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Chapter 2: Problem 31

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. $$f(x)=x^{3}+7 x^{2}-4 x-28$$

Short Answer

Expert verified
The zeros of the function \(f(x)=x^{3}+7 x^{2}-4 x-28\) are \(x = 1, x = -7\), and \(x = -4\) each with a multiplicity of 1 and the graph crosses the x-axis at these points.

Step by step solution

01

Factor the Polynomial

To find the zeros of the function, first factor the cubic power polynomial. Using the rational root theorem, possible factors of \(f(x)\) are \(\pm 1, \pm 2,\pm 4, \pm 7, \pm 14, \pm 28\), as those are the factors of 28. Upon trying those, it is easily found that 1 and -7 are real roots of the equation because \(f(1)=1+7-4-28=0\) and \(f(-7)=-343+343+28=0\). Towards this, \(f(x)\) can be factored in the following manner: \(f(x)=(x-1)(x+7)(x+4)\).
02

Find the Zeros

To find the zeros of the function, set each factor equal to zero and solve for \(x\). These zeros are \(x = 1, x = -7\), and \(x = -4\).
03

Determine Multiplicity

Here, as each factor occurs once, therefore the multiplicity of each zero is 1.
04

Analyze the Graph

When the zero has a multiplicity of 1, the graph will cross the x-axis at that point. Hence, the graph crosses the x-axis at \(x = 1, x = -7, and x = -4\).

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Most popular questions from this chapter

Use long division to rewrite the equation for \(g\) in the form $$\text {quotient }+\frac{\text {remainder}}{\text {divisor}}$$ Then use this form of the function's equation and transformations of \(f(x)=\frac{1}{x}\) to graph \(g.\) $$g(x)=\frac{2 x+7}{x+3}$$

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