The Fundamental Theorem of Algebra states precisely that:
Fundamental Theorem of Algebra. Every nonzero polynomial $p(x)$ with coefficients in $\mathbb{C}$ can be factored, in essentially a unique way, as a product of a constant and linear terms, in the form
$$p(x) = a(x-r_1)\cdots(x-r_n)$$
where $a$ is the leading coefficient of $p(x)$ and $n$ is the degree of $p(x)$.
This result does not hold in general. In some cases, you cannot factor them (you cannot in $\mathbb{R}$, for instance, where $x^2+1$ cannot be written as a product of linear terms; over $\mathbb{Q}$, there are polynomials of arbitrarily high degree that cannot be factored at all, let alone into a product of linear terms). In some cases, the factorization is not unique: in the quaternions, the polnomial $x^2+1$ can be factored in infinitely many essentially distinct ways as a product of linear terms, e.g., $x^2+1 = (x+i)(x-i) = (x+j)(x-j) = (x+k)(x-j) = \cdots$.
A field $F$ is said to be algebraically closed if and only if every nonconstant polynomial $p(x)$ with coefficients in $F$ has at least one root. It is then easy to verify that $F$ is algebraically closed if and only if every nonzero polynomial can be factored into a product of linear terms, as above. Subject to some technical assumptions (The Axiom of Choice), every field is contained in an algebraically closed field, and for every field there is a "smallest" algebraically closed field that contains it, so there is a "smallest" larger field $K$ where you can guarantee that every polynomial factors. This does not hold for arbitrary sets of coefficients (e.g., the quaternions, because they are non-commutative; or rings with zero divisors; and others).
Once you go beyond one variable, it is no longer true that a polynomial can always be factored into terms of degree one, even in $\mathbb{C}$. For example, the polynomial $xy-1$ in $\mathbb{C}[x,y]$ cannot be written as a product of polynomials of degree $1$, $(ax+by+c)(rx+sy+d)$. If you could, then $ar=bs=0$, and we cannot have $a=b=0$ or $r=s=0$, so without loss of generality we would have $b=0$ and $r=0$, so we would have $(ax+c)(sy+d) = xy-1$. Then $ad=0$, so $d=0$ (since $a\neq 0)$, but then we cannot have $cd=-1$. So no such factorization is possible.
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.
Best Answer
The expression 'complex coefficients' was not the best way of describing what they wanted. The better way to phrase the question would have been
In which case your final result is what they want.