/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 Find the zeros of each polynomia... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 45

Find the zeros of each polynomial function. If a zero is a multiple zero, state its multiplicity. $$P(x)=x^{3}-8 x^{2}+8 x+24$$

Short Answer

Expert verified
The zeros of the polynomial \(x^{3}-8 x^{2}+8 x+24\) are \(x = -1\), \(x = 2\) and \(x = 4\). Each zero has a multiplicity of 1.

Step by step solution

01

Setting the equation to zero

First, set the equation to zero to solve for \(x\), i.e., \(x^{3}-8 x^{2}+8 x+24 = 0\).
02

Factoring the equation

In order to find the roots of this polynomial, you should attempt factoring. However, factoring this polynomial directly seems complex, so the Factor Theorem will need to be used.
03

Applying the Factor theorem

The factor theorem states that if \(P(c) = 0\), then \(x-c\) is a factor of the polynomial. So, substitute different integer values of x into Polynomial until we find a number \(c\) that makes \(P(c) = 0\). After testing, you find that \(P(-1) = 0\) , \(P(2) = 0\) and \(P(4) = 0\). Thus, \(x + 1\), \(x - 2\) and \(x - 4\) are factors of the polynomial.
04

Solve for x

Setting each factor equal to zero and solving for x gives the solutions \(x = -1\), \(x = 2\), \(x = 4\).

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