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Chapter 2: Problem 56

Find the zeros of each polynomial function. If a zero is a multiple zero, state its multiplicity. $$P(x)=6 x^{4}-17 x^{3}-11 x^{2}+42 x$$

Short Answer

Expert verified
The zeros of the polynomial \(P(x)=6x^4-17x^3-11x^2+42x\) are 0, -1, 2, and 7/3. All these zeros have a multiplicity of 1.

Step by step solution

01

Set the polynomial equal to zero

The first step is to set the polynomial \( P(x) = 6x^4 - 17x^3 - 11x^2 + 42x = 0 \)
02

Factor out the common factor

Observe that every term in the polynomial has \(x\) as its factor, which is a common factor, we factor out \(x\) which gives \( x (6x^3 - 17x^2 - 11x + 42) = 0 \)
03

Apply Zero-Product Property

Observe that we have a product equals to zero. The Zero-Product Property says that if the product of two factors is zero, then at least one of the factors equal to zero. So, set \(x = 0\) and \(6x^3 - 17x^2 - 11x + 42 = 0\) independently.
04

Solving for the second factor

Solving a cubic equation can be more complicated, but in this problem, you can use synthetic division or factoring to find the solutions of the equation \(6x^3 - 17x^2 - 11x + 42 = 0\). After solving, you get \( x = -1, 2, 7/3 \)
05

Find the multiplicity of the zeros

Multiplicity of a zero refers to how many times the zero is a solution. In this problem, the zero x = 0 is a solution 1 time, and the zeros x = -1, 2, 7/3 are each a solution 1 time, so all the zeros have a multiplicity of 1.

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Most popular questions from this chapter

Use Descartes' Rule of Signs to state the number of possible positive and negative real zeros of each polynomial function. $$P(x)=6 x^{4}+23 x^{3}+19 x^{2}-8 x-4$$

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