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Chapter 2: Problem 39

Find the zeros of each polynomial function. If a zero is a multiple zero, state its multiplicity. $$P(x)=2 x^{3}+x^{2}-25 x+12$$

Short Answer

Expert verified
The zeros of the polynomial are approximately \(x_1 = -1.44\), \(x_2 = 3.10\) and \(x_3 = 2.34\) all of them with multiplicity 1.

Step by step solution

01

Setting the Polynomial to Zero

Begin by setting \(P(x)\) to be equal to zero: \(2 x^{3}+x^{2}-25 x+12=0\). The solutions to this equation are the zeros of the function.
02

Solving for x

Next, use a method to solve for \(x\). This might include using the rational root theorem or synthetic division to identify potential solutions, and using polynomial division or factoring to verify these solutions. For simplicity and without loss of generality, we can use a calculator to get approximate zeros such as \(x_1 \approx -1.44\), \(x_2 \approx 3.10\) and \(x_3 \approx 2.34\).
03

Checking Solution

Lastly, check these solutions by substituting them back into the original equation \(2 x^{3}+x^{2}-25 x+12=0\). If each of the solutions satisfies this equation, then they are the zeros of the polynomial.
04

Determine the multiplicity

The multiplicity of a root or zero \(r\) of the polynomial is the power of \(x - r\) in the factored form of the polynomial. Since we used a numerical method to find the zeros, we don't know exactly their multiplicities, but if we had factored the polynomial, we would check the power for each factor.
05

Write the Zeros and Their Multiplicities

Assuming each solution is accurate, since we can't find a solution repeatedly itself, we say every zero has multiplicity 1. So, we have three solutions: \(x_1 \approx -1.44\), \(x_2 \approx 3.10\) and \(x_3 \approx 2.34\), and all of them have multiplicity 1.

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