2x^5+12x^4+16x^3-12x^2-18x = 0
take out common factor
2x(x^4 + 6x^3 + 8x^2 - 6x - 9) = 0 so one root is x = 0.
x(x^4 + 6x^3 + 8x^2 - 6x - 9) = 0 by simple observation/substitution, two other roots are x = 1 and x = -1.
x(x - 1)(x^3 + 7x^2 + 15x + 9) = 0
x(x - 1)(x + 1)(x^2 + 6x + 9) - 0 the final quadratic is a perfect square.
x(x - 1)(x + 1)(x + 3)^2 = 0
The roots are: 0,1, -1, -3 (twice)