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Chapter 2: Problem 5

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

Short Answer

Expert verified
The zeros of the polynomial \(P(x)=(x^{2}-4)(x+3)^{2}\) are -3, 2, and -2. The zero -3 has multiplicity 2, while the zeros 2 and -2 each have multiplicity 1.

Step by step solution

01

Identify and set up the zeros

The zeros of the function are found by solving the equation \(P(x)=(x^{2}-4)(x+3)^{2}=0\). So set \(x^{2}-4=0\) and \((x+3)^{2}=0\) and solve these equations.
02

Solve for \(x^{2}-4=0\)

Solving for \(x^{2}-4=0\) gives \(x^2 = 4\), so \(x=\pm\sqrt{4}\). It leads to two solutions: \(x=2\) and \(x=-2\). This is because \((x-2)\) and \((x+2)\) are factors of \(x^{2}-4\), and each has multiplicity 1.
03

Solve for \((x+3)^{2}=0\)

Solving for \((x+3)^{2}=0\) gives \(x+3=0\), so \(x=-3\). Because the factor \((x+3)\) is squared, it has multiplicity of 2. Squaring a factor increases its multiplicity.

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