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$f(x)=x^4-16x^2+4$, 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.

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use the quadratic formula, twice if neessary. – i. m. soloveichik Nov 13 '12 at 17:41
Please don't yell at us. It's rude. – Cameron Buie Nov 13 '12 at 17:42
How do you mean, "the root"? This polynomial has more than one root. – joriki Nov 13 '12 at 18:06

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 $\mathbb Q$.

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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.

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