Roots of Cubic Polynomials over R

In summary: Just wanted to double check.In summary, if a polynomial over a subfield of ℝ has a root p+q\sqrt{r} in that field, then p-q\sqrt{r} is also a root.
  • #1
alexfloo
192
0
I'm trying to prove the following, which is left unproven in something I'm reading on ruler-and-compass constructions:

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

The theorem immediately before this made use of expanding [itex]a(x-r_1)(x-r_2)(x-r_3)[/itex] 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 [itex]g(t)=f(p-t\sqrt{r})-f(p+t\sqrt{r})[/itex], and noting that [itex]g(q)=f(p-q\sqrt{r})[/itex] and [itex]-g(-t)=g(t)[/itex].

Then we see that

[itex]f(p-q\sqrt{r})=g(q)=-g(-q)=-f(p+q\sqrt{r})=0[/itex].

However, I know this must be flawed, because I don't believe I ever used the assumption that [itex]\sqrt{r}\notin F[/itex]. Any thoughts?
 
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  • #2
alexfloo said:
[itex]-g(-q)=-f(p+q\sqrt{r})[/itex]

This step is wrong.

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

[tex]a(x-p-q\sqrt{r})(x-r_2)(x-r_3)[/tex]

What do you get when you work that out??

For example, you must get that

[tex]-(p+q\sqrt{r})r_2r_3\in F[/tex]

Do you see why?? Does that imply something for r2 and r3??
 
  • #3
I think I got it but I just want to check.

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

Then we use the middle equation to get

[itex]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}[/itex]

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

Of course, the unique pair of numbers with that sum and product are x and [itex]p-q\sqrt r[/itex]
 
  • #5


I would approach this problem by first understanding the definitions and properties of cubic polynomials and their roots. A cubic polynomial is a function of the form f(x) = ax^3 + bx^2 + cx + d, where a, b, c, and d are constants. The roots of a polynomial are the values of x that make the polynomial equal to zero. In this case, the polynomial has three roots, which can be real or complex numbers.

The given problem is asking us to prove that if p + q√r is a root of the polynomial f(x), then p - q√r must also be a root. This is known as the conjugate root theorem. To prove this, we can use the concept of complex conjugates, which states that if z = a + bi is a complex number, then its conjugate is z̅ = a - bi. This means that the real part of the complex number remains the same, but the sign of the imaginary part is changed.

In this case, we have a polynomial with coefficients in a subfield F of ℝ, and one of its roots is p + q√r, where √r is not in F. This means that q√r is a complex number, and its conjugate is q√r̅ = q√r. Therefore, if p + q√r is a root, then p - q√r must also be a root, as stated in the theorem.

To prove this, we can use the method suggested in the book, which is to expand the polynomial (x - (p + q√r))(x - (p - q√r))(x - r) and equate the coefficients to get a system of equations in the roots and coefficients. This will lead to the conclusion that p - q√r is indeed a root of the polynomial.

Alternatively, we can use the approach mentioned in the content, which is to define g(t) = f(p - t√r) - f(p + t√r). We can then show that g(q) = 0, which means that p - q√r is a root of the polynomial. This approach also uses the concept of complex conjugates, as g(-t) = g(t), which means that the imaginary parts cancel out and we are left with only real numbers, leading to the
 

1. What are the roots of a cubic polynomial?

The roots of a cubic polynomial over the real numbers (R) are the values of x that make the polynomial equal to zero. These values can be found by factoring the polynomial or by using the cubic formula.

2. How many roots does a cubic polynomial have?

A cubic polynomial can have up to three distinct roots, but it can also have less. The number of roots can be determined by the degree of the polynomial and the multiplicities of the roots.

3. How do you find the roots of a cubic polynomial?

To find the roots of a cubic polynomial, you can use the rational root theorem, synthetic division, or the cubic formula. The rational root theorem helps to narrow down the possible rational roots, while synthetic division helps to simplify the polynomial. The cubic formula is a direct formula that can be used to find the roots.

4. Can a cubic polynomial have complex roots?

Yes, a cubic polynomial can have complex roots. In fact, a cubic polynomial must have at least one complex root if it has real coefficients. This is due to the complex conjugate root theorem, which states that if a polynomial has real coefficients and a complex root, then its conjugate is also a root.

5. What is the relationship between the roots and coefficients of a cubic polynomial?

The roots of a cubic polynomial can provide information about its coefficients. For example, the sum of the roots is equal to the opposite of the second coefficient divided by the leading coefficient. The product of the roots is equal to the constant term divided by the leading coefficient. These relationships are known as Vieta's formulas.

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