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Chapter 2: Problem 92

Use a graphing utility to graph the function. Use the zero or root feature to approximate the real zeros of the function. Then determine the multiplicity of each zero. $$h(x)=\frac{1}{5}(x+2)^{2}(3 x-5)^{2}$$

Short Answer

Expert verified
The function \(h(x)=\frac{1}{5}(x+2)^{2}(3 x-5)^{2}\) has two real zeros or roots: \(x = -2\) and \(x = \frac{5}{3}\), both with multiplicity 2.

Step by step solution

01

Identifying Potential Zeros from the Function

First, look for potential zeros in the function \(h(x)=\frac{1}{5}(x+2)^{2}(3 x-5)^{2}\). These are the values of \(x\) that make the function equal to zero and they are found by setting each factor to zero and solving for \(x\). For this function, setting \((x+2)\) and \((3x-5)\) equal to zero yields two potential zeros, \(x = -2\) and \(x = \frac{5}{3}\), respectively.
02

Graphing the Function

Next, graph function \(h(x)=\frac{1}{5}(x+2)^{2}(3 x-5)^{2}\). An efficient way to graph the function is by using a graphing utility.
03

Approximating the Real Zeros

After graphing the function, use the zero or root feature of the graphing utility to find the x-values where the function intersects the x-axis. These are the real zeros or roots. In this case, the algorithm should return the values \(x = -2\) and \(x = \frac{5}{3}\).
04

Determine the Multiplicity of Each Zero

Finally, the multiplicity of roots is given by the power of the factor in the function. In this case, both \((x+2)^{2}\) and \((3x-5)^{2}\) have a power of 2 which means the roots \(x = -2\) and \(x = 5/3\) each have multiplicity 2.

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