The other tutor did not factor it completely into
linear factors.

f(x)=3x^4-10x^3+20x^2-40x+32

You say you know that the first zero is 2,
so we do this synthetic division:

2|3 -10  20 -40  32
 |    6  -8  24 -32
  3  -4  12 -16   0

So we have now partially factored f(x) as

f(x) = (x-2)(3x³-4x²+12x-16)

Now we will factor this

3x³-4x²+12x-16

by grouping.  Out of the first two terms 
factor out x²

x²(3x-4)+12x-16

Out of the last two terms factor out +4

x²(3x-4)+4(3x-4)

Factor out (3x-4) and get

(3x-4)(x²+4)

So we have now further partially factored f(x) as

f(x) = (x-2)(3x-4)(x²+4)

Now since i² = -1, -i² = 1, and since 4 = 4*1, 
4*1 also equals 4(-i²) or -4i²,
so replace 4 by -4i²

f(x) = (x-2)(3x-4)(x²-4i²)

Now we can completely factor f(x) by factoring that
last factor as the difference of two squares:

f(x) = (x-2)(3x-4)(x-2i)(x+2i)

To solve it we set each of the factors = 0

x-2=0, 3x-4=0,   x-2i=0,  x+2i=0
  x=2,   3x=4,      x=2i     x=-2i
          x=4/3

So the solutions are 2, 4/3, 2i, and -2i.

Edwin