[0] #(x²+3x)/(2x-14)*(x²-49) /(x²-2x-15)# first let's multiply.
[1]#((x²+3x)(x²-49)) /((2x-14)(x²-2x-15)) #, then, #a²-b²=(a-b) (a+b)#
[2]#(x(x+3)(x-7)cancel(x+7))/(2(cancel(x+7)) (x²-2x-15))# now, we see that #color(red)(-3 and 5)# are trivial roots of #(x²-2x-15)#
[3] #(x(cancel(x+3))(x-7))/(2(cancel(x+3))(x-5))#
[4]#(x(x-7))/(2(x-5))#
[5]#(x²-7x)/(2x-10)#