I must be tired, real tired

[Steven G]

For a polynomial I know that complex and irrrational roots occur in pairs.
In the equation x³-x-2=0 the only possible rational roots are +/- 2. In fact neither of them are roots.
If there are 2 complex roots, then when you multiply them you get a quadratic. You can only multiply that by a linear factor to get the cubic. Now linear factor yield a single root. In this case should we not have 2 complex roots and a rational root??
If two roots were irrational (and they do come in pairs) then the 3rd root must be rational!

What am I missing here?

 

[topsquark]

For a polynomial I know that complex and rational roots occur in pairs.
In the equation x³-x-2=0 the only possible rational roots are +/- 2. In fact neither of them are roots.
If there are 2 complex roots, then when you multiply them you get a quadratic. You can only multiply that by a linear factor to get the cubic. Now linear factor yield a single root. In this case should we not have 2 complex roots and a rational root??
If two roots were irrational (and they do come in pairs) then the 3rd root must be rational!

What am I missing here?

If we have two complex or irrational roots, why should the real root have to be rational?

-Dan

 

[Dr.Peterson]

For a polynomial I know that complex and [ir?]rational roots occur in pairs.
In the equation x³-x-2=0 the only possible rational roots are +/- 2. In fact neither of them are roots.
If there are 2 complex roots, then when you multiply them you get a quadratic. You can only multiply that by a linear factor to get the cubic. Now linear factor yield a single root. In this case should we not have 2 complex roots and a rational root??
If two roots were irrational (and they do come in pairs) then the 3rd root must be rational!

What am I missing here?

Why can't the real root be irrational?

It is not irrational roots that come in conjugate pairs, but roots of the form $a + b\sqrt{c}$. Other irrational numbers don't have conjugates in the first place.

Of course, your statements apply specifically to polynomials with real coefficients, and with rational coefficients, respectively.

 

[Steven G]

I am confused. You are saying that you can have a cubic polynomial with three roots where two are complex and one is irrational and the coefficients are rational????

x² + 4 has 2 complex roots. (x-sqrt2) has an irrational root. However, (x² + 4)(x-sqrt2) has irrational coefficients.

It is not irrational roots that come in conjugate pairs, but roots of the form a+bsqrt(c)
If a=0 and b=1, then sqrt(c) = 0 + 1sqrt(c) is in the form that you claim comes in conjugate roots

 

[Dr.Peterson]

x² + 4 has 2 complex roots. (x-sqrt2) has an irrational root. However, (x² + 4)(x-sqrt2) has irrational coefficients.

It is not irrational roots that come in conjugate pairs, but roots of the form a+bsqrt(c)
If a=0 and b=1, then sqrt(c) = 0 + 1sqrt(c) is in the form that you claim comes in conjugate roots

As I said, it is polynomials with rational, not irrational, coefficients, whose quadratic surd roots come in conjugate pairs.

Here are the solutions to your original cubic:



Those aren't square roots.

 

[topsquark]

Get some sleep Steve.

-Dan

 

[Steven G]

As I said, it is polynomials with rational, not irrational, coefficients, whose quadratic surd roots come in conjugate pairs.

Here are the solutions to your original cubic:

Sure, the polynomials I am thinking about and the example I supplied does have rational coefficients.
Yes, I did go to WA and saw that there are two imaginary roots. Now i know quite well that if x=a is a root IFF (x-a) is a factor. My concern is that the real root that WA found is irrational (if it were rational, then the rational root test would have found it). If r is this irrational root, then (x-r) would be a factor of this cubic polynomial. How can the product of all three factors have rational coefficient?

Can you show me an example??

 

[lookagain]

In the equation x³-x-2=0 the only possible rational roots are +/- 2.
. . .
What am I missing here?

You are missing other possible rational roots of $\displaystyle \ \pm 1.$

 

[Dr.Peterson]

If r is this irrational root, then (x-r) would be a factor of this cubic polynomial. How can the product of all three factors have rational coefficient?

Because cube roots can do that. Irrational roots don't have to involve square roots, which are needed for the theorem.

Not that I've tried verifying it by hand.

 

[Qwertyuiop[]]

so I'm not the only one confused here

 

[Dr.Peterson]

I've tried to find a good reference for the relevant theorem, but I don't find it in Wikipedia, a textbook, or other such places. (Google doesn't seem to distinguish between this and the complex conjugate root theorem.) But here is one reference (including a "duplicate" question with proofs:

The radical conjugate roots theorem

If the coefficients of $$p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$$ are rational, the Conjugate Radical Roots theorem states that if the equation $p(x)=0$ has a root of the form $s+t\sqrt{u}$ whe...

math.stackexchange.com

 

[JeffM]

so I'm not the only one confused here

Let's see if this will help.

There is no rule that says a cubic with rational coefficients must have a rational root.

An example is $f(x) = x^3 - 2$.

It has a REAL root at x = $\sqrt{2}.$ But that root is an irrational number, and that function's slope is non-negative everywhere. Consequently, that function necessarily has only one real root. It also has complex roots, and, because it is polynomial with real coefficients, any complex roots come in conjugate pairs.

There is no assurance that a polynomial with even integer coefficients has one or more rational roots. If, however, a polynomial with real coefficients has complex roots, they come in conjugate pairs.

I think students get confused by the rational root theorem, which is contingent on the existence of a rational root in the first place. It gives no assurance that such a beast exists.

Clear now?