Find solutions of the cubic equation:

`x^3-4x^2-3x+18=0`

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Using Real Zeros Theorem we will have a possible zeros: +/-1, +-/2,+/-3,+/-6,+/-9,+/-18.

Let us try x = -2. Let us use Synthetic Division.

-2| 1 -4 -3 18

+

-2 12 -18

--------------------------------------

1 -6 9 0

So, we will have (x^3 - 4x^2 - 3x + 18)/(x + 2) = x^2 - 6x + 9.

We can factor the result as (x - 3)^2.

Equating it to zero.

(x - 3)^2 = 0

Take the square root of both sides.

x - 3 = 0

Add 3 on both sides.

x = 3.

Therefore, **x = {-2, 3}**.

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