Who can answer this algebra problem

find the domain. x^2-5x+6/3x^2-2x-8
how do I solve this!! Looking for best answer!

Any number divided by zero gives you an undefined value, so this problem's domain cannot include the roots of the bottom polynomial.
To find the roots you use the equation (-b+(b^2-4ac)^.5)/2a
Plug in the values from 3x^2-2x-8;
(2+(4-4*3*-8)^.5)/2*3 -> (2+(100)^.5)/6 -> Since sqrt(100)=10&-10 you solve for both: (2+10)/6=2 and (2-10)/6= -4/3
Since the equation has an undefined value for the values of x they are not a part of it's domain, but all other numbers would give an answer so the domain is all real numbers except 2 and -4/3.

This equation will have vertical asymptotes anywhere that the denominator is equal to zero, so the domain of this will be everywhere except for where the denominator is 0. To find the zeroes of the denominator, you have to factor it.

3x^2-2x-8
3x^2-6x+4x-8
3x(x-2)+4(x-2)
(3x+4)(x-2)

Now, you find the zeroes of those factors to find the zeroes of the denominator, which are -4/3 and 2

Therefore, the domain is all real numbers except for -4/3 and 2, or {xeR | x not equal -4/3, 2}

f(x) = x^2-5x+6/3x^2-2x-8
= (x-2)(x-3) / (3x +4)(x -2)

= (x-3) / (3x +4)

Domain is all real numbers except x = - 4/3