R7

#$&*

course 158

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 + 4 x + 4) / (x^4 - 16)

(x+2)(x+2)

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

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

confidence rating #$&*: 1

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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 thought the 4 should reduce more but when I tried it that way it didn’t make sense either

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Self-critique Rating:1

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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

(x+2)(x-2) / 12x

12x(x-2) / 4x(x-2)(x-2)

3 / (x-2)

confidence rating #$&*: 2

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

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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:3

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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+x=3x

-5+4=-1

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

confidence rating #$&*: 2

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

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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):ok

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Self-critique Rating:3

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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-1)(x^2+1)

X^3*x^2+1=x^3(x^2+1)

X*x^3=x^4

(x^2 + 1)*x^3=x^3(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’m lost after I multiple to get the same denominator
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Self-critique Rating:1

@&
The idea here is that you need to have a common denominator in order to add two fractions.

For example if you want to find 2/3 + 3/4, you would identify your denominators as 3 and 4. The smallest number that is a common multiple of 3 and 4 is 12, so 12 will be your common denominator.

To express 2/3 with denominator 12, the only valid mathematical operation that accomplishes this is to multiply by 4 / 4, which is the same as multiplying by 1 and therefore doesn't change the value of the fraction. So

2 / 3 * 4 / 4 = 2 * 4 / (3 * 4) = 8 / 12

has the same value as 2/3, while

3/4 * 3 / 3 = 3 * 3 / (4 * 3) = 9 / 12

has the same value as 3/4.

Now we can write our sum

2/3 + 3/4

as

8/12 + 9/12.

In this form we can factor out 1/12 to get

1/12 * (8 + 9) = 1/12 * 17 = 17/12.

This shows that we can simply add the numerators to get

8/12 + 9/12 = 17/12.

In the current example our fractions are (x - 1) / x^3 and x / (x^2 + 1).

The simplest expression that is a multiple of both denominators is

x^3 * (x^2 + 1)

so this will be our common denominator.

To express (x - 1) / x^3 in terms of the common denominator, the simplest mathematical operation is to multiply by (x^2 + 1) / (x^2 + 1):

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

To express x / (x^2 + 1) in terms of the common denominator we multiply by x^3 / x^3:

x / (x^2 + 1) * x^3 / x^3 = x * x^3 / (x^3 * (x^2 + 1)) = x^4 / (x^3 * (x^2 + 1))
The new fractions are identical in value to the old, but they now have identical denominators. So

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

The numerator can be expanded to give us

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

*@

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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 and x^3 - 9x

x-3 = x-3

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

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

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Self-critique Rating:3

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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)

confidence rating #$&*: 1

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

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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 I understand you have to have a common denominator but I don’t understand when you responded to the students question that the denominator for the first fraction is x-1 how come it isn’t (x-1) -(x-4). What happened to the -(x-4)???

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Self-critique Rating:2

@&
Order of operations. (x - 1) is preceded by / and followed by -. The / takes precedence, so the first term of the expression is 3x / (x - 1).

(x - 4) is the numerator of the second term.

You might still need a little convincing, and it would probably reinforce the idea were you to take a minute and enter the expression into Wolfram Alpha (just search 'Wolfram Alpha' and you'll find it very quickly).
*@

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).

Self-critique (if necessary):

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Self-critique rating:

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).

Self-critique (if necessary):

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Self-critique rating:

#*&!

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).

Self-critique (if necessary):

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Self-critique rating:

#*&!#*&!

R7

&#Good responses. See my notes and let me know if you have questions. &#