# How do you find all the real and complex roots of # P(x) = x^5 + 3x^3 + 2x - 6#?

##### 1 Answer

Here's a partial answer. I may provide another (numerical) approach later...

#### Explanation:

First note that the sum of the coefficients is zero.

That is:

So

#x^5+3x^3+2x-6 = (x-1)(x^4+x^3+4x^2+4x+6)#

By the rational root theorem any rational zeros of

In addition, all of the coefficients of

#-1# ,#-2# ,#-3# ,#-6#

None of these work, so

It is possible to solve

Here's a quick sketch of how to do it:

First multiply

#t^4+58t^2+139t+1338=0#

Since this has no cubic term, it has a factorisation of the form:

#t^4+58t^2+139t+1338 = (t^2+At+B)(t^2-At+C)#

Multiply this out and equate coefficients to get three simultaneous equations in

Then use

#(A^2)^3+116(A^2)^2-1988(A^2)-19321 = 0#

To solve this, multiply by

#s^3-58260s+4675597=0#

This cubic has

So we have the choice of expressing them in terms of Complex cube roots (using Cardano's method or Vieta's substitution) or using the trigonometric cubic solution, expressing the roots in terms of

If we have expressions for the roots of our cubic in

This is way too messy, so I would recommend using a numeric method such as Newton's instead.