I have this problem:
$$9x^3 - 18x^2 - 4x + 8 = 0$$
However, I'm not sure how to find the values of $x$. I moved the 8 over and factor out an $x$, but the trinomial it created can't be factored. Could someone enlighten me?
Factor the function.
$x=2$ or $2/3$ or $-2/3$
The rational root theorem is your friend. It says all rational roots have numerators that are factors of the constant term and denominators that are factors of the leading term. Here the numerators can be $\pm 1, \pm 2, \pm 4, \pm 8$ and the denominators can be $1,3,9$. Not too many to try. When you find one, you can divide out that root to get a quadratic. It doesn't always work (as shown with the example with -8 for a constant term) but does sometimes, often in homework.
By trial and error method I'm solving this
Divide 9x3−18x2−4x+8 by x-2 reminder is 0 and quotient is 9x2-4x
x-2 = 0 and 9x2-4x = 0
Factorizing 9x2-4x = 0, we get (3x-2)(3x+2) = 0
ie. x = 2, x = 2/3, x = -2/3 are the solutions
(Sorry I'm new to this group and don't know how to write Square of a variable, But answer must be this)