Accordding to the rational root the following are potential roots f(x)= 6x^4 + 5x^3 -33x^2 -12x + 20 -5/2 , -2, 1, 10/3 what is an actual root of f(x)

Along with the roots you have listed as your possibilities, there are 28 additional root possibilities when you take into account you have both the positive and the negative of all of them!  Thankfully, a root is found within the 4 possibilities they gave us!  You have to use synthetic division to accomplish this.  I'm going to start easy and use the root x = 1.  Put 1 on the outside of the "box" and put the coefficients from the polynomial inside.  So we have 1  (6   5   -33   -12   20).  Start by bringing down the first coefficient which is 6.  Multiply that by the 1 to get 6.  Put that 6 up under the 5 and add to get 11.  Multiply 11 by 1 to get 11.  Put that 1 up under the -33 and add to get -22.  Multiply that -22 by the 1 to get -22.  Put that up under the -12 and add to get -34.  Multiply that by 1 to get -34.  Put that up under the 20 and add to get -14.  Because that last number there is not a 0, that means that the factor (x-1) doesn't go into the polynomial evenly; therefore, x = 1 is not a root or a solution to the polynomial.  Next I tried -2 and got a remainder of -32, so x = -2 is not a solution either.  As much as I didn't want to try a fraction, it is what it is, so I chose -5/2 next.  Bring down the 6 and multiply it by -5/2 to get -15.  Put that up under the 5 and add to get -10.  Multiply that by -5/2 to get 25.  Put that under the -33 and add to get -8.  Multiply that by -5/2 to get 20.  Put that under the -12 and add to get 8.  Multiply that by -5/2 to get -20.  Put that up under the 20 and add to get 0.  That means that -5/2 or (2x+5) is our first root.  The polynomial left over, the depressed polynomial is a degree lower than what we started with (it will be a 3rd degree), and its coefficients are made from the numbers we got when we added in our process of synthetic division.  It would be .  That would be factored then to find the remaining roots.  The number of roots a polynomial has is always equal to the degree of the original polynomial.  Our polynomial is a 4th degree so we will have 4 roots.  Factoring it finds all the roots.