Thread: Finding roots of a polynomial

Prove the polynomial x^6 + x^4 +5x^2 + 1 has at least four real roots.
This is what I have so far:
1. f(x)=x^6 + x^4 +5x^2 + 1 is continuous because it is a polynomial
2. f(0)= 1>0
3. f(-1)= -2<0
Since 1-3 are satisfied, there exists x in (0,1) such that for f(x)=0
Using f(-0.5), f(-0.75), f(-0.875) we can find a root.

My question is: Is there another way of find the roots besides the way I'm doing it since I not only have to find one root, i have to find four?

Originally Posted by summerset353

Prove the polynomial x^6 + x^4 +5x^2 + 1 has at least four real roots.
This is what I have so far:
1. f(x)=x^6 + x^4 +5x^2 + 1 is continuous because it is a polynomial
2. f(0)= 1>0
3. f(-1)= -2<0
Since 1-3 are satisfied, there exists x in (0,1) such that for f(x)=0
Using f(-0.5), f(-0.75), f(-0.875) we can find a root.

My question is: Is there another way of find the roots besides the way I'm doing it since I not only have to find one root, i have to find four?

First , $\displaystyle f(-1)=8\neq -2$ , second the polynomial you wrote has no real roots at all since all the powers of x are even and thus its minimal value is $\displaystyle f(0)=1$ ...

Tonio