If [tex]f(1)=0[/tex], then by the remainder theorem [tex]\dfrac{f(x)}{x-1}[/tex] has a remainder of 0, which means [tex]f(x)[/tex] is divisible by [tex]x-1[/tex]. In fact,
[tex]x^3+3x^2-x-3=x^2(x+3)-(x+3)=(x^2-1)(x+3)=(x-1)(x+1)(x+3)[/tex]
So we know 1 is a root, and the other two roots are -1 and -3.
