Cubic and Roots?

Question

Can anyone tell me if the following answer is the answer to the question below. Have I missed anything or can I add anything?:
Cubics - What form will the equations take (factorised) if they have one root, two roots, three roots?


ANSWER-




Cubic = ax3 + bx2 + cx + d

Therefore a cubic with real coeffiecients
ax3 + bx2 + cx + d = 0

If the cubic has one real root, call it t, then factored it looks like where f and g are new constants):
a(x-t)(x^2 + fx + g) = 0


If the cubic had 3 real roots, call them t, u, and v, then it can be written as:
(x-t)(x-u)(x-v) = 0.


If the cubic had 3 real roots it cannot be done because:

Any polynomial with real coefficients with nonreal roots has an even number of nonreal roots.

Answer 1

In your first case (one real root), you can add the condition that the descriminant (f/2)^2 - g is negative.

I think in your last case you meant can't have TWO real roots.

However, you can have three real roots of which two are the same, as in (x-1)(x-2)^2

Answer 2

Yes its correct. Any polynomial expression with rational co-efficients, will have irrational and complex roots in pairs which are conjugates of each other.