/*! 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 52 Find the zeros of each polynomia... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 52

Find the zeros of each polynomial function. If a zero is a multiple zero, state its multiplicity. $$P(x)=3 x^{4}-4 x^{3}-11 x^{2}+16 x-4$$

Short Answer

Expert verified
The roots of the polynomial function are \(x = 2\) (multiplicity 2), \(x = \frac{\sqrt{3}}{3}\) and \(x = - \frac{\sqrt{3}}{3}\) with multiplicity 1 for each.

Step by step solution

01

Write down the given polynomial equation

The initial polynomial equation is given as: \(P(x)=3 x^{4}-4 x^{3}-11 x^{2}+16 x-4\).
02

Factorize the polynomial equation

The polynomial can be factorized as: \(P(x) = (3x^2 - 1)(x - 2)^2\).
03

Equate each factor to zero

You then need to solve each factor for \(x\) by setting it equal to zero. You get two equations: \(3x^2 - 1 = 0\) and \(x - 2 = 0\).
04

Solve the resulting equations for \(x\)

Solving \(3x^2 - 1 = 0\) gives two roots: \(x = \frac{\sqrt{3}}{3}\) or \(x = - \frac{\sqrt{3}}{3}\). Solving \(x - 2 = 0\) gives \(x = 2\).
05

Determine the multiplicity of each root

The root \(x = 2\) is a solution of the second factor, which is of degree 2. Therefore, \(x = 2\) is a root with multiplicity 2. Both the roots \(x = \frac{\sqrt{3}}{3}\) and \(x = - \frac{\sqrt{3}}{3}\) are simple, so they have a multiplicity of 1.

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