We have the given polynomial as P(x) = -x(- x - 2)2(x + 2)

P(x) = -x(- x - 2)2(x + 2)

=> P(x) = -2x(-x - 2)(x + 2)

=> P(x) = 2x(x + 2)(x + 2)

=> P(x) = 2x(x + 2)^2

The zeros are 0 and -2

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We have the given polynomial as P(x) = -x(- x - 2)2(x + 2)

P(x) = -x(- x - 2)2(x + 2)

=> P(x) = -2x(-x - 2)(x + 2)

=> P(x) = 2x(x + 2)(x + 2)

=> P(x) = 2x(x + 2)^2

The zeros are 0 and -2

**Therefore the required zeros are **

**-2 with multiplicity 2 and 0 with multiplicity 1. **

Given the polynomial P(x) = -x (-x-2)^2 (x+2)

We need to determine the zeros of P(x).

We notice that P(x) is already written in its factors form.

Then the zeros of P(x) are the zeros of the factors.

The zero of -x is 0

The zero of (-x-2)^2 is -2

The zero of (x+2) is -2.

Then the zeros of P(x) are 0 and -2.

Since the factor (-x-2)^2 is a square , then the multiplication is 2.

Then the zeros are :

**x1 = 0 ==> multiplication is 1**

**x2= -2 => multiplication is 2.**