f(x)=5x4-7x³+17x²-21x+6

The possible rational zeros (if any) are 
positive or negative fractions whose numerators
are factors of 6 and whose denominators are
factors of 5

The factors of 6 are 1,2,3,6  (possible numerators)

The factors of 5 are 1,5  (possible denominators)

Possible fractions are 1/1, 1/5. 2/1, 2/5, 3/1, 3/5, 6/1, 6/5

They reduce to 1, 1/5, 2, 2/5, 3, 3/5, 6, 6/5

They could be positive or negative, so all possible rational
zeros are:

±1, ±1/5, ±2, ±2/5, ±3, ±3/5, ±6, ±6/5

We will try the easiest one first, which is 1, using synthetic division
to see if we get 0 remainder:

1|5 -7  17 -21  6
 |   5  -2  15 -6 
  5 -2  15  -6  0

We're in luck.  It worked.  So we have now

factored the original

f(x)=5x4-7x³+17x²-21x+6

as

f(x)= (x - 1)(5x³ - 2x² + 15x - 6)

Now we try to factor the polynomial in the second parentheses:

First we change the parentheses ( ) to brackets [ ] so we can put
parentheses inside with less confusion:

f(x)= (x - 1)[5x³ - 2x² + 15x - 6]

Out of the first two terms in the bracket we can factor out x²:

f(x)= (x - 1)[x²(5x - 2) + 15x - 6]

Out of the last two terms we can factor out 3:

f(x)= (x - 1)[x²(5x - 2) + 3(5x - 2)]

Now we have a common factor of (5x - 2)

f(x) = (x - 1)[(5x - 2)(x² + 3)]

We don't need the brackets anymore:

f(x) = (x - 1)(5x - 2)(x² + 3)

To find the zeros we set each factor = 0

x - 1 = 0;   5x - 2 = 0;   x² + 3 = 0
    x = 1;       5x = 2;       x² = -3 _
                  x = 2/5       x = ±i√3 (not real)

So the correct choice is D

Edwin