How do you find all the real and complex roots of # P(x) = x^5 + 3x^3 + 2x - 6#?
Here's a partial answer. I may provide another (numerical) approach later...
First note that the sum of the coefficients is zero.
#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:
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
#(A^2)^3+116(A^2)^2-1988(A^2)-19321 = 0#
To solve this, multiply by
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.