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Chapter 3: Problem 30

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. $$f(x)=x^{3}+4 x^{2}+4 x$$

Short Answer

Expert verified
The zeros of the polynomial are \(x=0\) with multiplicity 1 and \(x=-2\) with multiplicity 2. At \(x=0\) the graph crosses the \(x\)-axis and at \(x=-2\) it touches the \(x\)-axis and turns around.

Step by step solution

01

Set the Polynomial to Zero

First step is to set the given polynomial function to zero: \(x^{3}+4x^{2}+4x=0\).
02

Factor out the Common Factor

Next step is to take out the common factor which is \( x \) from this equation: \(x(x^{2}+4x+4)=0\).
03

Factor the Quadratic

The quadratic \(x^{2}+4x+4\) is a perfect square quadratic and can be factored as \((x+2)^{2}\). Substituting this gives us: \(x(x+2)^{2}=0\).
04

Use the Zero Product Property

The Zero Product Property states that if a product of factors equals zero, then at least one of the factors must be zero. Setting each factor to zero will give the solutions. So \(x=0\) and \(x+2=0\) which further simplifies to \(x=0\) and \(x=-2\).
05

Determine the Multiplicity

The root \(x=0\) appears once and so has a multiplicity of 1. \(x=-2\) is a repeated root or zero - it appears twice, hence it has a multiplicity of 2
06

Determine Where the Graph Crosses the x-axis

When the multiplicity is odd, the graph crosses the x-axis. Therefore, at \(x=0\) the graph crosses the x-axis because its multiplicity is 1 (which is odd). When the multiplicity is even, the graph just touches the x-axis and turns around. So, the graph touches the x-axis at \(x=-2\) and turns around because it has an even multiplicity of 2.

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