JKAsiege

Turky wrote:15/5 is 3, not 5, that's where I went wrong. This tends to happen when simple arithmetic is the least of your problems.

amen.

I could really yews some help on this particular problem; I'm stumped.
Solving for X...

Multiply both sides by (2x^2-8x) to eliminate the denominator.
Combine like terms by adding 2x^2 and -8x to both sides (algebraic addition, where adding a negative is the same as subtracting a positive).
Factor out the common terms leaving 2x*(x^2-16)=0.
You're dealing with a trinomial which will necessarily have three solutions for x; either one real and two imaginary, or three real.
By that theorem I forget the name of, you set the individual terms equal to 0 and solve. The x^2-16=0 I hope you recognize is difference of squares.
You should be able to get the answer from this point.
I left out a lot intentionally; from 5 years tutoring math (mostly algebra and stats, some calculus), I know that doing all the work for someone doesn't really teach them anything. If you want more assistance, ask me some time when I'm not at work...

That actually really helped a lot, I got 0,4,-4 for my answers.. would those be correct?

Also; I have one last problem that I'm having a hard time with and then I'm finally done. I drew out the picture in paint so you could see the problem.

Multiply both sides by (2x^2-8x) to eliminate the denominator.

2x^3 -2x^2 -24x= -2x^2 +8x
2x^3 -32x=0
x( 2x^2 -32)=0
x=0 V 2x^2 -32=0
2x^2=32
x^2=16
x=4 V x= -4


There we go, the full procedure.

dude trying to redeem himself with simple maths

v=a * b * c

a=x+4
b= this is the one we're looking for
c= x+1

a* c * b = v ( commutation when multiplying )

Since you have ( x + 1 ) in the denominator, you want to multiply both sides with it.

That will give you ( x + 4 ) * ( x + 1 )^2 * b = x^3 + 7x^2 +14x + 8

From there, put the (x+1) to second power, and then multiply it by (x+4), while rewriting the right side.

You should get ( x^3 + 6x^2 + 9x + 4 ) * b = x^3 + 7x^2 + 14x +8

From there,

x^3+7x^2+14x+8
b = ----------------------
x^3 +6x^2+ 9x + 4


A simple polynomial division and you're done.

Turky wrote:dude trying to redeem himself with simple maths

I know as much math as you do, it's just that I don't speak math in english, so when you say fractions, I have to check a dictionary and find out that is my 5th grade elem math

Helaku wrote:

Turky wrote:dude trying to redeem himself with simple maths

I know as much math as you do, it's just that I don't speak math in english, so when you say fractions, I have to check a dictionary and find out that is my 5th grade elem math

Because fractions are only used in 5th grade. Much like numbers are only used in kindergarden.

Helaku wrote:

Turky wrote:dude trying to redeem himself with simple maths

I know as much math as you do

As much as I do?

I hold a B.S. in Physics with a minor in math... not to even mention linear and abstract algebra, that's a whole different animal from classical algebra and calculus.