Showing $f(t) = t^4 + 2t^2 + 9$ is reducible over $\Bbb Q$ So, as a part of a homework problem, I'm tasked with showing that $f(t) = t^4 + 2t^2 + 9$, despite having no roots in the rationals, is reducible over them. The former task - showing a lack of rational roots - is pretty easy, but the latter is proving an absolute pain. 
And I suspect it's actually not reducible.

It is obvious that since $f$ has no rational roots, we know that, if $f$ is reducible, it must have a factorization $f = gh$ where $g,h$ are quadratics. 
If we made the substitution $u = t^2$ then $f(u) = u^2 + 2u + 9$. This doesn't seem immediately factorizable by elementary methods (i.e. the whole "sum of factors of constant term sum to the linear coefficient" thing we learn in middle school). Okay so what now?
My immediate thought was to play around with the fundamental theorem of algebra just to simplify matters. So we find roots for $f$ in terms of $u$, which are $u = -1 + 4i \sqrt 2, \overline u = -1 - 4i \sqrt 2$.

Edit: As noted in one of the answers, by egreg, I actually did not find the correct roots in terms of $u$: I forgot to divide the coefficient of the root by two. And, of course, other answers have also noted I overlooked some factoring techniques. Oh well, such is life. In any event, note that this error rippled forward from here in my work and made everything more problematic.

Then by our substitution, the roots to $f$ in terms of $t$ are $\sqrt u, -\sqrt u, \sqrt {\overline{u}}, - \sqrt{\overline{u}}$. Thus by the fundamental theorem of algebra,
$$f(t) = (t-\sqrt u)(t-\sqrt {\overline{u}})(t+\sqrt u)(t+\sqrt {\overline{u}})$$
Surely, some pairing of these four factors in two groups of two will multiply together to give a quadratic in rational coefficients, yielding that $f$ is reducible. However, no matter how I pair them, I can't seem to make this work.
Consider our first factor.


*

*If you multiply it by the second, the linear coefficient is $-(\sqrt u + \sqrt{ \overline {u}})$. Wolfram gives this as $\sqrt{2\sqrt{33} - 2}$. Obviously not rational.

*If you choose the third, then you get a square root of a nonreal complex number as the constant term. Thus, it is not rational.

*If we choose the fourth, then we get the linear coefficient to be $\sqrt{ \overline {u}} - \sqrt u$. Wolfram gives this to be a purely imaginary number, same coefficient as before otherwise.


So no matter which way we pair the first factor it seems we cannot generate a quadratic with rational coefficients.
Does this mean there's some sort of mistake on the assignment, that this polynomial is irreducible? Am I overlooking something, making an error somewhere? (I've been dealing with this for a few hours now and it's not unrealistic, given how late the hour is, that I might have overlooked a flaw somewhere.)
 A: The roots of $z^2+2z+9$ are $-1\pm i\sqrt{8}$. Let's find the square roots of both.
$$
-1+i\sqrt{8}=(x+iy)^2
$$
yields
\begin{cases}
x^2-y^2=-1 \\[4px]
2xy=2\sqrt{2}
\end{cases}
that solves as $x=1$ and $y=\sqrt{2}$ or $x=-1$ and $y=-\sqrt{2}$. Similarly the other root yields $x=1$ and $y=-\sqrt{2}$ or $x=-1$ and $y=\sqrt{2}$.
Now the pairing should be quite obvious: for $1+i\sqrt{2}$ with $1-i\sqrt{2}$ the sum is $2$ and the product is $3$.
Thus you get the factor $t^2-2t+3$. The other factor is $t^2+2t+3$.
A: $$t^4+2t^2+9=\big(t^4+6t^2+9\big)-4t^2=$$
$$(t^2+3)^2-4t^2=$$
$$(t^2-2t+3)(t^2+2t+3)$$
A: Note that
$$
t^4+2t^2+9=
(t^2 + 2t + 3)(t^2 - 2t + 3)$$
is reducible.
A: Here's a rather general approach for finding irreducible rational factors, which is doable by hand if the coefficients and the degree aren't too large (say $\deg f\leq 8$). It is not much more than brute force.
As you note, the polynomial has no rational roots. Indeed, by the rational root theorem, if it has a rational root then it must be a divisor of $9$, and the $6$ divisors are easily checked not to be roots. Then if it is reducible, it must be the product of two irreducible quadratics. So suppose that
$$t^4+2t+9=(t^2+at+b)(t^2+ct+d),$$
where I have already assumed (without loss of generality) that the quadratics are monic. Moreover, by Gauss' lemma the coefficients of the quadratics are integers. Expanding the product yields the equations
$$a+c=0,\qquad ac+b+d=2,\qquad ad+bc=0,\qquad bd=9.$$
Then $c=-a$ and so $0=ad+bc=a(d-b)$ so either $a=0$, which leads to  $b+d=2$ and $bd=9$ which is impossible, or $d-b=0$ which leads to $b^2=9$ and $2b=a^2+2>0$. Then $b=3$ and $a=2$, and we see that
$$(t^2+2t+3)(t^2-2t+3)=t^4+2t+9,$$
so indeed the polynomial is reducible.
A: This kind of factorisation problem is rather common in high school Olympiads. The way I was taught is to think of the method as completing the square, but trying to preserve the constant term rather than the middle term. For example, say we wanted to factor $x^4+x^2+1$. The idea is to find a quadratic whose square agrees with our quartic polynomial in the leading coefficient and the constant term, then subtract the extra term, hoping that things work out. In this case, $x^4+x^2+1=(x^2+1)^2-x^2=(x^2+x+1)(x^2-x+1)$, and in your case, $t^4+2t^2+9=(t^2+3)^2-4t^2=(t^2+2t+3)(t^2-2t+3)$.
The reason I compare this to completing the square is that usually, completing the sqaure asks you to find a perfect square which agrees with the polynomial you want in the leading coefficient and the middle term. For example, you would write $x^2+4x+3=(x+2)^2-1=(x+3)(x+1)$ with $(x+2)^2$ being the desired perfect square. Completing the square doesn't work here, but we're fortunate that a close variant of it does!
