Question 970091
The potential factors of x^3-3x^2-x-3  are +/- 3  and +/- 1; I need 3 roots or 2 roots with one a square.


Trying -3 (factor x+3) 

-3   1 ; 3  ;-1  ;;-3
;;;  1; -3;;  0 ;;  3 (+3)  (difficult to format synthetic division)
;;;;;;;;0;; -1;;0
This works with no remainder, so  (x+3) and (x^2-1) are factors.

Difference of squares noted:
Factors are (x+3) (x+1) (x-1)

Each of those =0 will give me zeros.

-3, -1, 1

For -3:   -27  +27  +3   -3  =0

For 1:    1+3-1-3=0

For -1:   -1+3 +1 -3=0

To graph it 
y=a (x+3) (x-1)(x+1)
point (0,-3) is on the graph (the constant).

-3= a (3)(-1) (1)
a=1
y=(x+3)(x+1)(x-1)
To the left of minus 3, the graph goes negative, driven by the first negative cubic term.

It rises through the root -1 and drops to 0,-3, when x=0.  It then rises for good, passing through (1,0) and becoming driven by the positive cube.

{{{graph (300,300,-5,5,-20,20,x^3+3x^2-x-3)}}}