# How to find GCD of those two complex polynomials?

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?

• Euclidean algorithm for gcd of polynomials. – Robert Israel Jul 3 at 14:15
• @RobertIsrael I have no clue how to apply it on those. – Melz Jul 3 at 14:16
• How to do this using long polynomial division? – Melz Jul 3 at 14:48

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 $$f(x)$$.

• How to prove those algebraically? For instance, long polynomial division? – Melz Jul 3 at 14:21
• @Enzo : That would certainly work, but you would have to group all of the terms of the same degree together and you would be faced with division of complex numbers which is easy but unwieldy. – MPW Jul 3 at 14:24
• The thing I'm still confused about is where to stop dividing those polynomials. – Melz Jul 3 at 14:25
• @Enco Note that this isn't a complete answer since we still need to check if the remaining quadratic factor has repeated roots. Did you intend to use only the GCD method, or are other ad-hoc methods admissable? The GCD method is the best way to proceed in general. – Bill Dubuque Jul 3 at 14:25

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.