Assignment R7

course MTH 158

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I don't show any work associated with your email address since the work posted on 6/11. If you have completed assignments that aren't posted, they should still be in your SEND files and you should resubmit them, along with a note giving me an idea of when they were originally submitted so I can check and see if anything else is missing.

007. `query 7

College Algebra

06-16-2008

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18:41:17

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 -->

Factor the numerator...

x^2 + 4x + 4 = (x + 2)(x + 2)

Factor the denominator

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

(x + 2)(x + 2)(x + 2)(x - 2)

Now together they make...

(x + 2)(x + 2) / (x + 2)(x + 2)(x + 2)(x - 2)

Which cancels out to...

1 / (x + 2)(x - 2)

confidence assessment: 2

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18:41:54

** 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 -->

self critique assessment: 2

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18:47:33

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 -->

First turn the denominator into its reciprocal...

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

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

now factor the denominator of the second fraction...

x^2 - 4x + 4 = (x - 2)(x + 2)

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

Now cancel out what can be..

3 / x + 2

confidence assessment: 2

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18:48:19

** [ ( 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 -->

self critique assessment: 2

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18:50:12

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 -->

Since the denominators are the same, you can put the two fractions together

2x - 5 + x + 4 / 3x + 2 = 3x - 1 / 3x + 2

confidence assessment: 2

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18:50:23

** 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 -->

self critique assessment: 3

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19:03:26

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 -->

Multiply both of them by each other's denominator...

[(x - 1) / x^3] * [x^2 + 1 / x^2 + 1], and

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

now join since they have the same denominator now..

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

This cancels out to give us

x (x - 1)

confidence assessment: 2

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19:04:59

** 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 -->

not understood

self critique assessment: 1

[(x - 1) / x^3] * [x^2 + 1 / x^2 + 1], and
[x / (x^2 + 1)] * [x^3 / x^3] is correct. But when you multiply these out and add numerators you don't get
(x - 1)(x + 1)(x + 1)( x )^4 / (x^3)(x + 1)(x + 1).

Can you explain how you got that numerator?

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19:18:01

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 -->

x - 3 = x - 3

x^3 + 3x = (x)(x + 1)(x + 3)

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

All together...

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

confidence assessment: 2

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19:18:49

** 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 -->

Why did I not need to factor x(x^2 + 3)?

self critique assessment: 1

x^2 + 3 doesn't factor. There is no way to express this in the form (x + a) ( x + b), with a and b both real numbers.

It could be factored using complex numbers, but that doesn't help here.

Note that (x+1)(x+3) = x^2 + 4 x + 3, not x^2 + 4.

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19:34:12
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 -->
First factor x^2 - 2x + 1 = (x - 1)(x - 1)
which gives us
[3x / x-1] - [x - 4 / (x-1)(x-1)]
multiply 3x / x-1 times the only LCD lacking and join numerators
3x (x-1) -x +4 / (x-1)(x-1) and cancel out to
2x + 4 / x - 1
confidence assessment: 2

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19:35:46
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 -->
not understood
self critique assessment: 1

You aren't using signs of grouping to indicate numerators and denominators; you need to do this.

However I can tell on this problem what you intend. You write

3x (x-1) -x +4 / (x-1)(x-1), by which you mean

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

The numerator (3x (x-1) - (x +4) ) is correct.

However when you simplify this you don't get 2x + 4.

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

Thus the final solution is (3x^2 - 4x - 4) / (x-1)^2.

The numerator can be factored into (3x + 2) ( x - 2) so the factored form of the solution would be

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

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19:37:44

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 -->

understood

self critique assessment: 2

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You're doing well, but there appear to be one or two things you're not seeing. See my notes and see if you can identify those things.

You need to be specific in your self-critiques.

&#

You need to tell me exactly what you do and do not understand about that given solution. You need to be specific in a self-critique.

If you have questions, discussions, comments or revised solutions you are welcome to insert them into a copy of this document, mark you insertions with &&&&, and submit.

Assignment R7

[(x - 1) / x^3] * [x^2 + 1 / x^2 + 1], and [x / (x^2 + 1)] * [x^3 / x^3] is correct. But when you multiply these out and add numerators you don't get (x - 1)(x + 1)(x + 1)( x )^4 / (x^3)(x + 1)(x + 1). Can you explain how you got that numerator?

x^2 + 3 doesn't factor. There is no way to express this in the form (x + a) ( x + b), with a and b both real numbers. It could be factored using complex numbers, but that doesn't help here. Note that (x+1)(x+3) = x^2 + 4 x + 3, not x^2 + 4.

You're doing well, but there appear to be one or two things you're not seeing. See my notes and see if you can identify those things. You need to be specific in your self-critiques.

&#

You need to tell me exactly what you do and do not understand about that given solution. You need to be specific in a self-critique.

If you have questions, discussions, comments or revised solutions you are welcome to insert them into a copy of this document, mark you insertions with &&&&, and submit.

[(x - 1) / x^3] * [x^2 + 1 / x^2 + 1], and
[x / (x^2 + 1)] * [x^3 / x^3] is correct. But when you multiply these out and add numerators you don't get
(x - 1)(x + 1)(x + 1)( x )^4 / (x^3)(x + 1)(x + 1).

Can you explain how you got that numerator?

x^2 + 3 doesn't factor. There is no way to express this in the form (x + a) ( x + b), with a and b both real numbers.

It could be factored using complex numbers, but that doesn't help here.

Note that (x+1)(x+3) = x^2 + 4 x + 3, not x^2 + 4.

You're doing well, but there appear to be one or two things you're not seeing. See my notes and see if you can identify those things.

You need to be specific in your self-critiques.