Cubic Polynomials: Solving w/o Rational Roots

[Jack Diamond]

Homework Statement

I think I saw another thread answer this question, but I was a little lost whilst reading it.
I have just recently learned of the rational root theorem and was using it quite happily; figuring out what possibly answers went with cubic and quartic polynomials gave new meaning to guess and check.
But then I realized something strange, I am aware that, because of conjugates, complex and imaginary solutions to polynomials come in pairs. I am also aware that the amount of roots in a polynomial stem from its degree. This became confusing to me when I came across an equation that was not solved using one of the roots found with the rational root theorem.
I am confused on how it is possible for a cubic polynomial to have no rational roots, and thus, three imaginary or complex roots - even though complex and imaginary numbers must come in pairs.
I asked my teacher, but he did not know.

Homework Equations

This was the equation that spurred the whole confusion:
2x$$^{3}$$-5x$$^{2}$$-9x+13=0

this is a different one my teacher showed the class after I asked him about it:
2x$$^{3}$$-9x$$^{2}$$-11x+8=0

The Attempt at a Solution

I thought it was as simple as a multiplicity. But that wouldn't work. I am really at a lost here.

 

[rock.freak667]

You are confusing rational with real. Rational and irrational numbers are both real numbers. So you can have a cubic such as (x-√2)(x-√3)(x-√5)=0, the roots are not rational but all are real.

 

[Dick]

You are confusing rational with real. Rational and irrational numbers are both real numbers. So you can have a cubic such as (x-√2)(x-√3)(x-√5)=0, the roots are not rational but all are real.

Right. But that also doesn't lead to a polynomial with rational coefficients. How about (x-(-1/2+i*sqrt(3)/2))*(x-(-1/2-i*sqrt(3)/2))*(x-1)=x^3-1? If you want an even more complicated example with three real roots try https://www.physicsforums.com/showthread.php?t=368349

 

[Dick]

Ooops. Forget my first example. That does have a rational root. Sorry. If you change it to x^3-2 and scale them all by 2^(1/3), then it should be ok.