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Chapter 3: 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 given polynomial are at \(x=-4, x=2, x=-3.5\), each with a multiplicity of 1. The graph crosses the x-axis at each of these points.

Step by step solution

01

Identify the zeros

Set the polynomial equal to zero and solve for x. This will give you the zeros of the polynomial. In this case, set \(x^{3}+7 x^{2}-4 x-28 = 0\). This cubic polynomial can be factored using synthetic division or factoring techniques (grouping, completing the square, etc.). By factoring, we find that \((x+4)(x-2)(x+3.5)=0\). Therefore, the zeroes are at \(x=-4, x=2, x=-3.5\).
02

Determine the multiplicity for each zero

The multiplicity of a zero is dictated by its corresponding exponent in the factored form of the polynomial. Here, all exponents are 1, so the multiplicity of each zero is 1.
03

Determine whether the graph crosses or touches the x-axis at each zero

When the multiplicity of a zero is an odd number, the graph crosses the x-axis at that point. When it is an even number, the graph touches the x-axis and turns around. In our case, all our zeros have an odd multiplicity (1), hence the graph crosses the x-axis at \(x=-4, x=2, x=-3.5\).

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