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

4 0

3 years ago

I WILL PICK BRAINLIST

eimsori [14]

Step-by-step explanation:

π/9 is less than sqrt(9) / 3 which is 1. Also, this question is quite ambiguous, and the answer could be none, because sqrt(9)/3 can also be -1

5 0

3 years ago

*TWO QUESTIONS IN ONE*

TWO QUESTIONS IN ONE