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Chapter 2: Problem 26

Find the zeros for each polynomial function and give the multiplicity for each zero. State whether the graph crosses the \(x\) -axis, or touches the \(x\) -axis and turns around, at each zero. $$f(x)=3(x+5)(x+2)^{2}$$

Short Answer

Expert verified
Based on the steps followed, the function \(f(x)=3(x+5)(x+2)^{2}\) has zeros at \(x = -5\) and \(x = -2\). The zero -5 has a multiplicity of 1 (odd), meaning that the graph crosses the x-axis at -5. The zero -2 has a multiplicity of 2 (even), meaning that the graph touches the x-axis at -2 and turns around.

Step by step solution

01

Find the Zeros

Set each factor equal to zero and solve for \(x\). So, for \(x+5=0\), we get \(x=-5\) and for \(x+2=0\) we get \(x=-2\).
02

Determine the Multiplicity

The multiplicity of the zeros is the power of their corresponding factor in the equation. The zero -5 has a multiplicity of 1 since \((x+5)\) has a power of 1 in the equation. The zero -2 has a multiplicity of 2 since \((x+2)^{2}\) has a power of 2 in the equation.
03

Analyze the Graph Behaviour

If the multiplicity of a zero is odd, then the graph will cross the x-axis at the zero. On the other hand, if multiplicity is even, then the graph will touch the x-axis at that zero and turn around. Here, the zero -5 has an odd multiplicity so the graph will cross the x-axis at -5. The zero -2 has an even multiplicity so the graph will touch the x-axis at -2 and turn around.

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