perform the indicated operation (rtnl expressions)

susan pearson

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Aug 16, 2006
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Once again, I've gotten lost. I need to simplify:

. . .(2x<sup>2</sup> - 2x)/(x<sup>3</sup> - 4x<sup>2</sup> - x + 4) - (16)/(2x<sup>3</sup> - 6x<sup>2</sup> - 8x)

I realize I need to find the LCD (the Least Common Denominator) and, before doing this, I need to factor out my denominators. I came up with:

. . .[2x(x - 1)] / [(x<sup>2</sup> - 1)(x - 4)] . . .-. . . 4<sup>2</sup> / [2x(x - 4)(x + 1)]

But after this, I got lost.
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Edited by stapel -- Reason for edit: clarifying formatting
 
Re: perform the indicated operation

Hello, Susan!


\(\displaystyle \L\frac{2x^2\,-\,2x}{x^3\,-\,4x^2\,-\,x\,+\,4}\:-\:\frac{16}{2x^3\,-\,6x^2\,-\,8x}\)

i realize i need to find the LCD
\(\displaystyle \;\;\)and before doing this i need to factor out my denominators.

i came up with: \(\displaystyle \L\,\frac{2x(x\,-\,1)}{(x^2\,-\,1)(x\,-\.4)}\:-\:\frac{16}{2x(x\,-\,4)(x\,+\,1)}\;\;\) . . . good!

The first denominator factors again . . .

Then: \(\displaystyle \L\,\frac{2x(\sout{x\,-\,1})}{(\sout{x\,-\,1})(x\,+\,1)(x\,-\,4)}\:-\:\frac{\not{16}^8}{\not{2}x(x\,-\,4)(x\,+\,1)} \;= \;\frac{2x}{(x\,-\,4)(x\,+\,1)}\:-\:\frac{8}{x(x\,-\,4)(x\,+\,1)}\)


The LCD is: \(\displaystyle \,x(x\,-\,4)(x\,+\,1)\)

\(\displaystyle \L\;\;\frac{x}{x}\cdot\frac{2x}{(x\,-\,4)(x\,+\,1)} \:-\:\frac{8}{x(x\,-\,4)(x\,+\,1)} \;= \;\frac{2x^2}{x(x\,-\,4)(x\,+\,1)}\:-\:\frac{8}{x(x\,-\,4)(x\,+\,1)}\)

\(\displaystyle \L\;\;= \;\frac{2x^2\,-\,8}{x(x\,-\,4)(x\,+\,1)} \;= \;\frac{2(x^2\,-\,4)}{x(x\,-\,4)(x\,+\,1)}\)

Answer: \(\displaystyle \L\,\frac{2(x\,-\,2)(x\,+\,2)}{x(x\,-\,4)(x\,+\,1)}\)

 
Re: perform the indicated operation

susan pearson said:
2x^2-2x/x^3-4x^2-x+4 - 16/2x^3-6x^2-8x =
susan, use BRACKETS (necessary here); and spaces improve readability:

(2x^2 - 2x) / (x^3 - 4x^2 - x + 4) - 16 / (2x^3 - 6x^2 - 8x)


20 / (2 + 3) = 20 / 5 = 4
20 / 2 + 3 = 10 + 3 = 13 ...see why?
 
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