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Chapter 2: Problem 32

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. $$f(x)=x^{3}+5 x^{2}-9 x-45$$

Short Answer

Expert verified
The zeros of the function \(f(x)=x^{3}+5 x^{2}-9 x-45\) are \(x = 3, -3, -5\). Each zero has a multiplicity of 1. The graph of the function will cross the x-axis at these points.

Step by step solution

01

Find the zeros

The first step is finding the zeros of the polynomial. This implies solving the equation \(f(x) = 0\), or in the provided example, \(x^3 + 5x^2 - 9x - 45 = 0\). One effective way of solving cubic equations like this is by factoring. Factoring by the rational root theorem, yields: \((x-3)(x+3)(x+5)=0\). Thus, the polynomial has three roots, \(x = 3\), \(x = -3\), and \(x = -5\).
02

Determine the multiplicity

The second step is determining the multiplicity of each zero. The multiplicity of a zero refers to how many times it occurs as a solution to the equation \(f(x) = 0\). Here, each factor of the equation appears to the power of one. Therefore, each zero occurs once, and their multiplicities are 1.
03

Graph behavior at zeros

The last step is determining the behavior of the graph at each zero. If the multiplicity of a root is odd, the graph will cross the x-axis at that zero. If the multiplicity of a root is even, the graph will touch the x-axis and turn around at that zero. Here, all zeros have odd multiplicity (1), so the graph of the function will cross the x-axis at \(x = 3, -3, -5\).

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