SOLUTION: Please help me solve the equation (x^2-3x)/(4x^2-3x)*(4x^2-16) I have tried to follow the instructions in my book by factoring it to: x(x-3)/4x(x-2)*(2x+4)(2x-4) can you please

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Question 29099: Please help me solve the equation (x^2-3x)/(4x^2-3x)*(4x^2-16)
I have tried to follow the instructions in my book by factoring it to:
x(x-3)/4x(x-2)*(2x+4)(2x-4)
can you please show the problem step by step since i have others like it that i would like to know how to solve
Thank You,
Alyssa
Found 2 solutions by sdmmadam@yahoo.com, ms.jennifer:
Answer by sdmmadam@yahoo.com(530)   (Show Source): You can put this solution on YOUR website!
Please help me solve the equation (x^2-3x)/(4x^2-3x)*(4x^2-16)
I have tried to follow the instructions in my book by factoring it to:
x(x-3)/4x(x-2)*(2x+4)(2x-4)
can you please show the problem step by step since i have others like it that i would like to know how to solve
Thank You,
Alyssa
If the problem is [(x^2-3x)/(4x^2-8x)]*(4x^2-16)
=[x(x-3)/4x(x-2)]X4(x^2-4)
=[(x-3)/4(x-2)]X4(x+2)(x-2) (cancelling out x)
=(x-3)(x+2)
Answer by ms.jennifer(4)   (Show Source): You can put this solution on YOUR website!
First you want to factor the top of the divison part (x^2-3x) to x(x-3) and the bottom of the divison part (4x^2-3x) to 4x(x-3).
So you have x(x-3)/4x(x-3)* (4x^2-16).
so the x-3 cancel and you are left with x/4x* (4x^2-16)
Next you will set 4x^2-16 over one to make it at fraction
The mulitply x/4x with 4x^2-16/1
So you get x(4x^2-16)/4x
Simiply 4x^3-16x/4x
there is a common Factor of 4x
4x(x^2-4)/4x
the 4xs cancel so you are left with x^2-4
which factors to be (x-2)(x+2)
so the roots are {2,-2}
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