[tex] x-a [/tex] is a polynomial of degree 1, and as such it cannot be factorized.
In fact, when you factor a polynomial [tex] p(x) [/tex] of degree [tex] n [/tex], you write it as
[tex] p(x) = r(x)\cdot s(x),\quad \deg(r)=a,\quad \deg(s)=b,\quad a+b=n[/tex]
So, if a polynomial is already of degree one, you should write it as a product of two polynomials, whose degrees sum to 1.
So, the only option would be
[tex] p(x) = r(x)\cdot s(x) [/tex]
with [tex] r(x) [/tex] of degree 1 and [tex] s(x) [/tex] of degree 0, i.e. a constant polynomial, i.e. a simple number.
But this factorization is trivial, because it only allows you to write expressions like
[tex] x-a = 1\cdot(x-a),\qquad\text{or}\qquad 2\cdot\dfrac{x-a}{2} [/tex]
which are not actual polynomial factorizations.
