A mass of confusion.
First, what you have provided is a polynomial expression rather than a rational expression.
Second, any polynomial of degree n > 0 can be factored into n polynomials of degree 1, perhaps by using complex coefficients. Any polynomial of degree 2n > 0 and real coefficients can be factored into n polynomials of degree 2 with real coefficients. Any polynomial of degree 2n - 1 > 1 and real coefficients can be factored into a polynomial of degree 1 and n polynomials of degree 2. Polynomials of degree 2 (quadratics) with real coefficients may or may not be factorable in turn into polynomials of degree 1 with real coefficients.
Third, finding factorizations, even approximately, for polynomials of degree > 4 will usually require advanced methods. Indeed, factoring even polynomials of degree 3 or 4 may be arduous. If the coefficients are rational, you can use the rational or integer root theorems to find a "nice" factorization if there is one. In other words, factoring ranges from the obvious to the virtually impossible without a computer program.
Fourth, your specific problem involves a quadratic. You can factor any quadratic by using the quadratic formula although in some cases the factoring is more quickly done by reversing the FOIL method. However, the quadratic formula works every time and does not take that long.
[MATH]ax^2 + bx + c = a(x^2 + px + q) = a * \left (x + \dfrac{p + \sqrt{p^2 - 4q}}{2} \right ) \left (x + \dfrac{p - \sqrt{p^2 - 4q}}{2} \right ).[/MATH]
If [MATH]p^2 < 4q[/MATH], there is no factoring with real coefficients.
Finally, if you are trying to determine whether a rational function can be simplified and do not see a factorization, try dividing the polynomial of higher degree by the one of lower degree.