Can $x^4+x^2-12$ be factored over the rationals?

Question

From my textbook: $f(x)=x^4+x^2-12$

(a) Write the polynomial as the product of factors that are irreducible over the rationals
(b) Write the polynomial as the product of linear and quadratic factors that are irreducible over the reals
(c) Write the polynomial in completely factored form

My first step was to determine possible rational roots using the rational roots theorem: $\pm 1, \pm2,\pm3,\pm4,\pm6,\pm12$. I noticed a pattern: as I moved along the list in ascending order (ignoring the signs since every term is of an even degree), $x^4 + x^2$ got larger and larger than 12. I concluded then that there is no factorization irreducible over the rationals. This was doubly confirmed by Descarte's rule of signs. There can only possibly be 1 positive or negative root, so they must be irrational conjugates.

Next, I factored: $f(x)=(x-\sqrt3)(x+\sqrt3)(x^2+4)$. This is part (b)

The correct solution to part (c) is $f(x)=(x-\sqrt3)(x+\sqrt3)(x-2i)(x+2i)$, Obviously, I've found all four of the roots guaranteed by the fundamental theorem of algebra. I entered the original polynomial as my answer for part (a) but it was marked wrong (by the automated textbook) with no explanation. What did I do wrong?

Answer 1

Let $y=x^2$ so that the polynomial is $y^2+y-12$. This makes it easier to notice that we can factorize as $(y-3)(y+4)$, i.e., $(x^2-3)(x^2+4)$. Now each of the quadratic factors is irreducible since $\pm\sqrt 3$ and $\pm 2i$ are not rational. Thus for part (b), we have $(x^2-3)(x^2+4)$.

For part (b), we can factorize as $(x-\sqrt 3)(x+\sqrt 3)(x^2+4)$, but again, $x^2+4$ is irreducible since it has no real root.