## Rational Expression Multiplications Negatives Confusion

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
dark808bb
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### Rational Expression Multiplications Negatives Confusion

I'm working out the multiplication of rational expressions and it is all clicking logically like it should except for these negatives.
$\left(\frac{8x-16}{44x^2-99x}\right)*\left(\frac{891x-396x^2}{2-x}\right)$

eventually I come to $\left(\frac{8(x-2)}{11x(4x-9)}\right)*\left(\frac{99x(9-4x)}{2-x}\right)$ and then ----> $\left(\frac{8(x-2)}{11x(4x-9)}\right)*\left(\frac{9*11x(9-4x)}{2-x}\right)$
This is as far as I understood.
The rest of the solution was to bring out a negative of the factors on the right fraction in order to be able to cancel out the $(x-2)$ and $(4x-9)$

Although it seemed difficult to understand how they slinged the negatives onto everything but the $11x$
While writing up this post I thought "since they factored the 99x into 9 and 11x first and then brought out the negative of 9 and (9-4x) eleven x was able to ligitimately escape becoming the opposite"

Before this thought, I always assumed they factored out -9 from 99x which would also give a -11, but it seems they factored out 9 from 99x and then took the opposite of 9(9-4x) and left the 11x alone.

Am I correct in thinking this? Can somebody give me some direction on this?

its confusing to me how this comes out to 72 instead of -72. Hopefully this post wasn't too confusing

Ahh. EDIT. They probably factored 99x, canceled out 11x then brought the opposite of that fraction.... Mathzone and my text book both sometimes... actually most the time are very brief in the explanation of some of the more confusing operations that they perform

FWT
Posts: 153
Joined: Sat Feb 28, 2009 8:53 pm

### Re: Rational Expression Multiplications Negatives Confusion

I'm working out the multiplication of rational expressions and it is all clicking logically like it should except for these negatives.
$\left(\frac{8x-16}{44x^2-99x}\right)*\left(\frac{891x-396x^2}{2-x}\right)$

eventually I come to $\left(\frac{8(x-2)}{11x(4x-9)}\right)*\left(\frac{99x(9-4x)}{2-x}\right)$ and then ----> $\left(\frac{8(x-2)}{11x(4x-9)}\right)*\left(\frac{9*11x(9-4x)}{2-x}\right)$
This is as far as I understood.
The rest of the solution was to bring out a negative of the factors on the right fraction in order to be able to cancel out the $(x-2)$ and $(4x-9)$
Look at the example in about the middle of the page here. It shows how to do the stuff with the negatives: if you reverse subtraction then a negative goes on in front. For the probelm, its like this:

9 - 4x = -1(-9) - 1(+4x) = -1(-9 + 4x) = -1(4x - 9) = -(4x - 9)
2 - x = -(-2) - (+x) = -1(-2 + x) = -1(x - 2) = 1(x - 2)

Then you can cancel.

btw: you don't have to show all that. Just turn it around and put the negative in front, like "2 - x = -(x - 2)".

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