How do I solve for complex roots of a polynomial function?

[azwraith69]

Homework Statement

f(x)=24x⁷-13x⁶-19x⁵+7x⁴-7x³+5x²+72x-54

Find all the roots..

Homework Equations

+-(p/q)...for rational roots

The Attempt at a Solution

i tried factor theorem,, and synthetic division for the possible roots..i've used uppe/lower bounds but i can't get a single root..

when i used a software to solve for this,, the roots i got are 6 imaginaries and .77547

i have no idea how to show the solution... how to solve for imaginary roots of such irreducible polynomial function??

thX.. really need this asap..

 

[HallsofIvy]

Since you mention "rational roots", I take it you know the "rational root theorem": If p/q is a rational root of the polynomial, then q is a factor of the leading coefficient and p is a factor of the constant term. In this case we can simplify the search a little by noting that both 24 and 54 are multiples of 6. If we were to divide the entire polynomial by 6, that wouldn't change the roots but would make the leading coefficient 4 and the constant term 9. The factors of 4 are $\pm 1, \pm 2, \pm 4$ and the factors of 9 are $\pm 1, \pm 3, \pm 9$ so the only possible rational roots are $\pm 1, \pm 1/2, \pm 1/4, \pm 3, \pm 3/2, \pm 3/4, \pm 9, \pm 9/2, \pm 9/4$.

I don't believe this has any rational roots!

Other than looking for rational roots and hoping to reduce, there is no general method of finding irrarational real or complex roots of such a polynomial.

 

[statdad]

You could try theorems that give upper and lower bounds for the sizes of real roots - these bounds are typically stated in terms of expressions involving the coefficients. (Assuming, of course, that you have had them discussed in your class.)

 

[azwraith69]

Since you mention "rational roots", I take it you know the "rational root theorem": If p/q is a rational root of the polynomial, then q is a factor of the leading coefficient and p is a factor of the constant term. In this case we can simplify the search a little by noting that both 24 and 54 are multiples of 6. If we were to divide the entire polynomial by 6, that wouldn't change the roots but would make the leading coefficient 4 and the constant term 9. The factors of 4 are $\pm 1, \pm 2, \pm 4$ and the factors of 9 are $\pm 1, \pm 3, \pm 9$ so the only possible rational roots are $\pm 1, \pm 1/2, \pm 1/4, \pm 3, \pm 3/2, \pm 3/4, \pm 9, \pm 9/2, \pm 9/4$.

I don't believe this has any rational roots!

Other than looking for rational roots and hoping to reduce, there is no general method of finding irrarational real or complex roots of such a polynomial.

ok,, i know the theorems,, upper/lower bounds,, etc..
and i know that this has no rational roots..

i just want to know methods to find irrational real, or imaginary roots for such irreducible equations... if there is really none,, ok... but i hope there is..

thx hallsofivy

 

[HallsofIvy]

Do you know that there exist polynomials of degree 5 or higher whose roots cannot be written in terms of combinations of radicals?

There cannot be a general way of giving exact solutions to such equations. You can, of course, use numerical algorithms to determine approximate solutions.