Solve:
$\displaystyle x^3-5x^2+8x-4 = 0$
$\displaystyle x^3-5x^2+8x-4 > 0$
by the remainder theorem, we see that $\displaystyle x = 1$ is a root. thus by the factor theorem, $\displaystyle (x - 1)$ is a factor.
dividing our cubic by $\displaystyle x - 1$, we see that:
$\displaystyle x^2 - 5x^2 + 8x - 4 = (x - 1)(x^2 - 4x + 4) = (x - 1)(x - 2)^2 = 0$
thus, we have $\displaystyle x = 1$ or $\displaystyle x = 2$
i leave this part to you. we found our roots to be $\displaystyle x = 1$ and $\displaystyle x = 2$, so we in fact can split up the real line into the intervals $\displaystyle (-\infty, 1)$, $\displaystyle (1,2)$, and $\displaystyle (2, \infty)$$\displaystyle x^3-5x^2+8x-4 > 0$
on which of these intervals is the cubic positive?
By inspection $\displaystyle (x-1)$ is a factor because $\displaystyle x = 1$ is a zero of the expression
then by use synthetic division
$\displaystyle x^3-5x^2+8x-4 = (x-1)(ax^2+bx+c)$
Compare coefficients of the highest order term to get a = 1 and lowest order to get c = 4 first because it is easy
$\displaystyle x^3-5x^2+8x-4 = (x-1)(x^2+bx+4)$ by comparing the coefficient of the term with x^2 or otherwise you should easily get b = -4
$\displaystyle x^3-5x^2+8x-4 = (x-1)(x^2-4x+4)$
$\displaystyle \Rightarrow x^3-5x^2+8x-4 = (x-1)(x-2)^2$
for the inequality you must sketch a graph.
You know be able to tell that for large positive values of x the function will be positive, and for large negative values of x the function will be negative.
You should also know the the sign of the function can only change at a root.
you should get a graph like this. the image below.
Remember the inequality is greater than zero not greater than or equal to, don't be lazy and just write x >1