given that f(x)= 3-7x+5x^2-x^3, factorise the polynomial completely, hence state the set of values of x for which the function is equal or less than zero .

-1 ( x^3 - 5 x^2 + 7 x - 3)

I can see right away that x = 1 makes this zero because 8-8 = 0
divide polynomial by (x-1)
******_x^2_-_4x____+_3_______________
x-1 / x^3 - 5 x^2 + 7 x - 3
*******x^3 - 1 x^2
*******---------------------
***********- 4 x^2 + 7 x -3
***********- 4 x^2 + 4 x
***********-----------------
*******************+ 3 x - 3
*******************+ 3 x - 3
remainder =0
so
-1 (x-1)(x^2 - 4 x + 3)
so
-1 (x-1)(x-1)(x-3)
zeros at 1, 1 (double,just hits axis) and 3
graph it remembering the - sign