Assmt 7

course Mth 158

I had considerable difficulty with #34. My answer came up with a zero denominator. Can you help me solve this? [( 9x^2 + 3x - 2)/(12x^2 + 5x - 2)]/[(9x^2 - 6x + 1)/8x^2 - 10x -3)].

Send me a copy of this question along with your factoring of each trinomial, and the remaining steps of your solution so I can see what you have done, and what you did and do not understand about the situation. The denominator would be a product of two trinomials, and will not be zero.

ùïÌþÆíløûìäÙ¸_}Ÿ»yèí©assignment #007

007. `query 7

College Algebra

09-05-2008

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12:53:51

Query 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 -->

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

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

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13:00:42

** 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 -->

I don't understand how the denominator was factored. I thought x^2-4 was the difference of 2 squares.

The given solution is for (x^2 + 4 x + 4) / (x^4 - 16).

Your solution is correct for (x^2 + 4 x + 4) / (x^2 - 4).

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13:04:14

Query 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 -->

x-2/4x * 12x/x^2 - 4x + 4

x-2/4x * 4x(3)/(x -2)(x - 2)=

3/(x-2)

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13:04:28

** [ ( 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 -->

ok

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13:07:18

Query 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 -->

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

3x - 1/3x + 2

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13:07:31

** 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 -->

ok

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13:13:38

Query 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 -->

First I found my common denominator of x^3(x^2 + 1).

Then I multiplied my numerators like so:

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

=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)

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13:14:00

** 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 -->

ok

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13:16:49

Query 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 -->

I factored each like so:

x - 3

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

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

My answer for LCM was x(x - 3)(x + 3)

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13:21:34

** 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 -->

My book has x^2 + 3x instead of x^3 + 3x for the second one. I think this would make my answer correct.

Your answer is correct for the expressions you used.

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13:27:16

Query 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 I found my LCM of (x - 1)^2.

Next, I multiplied the numerator like so:

3x(x - 1). The second numerator of x - 4 didn't need to be multiplied.

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

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

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13:28:03

** 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 -->

ok

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&#Your work looks very good. Let me know if you have any questions. &#