This has no sign changes, so there are no positive zeros.

The rational zeros, then if, any, are among the negatives common 
fractions whose numerators are divisors of the constant 
term and whose denominators are divisors of the leading coefficient.

The factors of 6 are 1,2,3,6
The factors of 25 are 1,5,25

We make all possible negative fractions whose numerators are 
divisors of 6 and whose denominators are divisors of 25:

-1/1, -1/5, -1/25, -2/1, -2/5, -2/25, -3/1, -3/5, -3/25, -6/1,
-6/5, -6/25

Simplifying: 

-1, -1/5, -1/25, -2, -2/5, -2/25, -3, -3/5, -3/25, -6, 
-6/5, -6/25

If f(x)=25x^4+51x^3+77x^2+153x+6 has any zeros, they are in that
list.

We try the easiest first:

Try x=-1

-1 | 25  51   77   153    6
   |    -25  -26   -51 -102 
     25  26   51   102  -96

The remainder is -96, not 0.  So -1 is not a zero.

We try the next easiest x=-2

-2 | 25  51   77   153    6
   |    -50   -2  -150   -6 
     25   1   75     3    0

The remainder is 0, so x=-2 is a zero of f(x).
Since x=-2 is equivalent to x+2=0, (x+2) is a factor
of f(x) and f(x) is now partially factored using the
4 numbers on the bottom row left of the 0 remainder.
So:

f(x) = (x+2)(25x³+x²+75x+3)

Now we only need look for zeros of 25x³+x²+75x+3 

It has no sign changes, so 

if it has any rational zeros, they are among these

-1/5, -1/25, -3, -3/5, -3/25

We try the easiest -3

-3 | 25   1   75    3    
   |    -75  222 -891
     25 -74  297 -888

The remainder is not 0, so -3 is not a zero.

We try -1/5, and since its decimal equivalent
is simply -0.2, That will be easier:

-0.2 | 25   1  75      3
     |     -5   0.8  -15.16
       25  -4  75.8  -12.16

The remainder is not 0, so -1/5 or -0.2 is not
a zero.

We try -1/25, and since its decimal equivalent
is simply -0.04, That will be easier:

-0.04 | 25   1  75   3
      |     -1   0  -3
        25   0  75   0

The remainder is 0, so x=-0.04 is a zero of 
25x³+x²+75x+3 and also f(x).  Since x=-0.04 is 
equivalent to x+0.04=0, (x+0.04) is a factor
of f(x) and f(x) is now partially factored using the
3 numbers on the bottom row left of the 0 remainder.
So:

f(x) = (x+2)(x+0.04)(25x²+75)

We factor out 25 from the last parentheses

f(x) = (x+2)(x+0.04)25(x²+3)

Finally if we distribute the 25 into the factor (x+0.04)
we get the factor (25x+1), which has no decimals.

So the complete factorization of f(x) over the integers
and the real numbers is 

f(x) = (x+2)(25x+1)(x²+3)

Edwin