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- anonymous

simplify completely x^2 + 4x - 5 / 5x^2 -8x + 3 ( multiplied ) 20 x - 12 / x^2 - 6x - 55

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- anonymous

simplify completely x^2 + 4x - 5 / 5x^2 -8x + 3 ( multiplied ) 20 x - 12 / x^2 - 6x - 55

- chestercat

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- anonymous

\[\Large \frac{x^{2}+4x-5}{5x^{2}-8x+3}*\frac{20x-12}{x^{2}-6x-55}\]
Sidenote: This is what I put into Equation: Large frac{x^{2}+4x-5}{5x^{2}-8x+3}*frac{20x-12}{x^{2}-6x-55}

- anonymous

Ok, so same thing as before: Break down each section of the equation and factor them out.

- anonymous

im slowly trying to figure this out . Like I said before Idk Why but im better when There is a variable in there , from when its like this ...

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- anonymous

I dont think Im Doing this right.

- anonymous

diagnally add or multiply/divide?

- anonymous

There is a variable there, X.
Step 1: Factor the polynomials. That will help you simplify the equation.
\[\Large x^{2}+4x-5\]
What are the two factors of this?
(x- ? )(x+ ? )

- anonymous

You COULD multiply it all out first, but that would just get confusing quick. Always better to see if there's any factors you can cancel out first before it gets confusing.

- anonymous

how about how to simplify the first denominator

- anonymous

cause I got for the first numerator (x+5)(x-1) and then Idk how to do the first denominator becasue 8 and 5 dont have anything they can both be divided from. the second numerator 4(x-3) and the second denominator (x-11)(x+5) .. right so far?

- anonymous

Second Numerator would be 4(5x-3).

- anonymous

Crap I need to pay more attention that was.. just me not paying attention -_-

- anonymous

First Denominator?

- anonymous

I need to hurry got to get off in 10 mins -_- one more question on math

- anonymous

Quadratic equation for that gets you:
(x-1)(5x-3)
Quadratic equation:
\[ax^{2}+bx+c\]
\[\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\]

- anonymous

\[4 \div x-11\] is this correct?

- anonymous

|dw:1341907059174:dw|

- anonymous

\[\Large \frac{(x+5)(x-1)}{(x-1)(5x-3)}*\frac{4(5x-3)}{(x-11)(x+5)} = \frac{1}{1}*\frac{4}{(x-11)}\]
Yep.

- anonymous

;D thank you. so much ! and now I know to use the quadratic formula if I cant simplify like that. Thank you :D

- anonymous

No problem at all. Have you covered the quadratic equation yet?

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