1. ## reducing a polynomial

hi. i need to find the roots of $2x^3+5$ over the complexs, reals, and rationals. i don't know what to do other than trial and error.. i am starting with the complex. exactly the roots should be easy to find. but i am trying an online calculator and it is telling me the cube root of -5/2 has no real root.. this is not true.. so i am confused. i also don't understand how a polynomial with odd degree can have a complex root.. if someone can help me please give me a pointer for this. then thank you. what i don't understand is that with the odd degree, the imaginary part will not disappear as it is supposed to.

2. ## Re: reducing a polynomial

This is a trivial example of a cubic polynomial. The x term appears only once, therefore you can solve directly for x:

$2x^3 + 5 = 0$

$2x^3 = -5$

$x^3 = -5/2$

$x = -(5/2)^{\frac{1}{3}}$

i also don't understand how a polynomial with odd degree can have a complex root.. if someone can help me please give me a pointer for this. then thank you. what i don't understand is that with the odd degree, the imaginary part will not disappear as it is supposed to.
Good question. The answer is that the numbers will not simply be imaginary: they will be of the form $a + bi$, with form a real and an imaginary component. It is possible for such numbers to be cube roots of a real number. For example, the complex cube roots of 1 are:

$-\frac{1}{2} \pm i\frac{\sqrt{3}}{2}$

That is,

$\left (-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right )\left (-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right )\left (-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right ) = 1$

In addition, to get the complex cube roots of any real number, including the one in the problem, just multiply these numbers by the real cube root. This is because when you cube the product, you get the product of the cubes, one of which will simply be 1.

So the three solutions to your polynomial are:

$x = -(5/2)^{\frac{1}{3}}$

$x = -(5/2)^{\frac{1}{3}} \cdot \left ( -\frac{1}{2} + i\frac{\sqrt{3}}{2} \right )$

$x = -(5/2)^{\frac{1}{3}} \cdot \left ( -\frac{1}{2} - i\frac{\sqrt{3}}{2} \right )$

3. ## Re: reducing a polynomial

thank you. but sorry my mistake, i am not just looking to solve it. i wrote this because i had my mind on just the complex solution. i am looking to factor it to prime polynomials. over each of the complex, real and rationals. i wrote the question wrong. but as for that solution you give. yes, i was able to find the real value solution (i am surprised as i say the online calculator which is normally useful said no real roots). $-(5/2)^{1/3}$. i found this. now i understand, to reduce a polynomial over complex numbers, you just multiply (x - the roots), so i would just plug these in. so

$2x^3+5=(x + (5/2)^{\frac{1}{3}})(x + (\frac{1}{2} - i\frac{\sqrt{3}}{2}))(x + (\frac{1}{2} + i\frac{\sqrt{3}}{2}))$, does this look right?

4. ## Re: reducing a polynomial

No, the last two are not the roots of this polynomial. You would need the entire root. Also, multiply everything by the leading coefficient:

$2 \left (x + (5/2 )^{\frac{1}{3}} \right ) \left (x+(5/2)^{\frac{1}{3}} \cdot ( -\frac{1}{2} + i\frac{\sqrt{3}}{2}) \right ) \left (x+(5/2)^{\frac{1}{3}} \cdot ( -\frac{1}{2} - i\frac{\sqrt{3}}{2} ) \right )$

You might also want to simplify the individual roots.

5. ## Re: reducing a polynomial

hi, thanks a lot. please may i ask where this is coming from
$\left (-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right )\left (-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right )\left (-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right ) = 1$

i didn't know this. and i don't know how to show it myself. i have the solutions ( i can figure the other solutions now too, for reals and rationals), so i have the question solved, thank you very much. but i fear i have not learned it fully, i wish to see where this solution is coming from. if i can derive this result i can do a write up of the problem now thanks to your help. also does $\left (-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right )\left (-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right )\left (-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right ) = 1$ also must be true? because i think if this is true for the conjugate i can understand the problem clearer and i feel i have correctly interpreted what you wrote for me. if not i am back a step.