Finding three X values for y=ax^3+bx^2+cx+d

Hello,

I am trying to solve for the x's for y=ax^3+bx^2+cx+d.

It cuts the y-axis at (0,30). The gradient at this point is -1.

The gradient where x=1, is -17 and the f ``(x) at this point is -10.

I s.t the coordinates given; set f `(x) =-1 and

f ``(x)=-10.

I think d=30 and c=-1.

So I need to solve for a,b.

I didn't think solving for x would be this tricky? What have I missed?

Theory of equations solution?

Hi..

these are the numbers for a=2, b=-11, c=-1, d=30. st these into equation

y=ax^3+bx^2+cx+d gives:

y=2x^3-11bx^2-x+30

If I do the Rational Roots test then I get

+/- 1,__2__,3,__5__,10,15,30,1/2,__3/2__,5/2,15/2

I know the roots are -3/2,5,2 but what does the solution look like with the theory of equations method solving for x1, x2, x3?