(3x^2-3x-60)/(x^2+9x+20)=3(x^2-x-20)/(x+5)(x+4)=3(x-5)(x+4)/((x+5)(x+4)). The factor x+4 cancels as long as x is not -4 (restriction). The result is 3(x-5)/(x+5). The restriction is to avoid dividing by zero, which cannot be evaluated because the result is infinity. The process involved factorising the numerator and denominator which exposed the common factor. The number 3 was also a factor in the numerator so could be set apart outside the bracketed factors. The process of factoring was to see what factors of each of the constant terms differed by 1 in the numerator and whose was 9 in the denominator. 20 has factors 4 and 5 whose difference is 1 and whose sum is 9.