Find all real zeroes and multiplicity?

Question

equation: x^5+x^3-6x=y

Find the real zeroes and determine the multiplicity of each zero.

How do I do that? Help?

Answer 1

One way is to find the factors of the form (x-a), where +a is a 'zero' of the equation.

The first one I see is x (which is the same as (x-0), meaning that 0 is a zero):
(x^4 + x^2 - 6)x = y
If x=0, then y=0.

x^2 = [-1 +/- SQRT(1 - (-6*4*1) )] / 2
x^2 = -1/2 +/- 5/2
x^2 = -6/2 (x is not a real number), and
x^2 = 4/2 = 2
x = +/- √2

Let us try +√2 as a zero

(x^4 + x^2 - 6) / (x-√2) =
x^3 + √2x^2 + 3x + 3√2 (with no remainder)

so far we have;
y = x(x-√2)(x^3 + √2x^2 + 3x + 3√2)
if x=0, then y = 0 and
x = √2, then y=0

Divide the last term by (x + √2) and keep going.