How to find the double roots of $f(x) = x^6 -x^4 +x^2 - 1$

Question

How do I find the double roots of:

$$f(x) = x^6 -x^4 +x^2 - 1$$

I tried doing long polynomial divisions and got $-1$ as double root. (It gives a remainder as $0$ one step after getting $-1$).

Answer 1

Hint: Find the roots of $\gcd(f,f')$.

The $-1$ you have found using the Euclidean algorithm is a polynomial, not a root. Hence, $\gcd(f,f')=1$ and $f$ has no double roots.

Answer 2

The general method to find the multiple roots is to determine the $\gcd$ of the function and its first derivate. With the second derivate you can then verify whether the multiplicity is larger than $2$.

Answer 3

We have $f(x)=(x^4+1)(x^2-1)$. Then $f$ has no multiple roots. There are two real roots $\pm 1$ and four complex (single) roots $\sqrt[4]{-1}.$

Answer 4

$f(x) = x^{6} -x^{4} +x^{2} -1 $

$= x^{4}(x^{2} -1) + (x^{2} -1) = (x^{2} -1) (x^4 + 1)$

Therefore the roots are: $x_{1} = 1, \ x_{2} = -1$

Answer 5

Well, you can see by inspection that $\pm 1$ are roots. Thus the polynomial is divisible by $x^2-1 =(x-1)(x+1)$. The division yields

$x^6-x^4+x^2-1 = (x^2-1)(x^4+1)$.

Moreover, $x^4+1 = (x^2-i)(x^2+i)$ where $i$ is the imaginary unit with $i^2=-1$.

Now $x^8-1=(x^4+1)(x^4-1)$ where $x^4-1= (x-1)(x+1)(x-i)(x+i)$, i.e., the zeros are the 4th roots of units.

It follows that $x^4+1$ has as zeros the primitive 8th roots of unity and so have $x^2-i$ and $x2+i$.