Showing $f(t) = t^4 + 2t^2 + 9$ is reducible over $\Bbb Q$

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!

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 $t^2+2t+3$.

$$t^4+2t^2+9=\big(t^4+6t^2+9\big)-4t^2=$$ $$(t^2+3)^2-4t^2=$$ $$(t^2-2t+3)(t^2+2t+3)$$