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Chapter 2: Problem 6

Find the zeros of the polynomial function and state the multiplicity of each zero. $$P(x)=(x+4)^{3}\left(x^{2}-9\right)^{2}$$

Short Answer

Expert verified
The zeros of the polynomial function \( P(x)=(x+4)^{3}(x^{2}-9)^{2} \) are \( x = -4 \) with multiplicity 3, \( x = -3 \) with multiplicity 2, and \( x = 3 \) with multiplicity 2.

Step by step solution

01

Identify and Write Down the Factors

Identify and write down the factors of the function. Factors are \( (x+4) \) and \( (x^2-9) \).
02

Set Each Factor to Zero and Solve for x

Set the factor \( (x+4) \) to zero and solve for \( x \):\n\[(x+4) = 0\]The solution for this equation is \( x = -4 \). \nNow, set the factor \( (x^2-9) \) to zero and solve for \( x \):\n\[(x^2-9) = 0\]The solution for this equation using the difference of squares formula: \( a^2-b^2 = (a+b)(a-b) \), we find that \( x = -3 \) and \( x = 3 \).
03

Determine the Multiplicity of Each Root

The multiplicity of a root is determined by the power on its factor in the polynomial function. In this case, the factor \( (x+4) \) has a power of 3, so the root \( x = -4 \) has a multiplicity of 3. The factor \( (x^2-9) \) has a power of 2, meaning the roots \( x = -3 \) and \( x = 3 \) both have a multiplicity of 2. Therefore, the roots of the function are \( x = -4 \) with multiplicity 3, \( x = -3 \) with multiplicity 2, and \( x = 3 \) with multiplicity 2.

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