$f(x)=x^4-16x^2+4$, the root of $f(x)$ is $a= \sqrt{3} + \sqrt{5}$
Factorise $f(x)$ as a product of irreducible polynomials over $\mathbb{Q}$, over $\mathbb{R}$ and over $\mathbb{C}$.
I am really confused as to how to start.
$f(x)=x^4-16x^2+4$, the root of $f(x)$ is $a= \sqrt{3} + \sqrt{5}$
Factorise $f(x)$ as a product of irreducible polynomials over $\mathbb{Q}$, over $\mathbb{R}$ and over $\mathbb{C}$.
I am really confused as to how to start.
Since the polynomial has coefficients in $\mathbb Q$ we must have all algebraic conjugates also roots, thus polynomial factored in $\mathbb R$ and $\mathbb C$ to
$(x - (\sqrt{3} + \sqrt{5}))(x - (\sqrt{3} - \sqrt{5}))(x - (-\sqrt{3} + \sqrt{5}))(x - (-\sqrt{3} - \sqrt{5}))$
and irreducible over $\mathbb Q$.
Use the substitution $u=x^2$, to get $u^2-16u+4$. Completing the square makes this $(u-8)^2-60.$ The roots of this are then $u=8\pm\sqrt{60}=8\pm 2\sqrt{15}=3\pm 2\sqrt{3}\sqrt{5}+5=\left(\sqrt{3}\pm\sqrt{5}\right)^2.$ I chose to do this because we know that $x=\sqrt{3}+\sqrt{5}$ is a root of the quartic in $x$, so we needed at least $u=x^2=\left(\sqrt{3}+\sqrt{5}\right)^2$ as a root of the quadratic in $u$.
Resubstituting gives us $x^2=\left(\sqrt{3}\pm\sqrt{5}\right)^2,$ so we have $4$ roots, namely: $x=\sqrt{3}\pm \sqrt{5}$ and $x=-\sqrt{3}\mp\sqrt{5}$. Thus, we have a real (and complex) irreducible factorization into $\left(x-\sqrt{3}-\sqrt{5}\right)\left(x-\sqrt{3}+\sqrt{5}\right)\left(x+\sqrt{3}-\sqrt{5}\right)\left(x+\sqrt{3}+\sqrt{5}\right).$ Now, none of these is a polynomial with rational coefficients, so to get our rational irreducible factorization, there are only two possibilities: (1) the original quartic is irreducible over $\Bbb Q$, or (2) we can pair the linear terms in such a way that both products give us irreducible quadratics over $\Bbb Q$. I leave it to you to determine which is the case.