Question

given p(x) = -(x + 2)²(x - 1)(x + 3)(x - 2), which list shows all zeros including multiplicities?
-3, -2, 1, 2 (with -2 multiplicity 2)
-3, -1, 1, 2 (with 2 multiplicity 2)
-3, -2, -1, 1, 2
-3, -2, -2, 1, 2

Explanation:

Step1: Recall Zero Product Property

To find the zeros of a polynomial \( p(x) \), we set \( p(x) = 0 \). For a factored polynomial \( p(x)=-(x + 2)^2(x - 1)(x + 3)(x - 2) \), we use the zero - product property: if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \).

Step2: Find each zero and its multiplicity

• For the factor \( (x + 2)^2 \): Set \( x+2 = 0 \), then \( x=-2 \). The exponent of \( (x + 2) \) is 2, so the multiplicity of \( x = - 2 \) is 2.
• For the factor \( (x - 1) \): Set \( x - 1=0 \), then \( x = 1 \). The exponent of \( (x - 1) \) is 1, so the multiplicity of \( x = 1 \) is 1.
• For the factor \( (x + 3) \): Set \( x+3 = 0 \), then \( x=-3 \). The exponent of \( (x + 3) \) is 1, so the multiplicity of \( x=-3 \) is 1.
• For the factor \( (x - 2) \): Set \( x - 2=0 \), then \( x = 2 \). The exponent of \( (x - 2) \) is 1, so the multiplicity of \( x = 2 \) is 1.

When we list the zeros including multiplicities, we have to list the zero \( - 2 \) twice (because of multiplicity 2) and the zeros \( - 3 \), \( 1 \), \( 2 \) once each. So the list of zeros including multiplicities is \( - 3,-2,-2,1,2 \).

Answer:

The correct option (assuming the cyan - colored option has the list \( - 3,-2,-2,1,2 \)) is the one with the list \( - 3,-2,-2,1,2 \) (the cyan - colored option in the given choices). If we consider the options as per the description: the option with the list \( - 3,-2,-2,1,2 \) (the last option with the cyan background).