Writing a polynomial as a product of irreducible polynomials over another field One root of the polynomial $f(x) = 2x^5 + x^4 + 44x^3 + 22x^2 - 150x - 75$ over $\Bbb{C}$ is 
$ -5i$.

How can I write this as a product of irreducible polynomials over $\Bbb{R}$ and $\Bbb{C}$?

I have no clue how to go about this. I used grouping to factor out products to write it as a product of irreducible polynomials over $\Bbb{Q}$ however I don't know if that is correct either.
Let $f(x) = 2x^5 + x^4 + 44x^3 + 22x^2 - 150x - 75$.

Since the coefficients of $f$ are real and $-5i$ is a root of $f$, it follows that $5i$ is also a root of $f$.

Since $5i$ and $-5i$ are roots of $f$, it follows that $x-5i$ and $x+5i$ are factors of $f$, hence $x^2 + 25$ is a factor of $f$.

By long division, the other factor comes out as
$$2x^3+x^2-6x-3$$
The rational root test yields the root $x = -\frac{1}{2}$, which yields the factor $2x+1$.

By long division, the remaining factor is $x^2-3$.

Hence:

$\;\;\;{\small{\bullet}}\;$Over $\mathbb{Q}$, $f$ can be factored as
$$(x^2+25)(2x+1)(x^2-3)$$
$\;\;\;{\small{\bullet}}\;$Over $\mathbb{R}$, $f$ can be factored as
$$(x^2+25)(2x+1)(x+\sqrt{3})(x-\sqrt{3})$$
$\;\;\;{\small{\bullet}}\;$Over $\mathbb{C}$, $f$ can be factored as
$$(x-5i)(x+5i)(2x+1)(x+\sqrt{3})(x-\sqrt{3})$$
by conjugate root theorem, 5i is also a root since 5i is conjugate of -5i.

Then we can have (x-5i) and (x+5i) as factors of the polynomial. 

(x-5i)(x+5i)is (x^2+25)

Finally you could do long division using this.

answer should be (x-5i)(x+5i)(2x+1)(x^2-3)

Hope this helps ;)