Prove that x6 + x4 + x2 + x + 3 = 0 has positive real roots - Maths - Complex Numbers and Quadratic Equations

Question

Prove that x6 + x4 + x2 + x + 3 =
0 has positive real roots

Mayur Pisode · Accepted Answer

Dear Student,

Using Descartes's Rule, there are no sign changes in the given
polynomial hence there are no positive roots

Now, f(-x) = (-x)6 + (-x)4 + (-x)2
+ (-x) + 3 = 0
⇒f(-x) = x6 + x4 + x2 - x +
3 = 0
Using â€‹ Descartes's Rule, there are two sign
changes in f(-x) therefore there are 1 are 2 negative rots

Regards

Prove that x6 + x4 + x2 + x + 3 = 0 has positive real roots.

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