Roots of Cubics

Do you want a condition for three roots?

$y = x(x-1)(x-2)$

In fact, given three roots, there is a cubic with those roots

Do you want a condition for three roots?

Let the polynomial in x of degree 3 be of the form;

$f(x) = ax^3 + bx^2 + cx + d$ ; a =/= 0




We now consider the derivitive of the derivitive of f(x).

$f'(x) = 3ax^2 + 2bx + c$

Solve the quadratic (assuming we have real solution(s) as if not the cubic has no turning points and hence one root)

Let the solutions to the quadratic be $p$ and $q$

These are the coordinates of the turning points. Substitute them back into f(x) to obtain two pairs of coordinates (p,s) and (q,t)

Can you see how we can go on from there ? What about the signs? Are there any special considerations we need to look into?

So for one root f'(x)>0 if a>0 and f'(x)<0 if a<0
For two roots s=0 and t&#8800;0 regardless of a's value
For three roots s<0 and t>0 regardless of a's value

I can't really think of any special considerations yet.