Factor SYMBOLIC third degree polynomial

Question

I apology in advance for the long post...

Say I have a very general 3rd-degree polynomial

$$x^3 + a_1x^2 + a_2x+a_3$$

I would really like to factor that polynomial as a product of a 1st order and a 2nd order polynomial, i.e.

$$(x-b_1)(x^2+b_2x+b_3)$$

And do all of that with the symbolic equation: finding the values of $b_i$ as a function of the $a_i$.

I have tried 3 things:

1. My first idea was to use a symbolic computation software (Wolfram Mathematica) and finds the symbolic roots of the first equation, let's call them $\alpha_i$. The roots are awful, but it works ...

Then I tried to keep one root alone to build up $(x-\alpha_i)$ and identify $b_1$ (easy so far). And next compute $(x-\alpha_2)(x-\alpha_3)$ and identify $(x^2 + b_2x +b_3)$ (or equivalently computing $b_2 = -(\alpha_2+\alpha_3)$ and $b_3 = \alpha_2\alpha_3$).

That is where things go wrong... Wolfram Mathematica just does not want to "simplify" $-(\alpha_2+\alpha_3)$ and $\alpha_2\alpha_3$ (it just literally writes that answer).

2. Alternatively, I tried to keep one of the root, and compute

$$ \frac{x^3+a_1x^2+a_2x+a_3}{x-\alpha_1}$$

In this case, $b_1$ is simply $\alpha_1$ (as before) and the result of the division would be $x^2+b_2x+b_3$. But here again, Mathematica does not explicitly compute or "simplify" the division and leaves me with $ \frac{x^3+a_1x^2+a_2x+a_3}{x-\alpha_1}$ as the answer.

3. Another way around my problem (too long to explain, just believe me) would be to compute the magnitude and phase of the two complex roots $\alpha_i$ (I know I have a complex conjugate pair and a real root). But again, Whenever I enter Abs[$\alpha_i$], Wolfram Mathematica gives me ... Abs[$\alpha_i$] as an answer and does not "simplify" or "compute" explicitly what the magnitude would be.

Does anyone have an idea of how I could "solve" that?

I realize that I am asking for a very "theoretical" thing and that if the exact answer would exist, it would probably be given somewhere in Wikipedia or in textbooks... Maybe Mathematica does not simplify or compute the product because things start to be too complex even for a computer...

Thanks so much for all of you that could read this post to the end. Any help, advice or hint would be very much appreciated!

Anthony.

Answer 1

Consider $$\Delta=(x^3 + a_1x^2 + a_2x+a_3)-(x-b_1)(x^2+b_2x+b_3)$$ Expand and group terms to have $$\Delta=(a_3+b_1 b_3)+ (a_2+b_1 b_2- b_3)x+ (a_1+b_1-b_2)x^2$$ So, we need to solve $$a_3+b_1 b_3=0 \tag 1$$ $$a_2+b_1 b_2- b_3=0 \tag 2$$ $$a_1+b_1-b_2=0 \tag 3$$

Using $(2)$ and $(3)$, you can have $$b_2=a_1+b_1 \qquad \text{and} \qquad b_3=a_2+a_1b_1+b_1^2$$ and, replacing in $(1)$ get $$a_3+a_2b_1+a_1b_1^2+b_1^3=0$$ which is a cubic in $b_1$ !!!