I was wondering if anyone could see if I am doing this correctly?

Question

The problem is...
(X-2)/(X^3-8)

and it needed to be simplified so I...

(x-2)/(x^3-2^3)... I factored
(x-2)/x^2-4(x-2)... pulled out the x^2-4
1/x^2-4... cancelled the "x-2"s
1/(x-2)(x+2)... factored again

If x doesn't equal 2, or -2.

So ... Is that right, if it isn't I kinda wanna know what I am doing wrong so I can finish my assignment.

Thanks in advance. :)

Answer 1

x³ - 8 can be factored as follows according to the difference of cubes** formula:

(x-2)(x² +2x + 4)

So your fraction is

(x-2) / [(x-2)(x² +2x + 4)]
= 1 / (x² +2x + 4)

**The factorization formula for a sum or difference of two cubes:

a³ - b³ = (a-b)(a²+ab + b²)
a³ + b³ = (a+b)(a²-ab + b²)

The trinomials in these formula are always prime, so don't even waste your time factoring.

Answer 2

(x^3 - 8) is not the same as (x - 2)^3.

you will get x^3 -6x^2 +12x -8.

are you trying to find the domain?

then set the denominator to zero. and you will get 2.
so for the domain...(all numbers on the x-axis, except 2)
because then you would have a zero...and you can't have a zero in the denominator.

is that what you are looking for?

if not...sorry.

Answer 3

every thing is right except you should have pulled out x^2 + 4 to have
(x-2)/x^2+4(x-2)

Answer 4

I don't think the 3rd step is right..... can you just pull x^2 out??? i'm not sure...... sorry, but I would worry bout the third step right now..

Answer 5

Kathleen K has given you the correct answer. None of the other answers above mine know what they're talking about, so please ignore them.

Answer 6

Yeah that looks right to me.