There are many odd things about how math is taught. I am not sure that there are many real-world problems that require factoring. Factoring is usually a convenience rather than a necessity. And if finding a factoring takes a lot of time, then it is not even a convenience. Frequently (but not always) a common factor in each term provides the clue if there is a simple factorization. (There is no guarantee that there is any factorization of an expression unless it is a polynomial, and there is no guarantee that the factoring will be simple even if the expression is factorable.)
Let's start with \(\displaystyle x^3y - 4xy.\)
It should be obvious that both x and y are factors of every term. So
\(\displaystyle x^3y - 4xy = xy(x^2 - 4).\) Now you should see a difference of like powers.
\(\displaystyle x^3y - 4xy = xy(x^2 - 4) = xy(x - 2)(x + 2).\)
\(\displaystyle 3x^{3/2} - 9x^{1/2} + 6x^{-1/2}.\)
This one is a lot harder. It is obvious that 3 is a common factor in each term, but it may be harder to see that \(\displaystyle x^{-1/2}\) is also a common factor.
\(\displaystyle x^{3/2} = x^{4/2} * x^{-1/2} = x^2 * x^{-1/2} \text { and } x^{1/2} = x^{2/2} * x^{-1/2} = x * x^{-1/2}.\)
\(\displaystyle \therefore 3x^{3/2} - 9x^{1/2} + 6x^{-1/2} = 3(x^{3/2} - 3x^{1/2} + 2x^{-1/2})=\)
\(\displaystyle 3x^{-1/2}(x^2 - 3x + 2) = \dfrac{3(x - 2)(x - 1)}{\sqrt{x}} = \dfrac{3(x - 2)(x - 1)\sqrt{x}}{x}.\)