x^3 - 3x^2 - 9x - 5
It ends in -5. The factors of 5 are 1 and 5, so whatever might factor out of x^3 - 3x^2 - 9x - 5 should be (x-1), (x+1), (x-5), or (x+5) if the problem is going to factor nicely. It's entirely possible that x^3 - 3x^2 - 9x - 5 has some weird non-integer roots (something like x - third root of 5), but let's try the simple possible roots first.
Let's try (x-1)
x^3 - 3x^2 - 9x - 5
x^2: x^3 - x^2
-2x^2 - 9x - 5
-2x: -2x^2 +2x
-11x - 5
-11: -11x + 11
-16
Not 0, so (x-1) doesn't factor out of x^3 - 3x^2 - 9x - 5
.
Let's try (x+1)
x^3 - 3x^2 - 9x - 5
x^2: x^3 + x^2
-4x^2 - 9x - 5
-4x: -4x^2 - 4x
5x - 5
5: 5x - 5
0
It works. x^3 - 3x^2 - 9x - 5 = (x-1)(x^2 - 4x + 5)
x^2 - 4x + 5 factors to (x-1)(x-5)
Answer: x^3 - 3x^2 - 9x - 5 factors to (x-1)(x-1)(x-5)