How do I factor x^3-9x-440=0?

　
We could use rational root theorem to find a rational root, and from this find the linear factor, then use synthetic division to find quadratic factor. Unfortunately 440 has 16 positive divisors, so there are 32 possible rational (in this case integer) roots.

So maybe we can use some method other than trying to calculate value of function at all these possible values of x.

Rewrite function as follows:

x³ − 9x − 440 = 0
x³ − 9x = 440
x (x² − 9) = 440
x (x − 3) (x + 3) = 440

Now we find prime factorization of 440 = 2^3 × 5 × 11
Can we find 3 factors of 440 that differ by 3?
Yes, and quite easily too. We can clearly see that 8 × 5 × 11 = 440
So we get:
x (x − 3) (x + 3) = 8 (8 − 3) (8 + 3) = 8 × 5 × 11 = 440

So x = 8 is a root of x³ − 9x − 440, and (x − 8) is a factor.
We can now use synthetic division to find quadratic factor.

8 | 1 .... 0 .. −9 .. −440
.. | ....... 8 ... 64 ... 440
.. ---------------------------
... 1 ..... 8 ... 55 ..... 0

Quadratic factor is (x² + 8x + 55), which cannot be further factored

x³ − 9x − 440 = 0
(x − 8) (x² + 8x + 55) = 0

Let f(x) = x³ - 9x - 440
f(8) = (8)³ - 9(8) - 440 = 0
Then (x - 8) is a factor of x³ - 9x - 440

x³ - 9x - 440 = 0
x³ - 8x² + 8x² - 9x - 440 = 0
(x³ - 8x²) + 8x² - 64x + 64x - 9x - 440 = 0
(x³ - 8x²) + (8x² - 64x) + (55x - 440) = 0
x²(x - 8) + 8x(x - 8) + 55(x - 8) = 0
(x - 8)(x² + 8x - 55) = 0

f (8) = 512 - 72 - 440 = 0
Thus x - 8 is a factor
Find other factors by synthetic division :-
8 I 1_____0_____-9_____-440
_ I______ 8_____64_____440
_ I 1____ 8_____55______0

Factors are :-
[ x - 8 [ x² + 8x + 55 ] = 0

I like to use a graphing calculator to give a clue as to roots. It shows the one positive real root is x=8, and the remaining factor is (x+4)^2+39 = x^2 +8x +55.

x^3 - 9x - 440 = 0
x(x^2 - 9) - 440 = 0
x(x + 3)(x - 3) - 440 = 0

(x - 8) (x^2 + 8 x + 55) = 0