X^4-5x^3-52x^2-70x-24=0the solution set

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.  

I'd use synthetic div. here:

     
4  /  1   -5   -52    -70    -24
             4    -4    224     616
     
        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:

-4  /  1   -5   -52    -70    -24
             -4     36     64     24
      
        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:

-2  /  1   -5   -52    -70    -24
              -2    14    76      -12
     
        1   -7   -38       6    -36     Here we have a remainder of -36, so we 
                                                conclude that -2 is not a root of the original 
                                                polynomial.

Try again.  Let's see whether 3 is a root.  I'll use the 4 coefficients we found earlier:  1   -9   -16     -6

3   /   1   -9   -16     -6
               3    -18   -102
     
          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.