help! z-2/x^2+4z+4+5/4z^2-16-z+2/2x^2-4z

susan pearson

New member
Joined
Aug 16, 2006
Messages
9
This problem has me stumped. I need to find the LCD and I can't seem to go any farther than what I've listed.

z-2 + 5 - z + 2
------------- ------------- -------------
x^2 + 4z + 4 4z^2 - 16 2x^2 - 4z


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

Where am I going wrong or is this really as far as I can factor these and I need to find the LCD from these?
 

galactus

Super Moderator
Staff member
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Sep 28, 2005
Messages
7,216
It's rather hard to read your post, but is this it:

\(\displaystyle \L\\\frac{z-2}{x^{2}+2z+4}+\frac{5-z}{4z^{2}-16}+\frac{2}{2x^{2}-4z}\)
 

Denis

Senior Member
Joined
Feb 17, 2004
Messages
1,465
Re: help!

Susan, I suggest you experiment with the "code" button; to get something like this:
Code:
     z-2               5                z + 2
------------   +   ---------    -     --------- 
x^2 + 4z + 4       4z^2 - 16          2x^2 - 4z
I have a feeling the 2 x's that appear should be z's: please check...
 

susan pearson

New member
Joined
Aug 16, 2006
Messages
9
YOU'RE RIGHT

galactus said:
It's rather hard to read your post, but is this it:

\(\displaystyle \L\\\frac{z-2}{x^{2}+2z+4}+\frac{5-z}{4z^{2}-16}+\frac{2}{2x^{2}-4z}\)
You're right: I don't what I'm doing! you interpreted my scribble with everything correct but the 5 - z. The z should actually be z + 2 on the last fraction.
 

Denis

Senior Member
Joined
Feb 17, 2004
Messages
1,465
Re: help!

So, if z is the only variable, your expression is:

(z-2) / (z^2+4z+4) + 5 / (4z^2-16) - (z+2) / (2z^2-4z)

factoring a bit:
(z-2) / (z+2)(z+2) + 5 / 4(z-2)(z+2) - (z+2) / 2z(z-2)

to keep it manageable, let x = z-2 and y = z+2; then:
x / y^2 + 5 / 4xy - y / 2zx

applying the lcd 8zx^2y^3 leads to:
(4zx^2 + 5zy - 2y^3) / (4zxy^2)

substituting back in leads to:
(2z^3 - 23z^2 + 2z - 16) / [4z(z^3 + 2z^2 - 4z - 8)]

That checks out ok; I tested it.
 
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