2x^5+12x^4+16x^3-12x^2-18x=0

Question

Identify the roots of the equation. State the multiplicity of each root.

Answer 1

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)