (x + 3) (x - 3)

So, x (x^2 + 9) (x + 3) (x - 3) is the LCM, since all of the solutions were present.

confidence rating #$&*1

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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):

OK

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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)^2

{[(3x)*(x-1^2)] / [(x-1) * (x-1)^2]} - {[(x-4) * (x-1)] / [(x-1)^2 * x-1)]}

{[(3x)*(x-1)^2] / (x-1)^3} - {[(x-4) * (x-1)] / (x-1)^3}

(3x^2 - 3x - 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):

OK

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Self-critique rating #$&*OK

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)."

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

#$&*