# Give the zeros of polynomial P and list their multiplicities : P(x) = -x(- x - 2)^2(x + 2).

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### 3 Answers

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.**

The function P(x) = -x(- x - 2)^2(x + 2)is not polynomial, as (-x-2)^2(x+2) is exponential.

To find the zeros of P(x) = -x(- x - 2)^2(x + 2).

The zeros of P(x) are those values of x for which P(x) = 0.

=> the values of x for which -x(- x - 2)^2(x + 2) = 0.

=> (-x)*(-x-2)^2(x+2) = 0.

=> -x = 0, or (-x-2)^2(x+2) = 0.

(-x-2)^2(x+2) = 0 gives: (x+2)^2(x+2) * (-1)^2x = 0.

=> (x+2)^2(x+2) = 0.

=> y^2y = 0. There is no number for which y^2y =0, as 0^0 is undefined.

**Therefore x = 0 is the only zero of P(x).**