The answer to your questions is that it depends on how we view the polynomial. In particular,
[tex]c^2+5c=3\implies c^2+5c-3=0[/tex]
If the left hand side were factorizable, then we would be able to write it in the form
[tex](c-r_1)(c-r_2)=c^2+5c-3=0[/tex]
If we expand the leftmost expression, we'd get
[tex]c^2-(r_1+r_2)c+r_1r_2=c^2+5c-3=0[/tex]
and so for the two polynomials to be the same, the coefficients must match. In other words, the unknowns [tex]r_1,r_2[/tex] would have to satisfy
[tex]\begin{cases}-(r_1+r_2)=5\\r_1r_2=-3\end{cases}[/tex]
Suppose that, moreover, we want integer solutions for [tex]r_1,r_2[/tex]. For this to happen, they must be factor pairs of the constant term.
-3 only has two factor pairs. Either [tex]r_1=-1[/tex] and [tex]r_2=3[/tex], or [tex]r_1=1[/tex] and [tex]r_2=-3[/tex]. In the first case, we'd get a linear coefficient of [tex]-(-1+3)=-2\neq5[/tex], while in the second, we'd get [tex]-(1-3)=2\neq5[/tex].
There is no integer solution for this system, so the original quadratic is not factorizable - but only so over the integers.
If we change the scope of the coefficients, i.e. allow for any real numbers/complex numbers to appear in the factorization, then we always factorize a quadratic. The above system is easy to solve.
[tex]r_1r_2=-3\implies r_2=-\dfrac3{r_1}[/tex]
[tex]\implies-\left(r_1-\dfrac3{r_1}\right)=5[/tex]
[tex]\implies{r_1}^2+5r_1-3=0[/tex]
[tex]\implies r_1=\dfrac{-5\pm\sqrt{37}}2[/tex]
[tex]\implies c^2+5c-3=\left(c+\dfrac{5-\sqrt{37}}2\right)\left(c+\dfrac{5+\sqrt{37}}2\right)[/tex]
so the original quadratic is factorizable over the reals.
