an x-intercept happens when y=0
so you can try solving 0=f(x), which in this case is a bit more tedious than necessary
you have a multiplication of three terms, the brackets
for the multiplication to result in 0 one of the terms has to be 0
so the question is which of these terms can we make 0 by inserting some value for x:
(x + 1)=0
-> x=-1
(x - 3)=0
-> x=3
(x - 4)=0
-> x=4
so there are 3 x-intercepts, at x=-1, 3 and 4
