Show these two polynomials are irreducible over $\mathbb{Q}$

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Need to show both of these are irreducible:

$f(t) = t^4 -14t^2 + 44 = 0 $

and $g(t) = t^4 -24t^2 + 4= 0$. I've tried Eisensteins but can't get anywhere.

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If $f(t)$ is reducible, it factors as $f(t)=h(t)y(t)$. There are two possibilities: 1.) $h(t)$ has degree one and $y(t)$ has degree three or 2.) $h(t)$ and $y(t)$ have degree two.

Case 1.: $h$ has degree one if and only if $f(t)$ has a root $\alpha\in\mathbb{Q}$. by rational root theorem, the only possible rational roots are $\alpha=\frac{p}{q}$ where $p|44$ and $q|1$ i.e. plus or minus$\{1,2,4,11,44\}$. None of these should be a root of $f$, so case 1 can't happen

Case 2.: $h(t)y(t)=(at^2+bt+c)(\alpha t^2+\beta t+\gamma)=f(t)=t^4-14t^2+44$. Now foil and you'll solve a system of equations. You'll get a contradiction because of the restriction $a,b,c,\alpha,\beta,\gamma\in\mathbb{Q}$. So case 2 can't happen.

Thus, $f$ is irreducible over $\mathbb{Q}$. Same thing should work for $g$

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Let's try $f(t)$. $f(t)$ is of degree $4$ so it has at most $4$ roots. What are the possibilities? Either it has at least one linear factor, or it has no linear factors - two irreducible quadratic factors. These questions always go the same way.

You can use the rational root test to see if it has a linear factor.

Then if not, you have to rule out two quadratic factors. So, suppose it factored into two irreducible polynomials over $\mathbb{Q}$. Well, by Gauss' lemma, you only need to see if it factors over $\mathbb{Z}$, so just imagine the polynomials are instead in $\mathbb{Z}$.

Now, equate coefficients and do some algebra and try to come up with a contradiction. Usually, it'll be that one of the coefficients which you assumed was an integer will not be an integer. Try it.

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Another approach: first, in each case, divide the roots by $2$, to get the easier-to-handle polynomials $t^4-7t^2+11$ and $t^4-12t^2+1$.

For the first of these, $t^4-7t^2+11\equiv t^4-t^2-1\pmod3$. Has no roots modulo $3$, and since the only irreducible quadratic polynomials over $\Bbb F_3$ are $t^2+1$, $t^2+t-1$, and $t^2-t-1$, and none of these divides our quartic, it’s irreducible over $\Bbb F_3$, and thus over $\Bbb Z$, and thus over $\Bbb Q$. (Thanks to @GitGud.)

For the second polynomial, $g(t)=t^4-12t^2+1$, and if $\alpha$ is a root, then $\alpha^2$ is a root of $t^2-12t+1$, thus $\alpha^2=6\pm\sqrt{35}$. Now let’s use the arithmetic of real quadratic fields, in particular the field $\Bbb Q(\sqrt{35}\,)$. An integral basis is $\{1,\sqrt{35}\}$, and by standard methods involving continued fractions, you know that $6+\sqrt35$ is a fundamental unit. In particular, its square root is not in $\Bbb Q(\sqrt{35}\,)$, so that $\bigl[\Bbb Q(\alpha):\Bbb Q\bigr]=4$, and thus the polynomial $t^4-12t^2+1$ is the minimal polynomial for $\alpha$, and consequently irreducible.