One root of the polynomial $f(x) = 5x^5+x^4-5x^3-x^2-10x-2$ over $\mathbb C$ is $i$.
(a) Write $f(x)$ as a product of irreducible polynomials in $\mathbb Q[x]$.
(b) Write $f(x)$ as a product of irreducible polynomials in $\mathbb R[x]$.
(c) Write $f(x)$ as a product of irreducible polynomials in $\mathbb C[x]$.
How can I write factorize this polynomials or how can I figure out the product?
Since $i$ is a root of $x$, and it has real coefficients, thus $-i$ is a root as well, meaning that $x^2+1$ divides this polynomial: $$ 5x^5+x^4-5x^3-x^2-10x-2= $$ $$ =5(x^2+1)x^3+x^2(x^2+1)-10x(x^2+1)-2(x^2+1)= $$ $$ =(x^2+1)(5x^3-10x-2)=(x^2+1)(5x+1)(x^2-2) $$ In the last step we tried to find a rational root, and since $f\left(\frac{p}{q}\right)=0\Rightarrow p|2$ and $q|5$. Trying these out we get to the factorisation.
a) Since $\pm i,\pm \sqrt{2}\not\in\mathbb{Q}$ we have $(x^2+1)(5x+1)(x^2-2)$
b) Since $\pm i\not\in\mathbb{R}$, we have $(x^2+1)(5x+1)(x-\sqrt{2})(x+\sqrt{2})$
c) $(x+i)(x-i)(5x+1)(x-\sqrt{2})(x+\sqrt{2})$
(used the fact that $ax^2+bx+c$ must have linear factors, if not irreducible, and thus roots in the field)