Find all the roots of the equation x^4-4x^3+x^2+12x-12=0

Hi there!
The question here asks us to find all the roots, or solutions, of the quartic equation x⁴ - 4x³ + x² + 12x - 12 = 0. We can start off by using the Rational Root Theorem, which calls for the constant over the leading coefficient, which for us would be 12 over 1. Since we need to list all the factors of 12, they all become + or -. 
+12 / + 1
=> +1, +2, +3, +4, +6, +12 / +1
Now, just simplify each one. We get the possible roots 1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 12, -12. The next part is the most tedious. Plug in each possible factor for x and see if the resulting value is 0. We get that only root 2 makes the whole equation equal to zero. However, we are not done yet. The question asks for the ALL the roots of the equation, not just real, but also imaginary. Let's use synthetic division to find the other roots. 
We can set the divisor to 2, as (x - 2) = 0 is equal to x = 2. 
2| 1 -4 1 12 -12
       2 -4 -6 12
    1 -2 -3 6  0
We can now factor the whole original equation to (x - 2)(x³ - 2x² -3x + 6) = 0. Now, we have to factor the expression x³ - 2x² -3x + 6, and grouping seems like the most appropriate way.
x³ - 2x² -3x + 6 = 0
(x³ - 2x²) + (-3x + 6) = 0
x²(x - 2) - 3(x - 2) = 0
(x² - 3)(x - 2) = 0
Next, use the Zero Product Property to get the values of x.
x² - 3 = 0
x² = 3
x = √3
x ≈ + 1.73
x - 2 = 0
x = 2
Therefore, the roots of the equation x⁴ - 4x³ + x² + 12x - 12 = 0 are x = 2,x = 2 (repeated solution), x = 1.73, x = -1.73. Hope this helped and have a great day!

Um what? rephrase that ?