Question

How do you begin to solve x^3-3x^2 > x-3?

Answer 1

For any polynomial, first check to see if it factors. Then you can set each factor equal to zero.
$x^3 -3x^2 -x+3 \ \textgreater \ 0 \\ \\ x^2 (x-3) - (x-3) \ \textgreater \ 0 \\ \\ (x^2 -1)(x-3) \ \textgreater \ 0 \\ \\ (x-1)(x+1)(x-3) \ \textgreater \ 0$

The three zeros are -1,1,3

Next plug in values in each interval between zeros to test whether its greater than 0.

Try x values: -2,0,2,4

x = -2 yields a negative result, the interval x< -1 is Not a solution.

x = 0 yields a positive result, the interval -1
x = 2 yields a negative result, the interval 1
x = 4 yields a positive result, the interval x>3 IS a solution