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Chapter 2: Problem 44

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

Short Answer

Expert verified
The zeros of the polynomial function \(P(x)=3 x^{3}-x^{2}-6 x+2\) are \(x=-1\) of multiplicity 1 and \(x=2/3\) of multiplicity 2.

Step by step solution

01

Start with the Given Polynomial

Set the given polynomial \(P(x) = 3x^{3} - x^{2} - 6x + 2\) equal to zero. So, find \(x\) for which \(3x^{3} - x^{2} - 6x + 2 = 0\).
02

Factorize the Polynomial

Factorize the polynomial if possible, unfortunately in this case it's not easily factorizable. We will use the Rational Root Theorem. According to this theorem, a polynomial's rational roots, if any exist, are composed of a factor of the constant term (in this case, \(2\)) divided by a factor of the coefficient of the highest term (which in this case, is \(3\)). So, the possible rational roots of the equation can be \(\pm1\), \(\pm2\), \(\pm1/3\), \(\pm2/3\). Testing these values, we find that \(-1\) and \(2/3\) are the roots of the polynomial equation.
03

Find Factors

Thus, the factors of the polynomial are \((x+1)\) and \((3x-2)\). So, we can write the original equation as \(P(x) = 3x^{3} - x^{2} - 6x + 2 = 3(x+1)(x-2/3)\^2 = 0\). This implies that \(x=-1\) or \(x=2/3\).
04

Finding Multiplicity of the Roots

Multiplicity is the power of the factor in the polynomial. In this case, the factor \((x-2/3)\) has a power of \(2\), so \(x=2/3\) is a root of multiplicity \(2\). The other root \(x=-1\) is of multiplicity \(1\) as it appears only once as the factor.

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Most popular questions from this chapter

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