Question 285906
[(x-7)/(x-4)^2] * [(x^2-3x-28)/(x-7)^2]


[(x-7)/(x-4)(x-4)] * [(x-7)(x+4)/(x-7)(x-7)]


[1/(x-4)(x-4)] * (x+4)



Answer: [(x-7)/(x-4)^2] * [(x^2-3x-28)/(x-7)^2] = (x+4)/(x^2-8x+16)