f(x) = 2x^{3} - 21x^{2} + 57x - 18

I use the synthetic division to find a zero that will give me a remainder of zero based on the remainder theorem.

These are the possible solutions: ±1, ±2, ±3, ±6, ±9, and ±18.

Turns out 6 will work because of two things:

1) f(6) = 2(6)^{3} - 21(6)^{2} + 57(6) - 18 = 432 - 756 + 342 - 18 = 0

2) __6__| 2 -21 57 -18

^{↓} __ 12 -54 18 __

2 -9 3 |__0|__

**(x - 6)(2x**^{2}** - 9x + 3)**

If finding solutions, then we will use the quadratic formula to find the zeros for 2x^{2} - 9x + 3.

x = [-b ± √(b^{2} - 4ac)] ÷ 2a

x = [-(-9) ± √((-9)^{2} - 4(2)(3))] ÷ 2(2)

x = [9 ± √(81 - 24)] ÷ 4

x = [9 ± √(57)] ÷ 4

**x = [0.363 , 4.137]**

In this case, there are 3 solutions. x = 0.363, 4.137, and 6.