f(x) = 8x4 - 24x3 - 424x2 - 72x

We set the right side = 0

8x4 - 24x3 - 424x2 - 72x = 0

We can factor out 8x:

8x(x3 - 3x2 - 53x - 9) = 0

So by the zero factor principle:

8x = 0         x3 - 3x2 - 53x - 9 = 0
 x = 0

So we see that one rational solution is 0

Next we try to solve x3 - 3x2 - 53x - 9 = 0

If it has any rational solutions they must be
± some divisor of 9

Possible rational solutions of x3 - 3x2 - 53x - 9 = 0
are ±1, ±3, ±9

It has 1 sign change so we know it has one positive
solution.

Try x = 1

1 |1 -3 -53  -9
  |   1  -2 -55
   1 -2 -55 -64

No, 1 is not a solution, since the remainder is not 0

Try x = 3

3 |1 -3 -53   -9
  |   3   0 -159
   1  0 -53 -168

No, 3 is not a solution, since the remainder is not 0
 
Try x = 9

9 |1 -3 -53 -9
  |   9  54  9
   1  6   1  0

Hurray! 9 is a solution because the remainder is 0.

Now we know that 0 and 3 are rational zeros of f(x).

Now we have factored f(x) as

f(x) = 8x(x - 9)(x2 + 6x + 1)

So all the rational zeros of f(x) are 0 and 9

However there are two irrational solutions obtainable by
setting x2 + 6x + 1 = 0 and using the quadratic formula:

x =  

x =  

x = 

x = 

x = 

x = 

x =  

x =  

x =  

But you weren't asked for those two irrational zeros.

The only rational zeros are 0 and 9.

Edwin