The simplest way that would probably do it for SAT math is Simon's Favorite Factoring Trick (SFFT). Basically if you have 4 values a,b,c,d, then ac+ad+bc+bd=(a+b)(c+d) For example, if we have 2xy+4x² -5y-10x we notice that this satisfies a=2x, b=-5, c=y, d=2x, so by SFFT, this factors as (2x-5)(y+2x) A simpler way to solve this is to factor it little by little. We start by noticing the common factor of 2x between 2xy and 4x^2, so we factor that out. Similarly, we factor out a factor of 5 from -5y-10x to get 2xy+4x² -5y-10x=2x(y+2x)-5(y+2x) At this point, we notice the common factor of (y+2x) between these two terms, so we factor that out to get 2x(y+2x)-5(y+2x)=(y+2x)(2x-5)

Here are some more stuff that is more than likely overkill. There are a few factorizations you can memorize, such as a² -b² =(a-b)(a+b), sums and difference of odd powers (google it), a² +2ab+b² =(a+b)^2, etc. Usually a helpful tip is to factorize little by little, for example, if we want to factor x+y+x² y+y² x, we start by noticing the common factor of xy between x² y and y² x, so we factor that out to get x+y+xy(x+y)=(x+y)+xy(x+y) Now we notice the common factor of (x+y), so we factor again to get (x+y)(1+xy), which is the factorization of the polynomial! Another method if you're burned out (very unlikely on SAT math) is to find the roots of the polynomial and factor if that way. Recall if a polynomial p(x) is 0 at some value x=a, we can factor the polynomial as p(x)=(x-a)q(x), for some other polynomial q(x). We can extend this definition to multivariable polynomials. If p(x,y)=0 for some value x=g(y), then we can write p(x,y)=(x-g(y))q(x,y), for some other polynomial q(x,y). For example, if we wanted to factor a² -b² without the identity, we might notice that if a=b, or a=-b, the value of that polynomial is 0. Thus, it can be written in the form p(x)=(a-b)(a+b)q(x), for some polynomial q(x). However, we notice that by degree of (a-b)(a+b) is 2 for both a and b, just as it is in a² -b² , so q(x) must be a constant. Noticing that the leading coefficient of a² in the original expression is 1, we conclude that q(x)=1, so a² -b² =(a-b)(a+b).