Find a cubic polynomial whose graph passes through the points (1,,−,−,0)

The generic equation of a third degree polynomial is given by:
 y = ax ^ 3 + bx ^ 2 + cx + d
 We must make a system of equations to find the values of a, b, c, d.
 We have then:
 For (1, 3):
 3 = a (1) ^ 3 + b (1) ^ 2 + c (1) + d
 3 = a + b + c + d
 For (2, -2):
 -2 = a (2) ^ 3 + b (2) ^ 2 + c (2) + d
 -2 = 8a + 4b + 2c + d
 For (3, -5):
 -5 = a (3) ^ 3 + b (3) ^ 2 + c (3) + d
 -5 = 27a + 9b + 3c + d
 For (4.0):
 0 = a (4) ^ 3 + b (4) ^ 2 + c (4) + d
 0 = 64a + 16b + 4c + d
 We obtain the following system of equations:
 3 = a + b + c + d
 -2 = 8a + 4b + 2c + d
 -5 = 27a + 9b + 3c + d
 0 = 64a + 16b + 4c + d
 Whose solution is:
 a = 1
 b = -5
 c = 3
 d = 4
 The polynomial will then be:
 y = x ^ 3 - 5x ^ 2 + 3x + 4
 
 y = x ^ 3 - 5x ^ 2 + 3x + 4

The answer would he -4,-2 because the points are just on the opposite side. since the x axis is +4 it's going to he -4 and since the y axis is -2 it stays there because whaen rotated it stays in the third quadrant