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Chapter 2: Problem 38

Find the zeros of each polynomial function. If a zero is a multiple zero, state its multiplicity. $$P(x)=x^{3}-19 x-30$$

Short Answer

Expert verified
The roots of the polynomial \( P(x) = x^{3} - 19x - 30 \) are -5, -2, and 3. None of them are multiple roots.

Step by step solution

01

Apply the Rational Root Theorem

The Rational Root Theorem suggests that if \( p/q \) is a root of the polynomial \( P(x) = x^{3} - 19x - 30 \), then \( p \) divides the constant term and \( q \) divides the leading coefficient. In this problem, the constant term is -30 and the leading coefficient is 1. So, the possible rational roots can be \(卤1, 卤2, 卤3, 卤5, 卤6, 卤10, 卤15, 卤30.\)
02

Test the possible roots in the equation

Substitute the possible rational roots into the polynomial \( P(x) = x^{3} - 19x - 30 \). Examine until you find a value that makes the equation true, meaning \( P(x) = 0 \). From the list of possible roots, it is found that -2, -5 and 3 are roots of the polynomial. Thus, \( P(-2) = 0, P(-5) = 0 \) and \( P(3) = 0 \).
03

Check for multiplicity

Having found the roots, the corresponding factored form of the polynomial can be written as \( P(x) = (x+2)(x+5)(x-3) \). Since the exponents of all of these factors are 1, it can be stated that all roots are simple roots - none of them are multiple roots.

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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. $$P(x)=x^{4}-1$$

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