Start by listing "x = ↏" as many times as the multiplicity
of each zero, for every zero:

x = -2,  x = -2, x = 0, x = 3

[Notice that I listed x = -2 twice because it has 
multiplicity 2]

Get 0 on the right of each of those equations:

x + 2 = 0, x + 2 = 0, x = 0, x - 3 = 0

[Notice that x = 0 already had a 0 on the right side]

Now indicate the multiplication of the left sides
of all of those equations and set it = 0

(x + 2)(x + 2)(x)(x - 3) = 0

Now multiply that out.

Multiplying the first two factors:

(x² + 4x + 4)(x)(x - 3) = 0

Multiply the first two factor of this:

(x³ + 4x² + 4x)(x - 3) = 0

Multiply those:

x⁴- 3x³ + 4x³ - 12x² + 4x² - 12x = 0

Collect like terms:

x⁴+ x³ - 8x² - 12x = 0

The polynomial itself does not contain the = 0
on the end, so the answer is just the left side:

P(x) = x⁴+ x³ - 8x² - 12x

Edwin