There's no if about it,
[tex]f(x)=x^3+3x^2-x-3
[/tex]
has a zero [tex]f(1)=0[/tex] so [tex]x-1[/tex] is a factor. That's the special case of the Remainder Theorem; since [tex]f(1)=0[/tex] we'll get a remainder of zero when we divide [tex]f(x)[/tex] by [tex]x-1.[/tex]
At this point we can just divide or we can try more little numbers in the function. It doesn't take too long to discover [tex]f(-1)=0[/tex] too, so [tex]x+1[/tex] is a factor too by the remainder theorem. I can find the third zero as well; but let's say that's out of range for most folks.
So far we have
[tex]x^3+3x^2-x-3 = (x-1)(x+1)(x-r)[/tex]
where [tex]r[/tex] is the zero we haven't guessed yet. Again we could divide [tex]f(x)[/tex] by [tex](x-1)(x+1)=x^2-1[/tex] but just looking at the constant term we must have
[tex]-3 = -1 (1)(-r) = r[/tex]
so
[tex]x^3+3x^2-x-3 = (x-1)(x+1)(x+3)[/tex]
We check [tex]f(-3)=(-3)^3+3(-3)^2 -(-3)-3 = 0 \quad\checkmark[/tex]
We usually talk about the zeros of a function and the roots of an equation; here we have a function [tex]f(x)[/tex] whose zeros are
[tex]x=1, x=-1, x=-3[/tex]
