I have a polynomial
$$f(x) = i(x^2-1)^3+(x^2+1)^3-8x^3$$
I want to check if this has repeated roots. To do so, I'll find greatest common divisor (euclidean algorithm) of $f(x)$ and its derivative $f'(x)$.
$$f(x) = i(x^2-1)^3+(x^2+1)^3-8x^3 = i(x^6-3x^4+3x^2-1)+(x^6+3x^4+3x^2+1)-8x^3$$
$$f'(x) = i(6x^5-12x^3+6x)+(6x^5+12x^3+6x)-24x^2$$
I know I should find the GCD as $(x-i)(x-1)$ but this is where I'm stuck. What would the next step be?
On
It's best not to split the polynomials up using $i$, but collect terms in powers of $x$.
$$ \eqalign{f_0 = f(x) &= \left( 1+i \right) {x}^{6}+ \left( 3-3\,i \right) {x}^{4}-8\,{x}^{3}+ \left( 3+3\,i \right) {x}^{2}+1-i \cr f_1 = f'(x) &= \left( 6+6\,i \right) {x}^{5}+ \left( 12-12\,i \right) {x}^{3}-24\,{x }^{2}+ \left( 6+6\,i \right) x } $$ The remainder of $f_0$ on division by $f_1$ is $$ f_2 = f_0 - (x/6) f_1 = \left( 1-i \right) {x}^{4}-4\,{x}^{3}+ \left( 2+2\,i \right) {x}^{2}+ 1-i $$ Then take remainder of $f_1$ on division by $f_2$, etc.
By inspection, $f(1)=0$ so $x-1$ is a factor of $f(x)$.
Since $f'(x)= 3i(x^2-1)^2(2x)+3(x^2+1)^2(2x)-24x^2$, we also have that $f'(1)=0$, so $x-1$ is a factor of $f'(x)$.
This means $(x-1)^2$ is a factor of $f(x)$.
One also has that by inspection $f(i)=0$ so $x-i$ is a factor of $f(x)$.
We also have that $f'(i)=0$, so $x-i$ is a factor of $f'(x)$.
So we also have that $(x-i)^2$ is a factor of $f(x)$.