# Roots of Cubic Polynomials over R

1. Dec 28, 2011

### alexfloo

I'm trying to prove the following, which is left unproven in something I'm reading on ruler-and-compass constructions:

If $ax^3+bx^2+cx+d$ is a polynomial over a subfield F of ℝ, and $p+q\sqrt{r}$ is a root (with $\sqrt{r}\notin F$) then $p-q\sqrt{r}$ is also a root.

The theorem immediately before this made use of expanding $a(x-r_1)(x-r_2)(x-r_3)$ and equating coefficients to get a system of three equations in the roots and coefficients. The book suggested that this proof was similar, but I was't able to derive anything useful from that.

I also tried defining $g(t)=f(p-t\sqrt{r})-f(p+t\sqrt{r})$, and noting that $g(q)=f(p-q\sqrt{r})$ and $-g(-t)=g(t)$.

Then we see that

$f(p-q\sqrt{r})=g(q)=-g(-q)=-f(p+q\sqrt{r})=0$.

However, I know this must be flawed, because I don't believe I ever used the assumption that $\sqrt{r}\notin F$. Any thoughts?

2. Dec 28, 2011

### micromass

Staff Emeritus
This step is wrong.

Basically, you know that you can write your polynomial in the form

$$a(x-p-q\sqrt{r})(x-r_2)(x-r_3)$$

What do you get when you work that out??

For example, you must get that

$$-(p+q\sqrt{r})r_2r_3\in F$$

Do you see why?? Does that imply something for r2 and r3??

3. Dec 28, 2011

### alexfloo

I think I got it but I just want to check.

From the first and third coefficient equations, we have $\prod r_i\in F$ and $\sum r_i\in F$. Therefore, $r_1r_2=f(p-q\sqrt r)$ and $r_1 + r_2 = t-q\sqrt r$.

Then we use the middle equation to get

$r_{2}r_{3}+r_{1}r_{3}+r_{1}r_{2} = \left(r_{2}+r_{1}\right)r_{3}+r_{1}r_{2} = \left(t-q\sqrt{r}\right)\left(p+q\sqrt{r}\right)+x\left(p-q\sqrt{r}\right) = pt-q^{2}r+xp+\left(t-x-p\right)q\sqrt{r}$

Therefore, $t-x-p = f\sqrt r$. Since all three terms are in F, it must be that f = 0. This gives us t = x+p, so $r_1 + r_2 = x+p-q\sqrt r$.

Of course, the unique pair of numbers with that sum and product are x and $p-q\sqrt r$

4. Dec 28, 2011

### micromass

Staff Emeritus
I think it's ok.