Confirming Conjugate Factor of Cubic Equation

[Michael_Light]

Homework Statement

If a cubic equation, f(x) has a factor of (3+√2), then the conjugate of the factor, (3-√2) is also a factor for f(x).

Homework Equations

The Attempt at a Solution

Just to confirm is that statement correct? I read it else where but i not sure is it correct or wrong, if yes, can anyone explain why? Thanks..

 

[voko]

A factor would be something of the form (x - a), where a is a root. You have number. Please clarify what you mean.

 

[Michael_Light]

A factor would be something of the form (x - a), where a is a root. You have number. Please clarify what you mean.

Sorry.. i mean root.. not factor.. so is this statement correct then?

If a cubic equation, f(x) has a root of (3+√2), then the conjugate of the root, (3-√2) is also a root for f(x).

 

[voko]

Only complex roots are conjugate. Your root is not complex, so it cannot have a conjugate in the strict sense of the word (i.e, the one whose imaginary part has an opposite sign), nor can it have a "conjugate" in the sense you are using it.

Consider x³ = (3+√2)³. Then (3+√2) is its only root (more technically, it has three identical roots).

 

[Curious3141]

Homework Statement

If a cubic equation, f(x) has a factor of (3+√2), then the conjugate of the factor, (3-√2) is also a factor for f(x).

Homework Equations

The Attempt at a Solution

Just to confirm is that statement correct? I read it else where but i not sure is it correct or wrong, if yes, can anyone explain why? Thanks..

Not true in general.

It is true, however, if the following conditions are met:

1) Your cubic has rational coefficients.

2) It has 3 real roots.

3) One root is rational.

If those conditions are met, then the other two real roots will either both be rational, or conjugate surds (Technically, the other roots will be algebraic numbers of degree $\leq 2$).

 

[Mentallic]

Homework Statement

If a cubic equation, f(x) has a factor of (3+√2), then the conjugate of the factor, (3-√2) is also a factor for f(x).

Homework Equations

The Attempt at a Solution

Just to confirm is that statement correct? I read it else where but i not sure is it correct or wrong, if yes, can anyone explain why? Thanks..

If the cubic equation has a rational constant term, then you know that there exists another root that must rationalize that root of $3+\sqrt{2}$ because when we multiply all the roots together, we will get the constant term of the cubic equation.

So if for example we have a general cubic

$$ax^3+bx^2+cx+d=0$$ where d is rational, another form of the cubic will be

$$a(x-\alpha)(x-\beta)(x-\gamma)=0$$ where $\alpha, \beta,\gamma$ are the roots of the cubic. So the constant term, d, must be equal to the constant term of the factored cubic, hence
$$a(-\alpha)(-\beta)(-\gamma)=-a\alpha\beta\gamma=d$$
And since $\alpha = 3+\sqrt{2}$ then
$$-(3+\sqrt{2})\beta\gamma=d$$

Now from this we can see that for this expression to be equal to a rational constant d, $\beta\gamma$ must be irrational such that it can cancel the $3+\sqrt{2}$ factor. Now if we assume that one of the roots is rational, then the other root must be of the form $c(3-\sqrt{2})$ for some constant c. However, if we don't make that assumption, then the only condition we need is to satisfy $\beta\gamma = 3-\sqrt{2}$ which there are an infinite number of combinations for.

For example,
$$\beta = \gamma = \sqrt{3-\sqrt{2}}$$
$$\beta = \sqrt{2}, \gamma = \frac{3\sqrt{2}}{2}-2$$

etc.