Wow.
Since the coefficient of x^4 is 1, we need only worry about factors of -24 in searching for possible roots of the given polynomial. Â
I'd use synthetic div. here:
  Â
4Â /Â 1Â Â -5Â Â -52Â Â -70Â Â -24
       4  -4  224   616
  Â
    1  -1  -56  154  592
There is a non-zero remainder (592) which is a long way frorm zero, so 4 is NOT a root. Try synth. div. using other factors of -24: {1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 8, -8, 12, -12, 24, -24} until your syth. div. produces a zero remainder.
What I did next was to graph X^4-5x^3-52x^2-70x-24, I immediately found that -4 is a zero (root), so I did the synth. div. again using -4 as divisor:
-4Â /Â 1Â Â -5Â Â -52Â Â -70Â Â -24
        -4   36   64   24
   Â
    1  -9  -16   -6    0
Thus, -4 has been confirmed to be a root of the given polynomial.
Next, look at the 4 coefficients in front of that 0 remainder. What are factors of -6? {1, -1, 2, -2, 3, -3, 6, -6)
Is -2 a root? Let's try it:
-2Â /Â 1Â Â -5Â Â -52Â Â -70Â Â -24
        -2  14  76   -12
  Â
    1  -7  -38    6  -36   Here we have a remainder of -36, so weÂ
                        conclude that -2 is not a root of the originalÂ
                        polynomial.
Try again. Let's see whether 3 is a root. I'll use the 4 coefficients we found earlier: 1  -9  -16   -6
3Â Â /Â Â 1Â Â -9Â Â -16Â Â Â -6
        3  -18  -102
  Â
     1  -6  -34  -108      No, 3 is not a root because the remainder                             is not zero.
Continue in this manner until you have identified the three roots of the polynomial whose coefficeints are 1  -9  -16   -6.