complex numbers solve for all values of x?

Question

2x^5-9X^4+10x^3+20x^2-36x-35=0
how do u solve for all values of x over the complex numbers?

Answer 1

First, check for rational roots. If this doesn't have
any we are in trouble, but fortunately it does have some.
If r/s is a rational root of this equation, then r must
divide -35 and s must divide 2.
By trial we find that -1 is a root of this equation,
and so x + 1 is a factor. But x+1 is also a root
of the quotient, so x+1 is a double root.
Now work on the resulting cubic.
We find that x = 5/2 is one of its roots.
The final quadratic is
x² - 4x + 7 = 0
which you can now solve by the quadratic formula.
Hope that helps a bit!