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Chapter 3: Problem 80

Find the rational zeros of the polynomial function. $$f(z)=z^{3}+\frac{11}{6} z^{2}-\frac{1}{2} z-\frac{1}{3}=\frac{1}{6}\left(6 z^{3}+11 z^{2}-3 z-2\right)$$

Short Answer

Expert verified
The rational root of the function \( f(z) = 6z^{3} + 11z^{2} - 3z - 2 \) is -1. The other roots are irrational.

Step by step solution

01

Simplify the function

First, notice that the original function can be simplified to \( 6z^{3} + 11z^{2} - 3z - 2 \). Essentially, the function was multiplied by 6 to get rid of fractions.
02

Identify the coefficients of the polynomial

The coefficients of the polynomial are 6, 11, -3, and -2 (which are the numbers that the terms of the polynomial are multiplied by). The leading coefficient is 6 and trailing constant is -2.
03

Apply the Rational Root Theorem

List all factors of the trailing constant (-2) which are 卤1, 卤2. Then, list all factors of the leading coefficient (6) which are 卤1, 卤2, 卤3, 卤6. According to the Rational Root Theorem, the rational roots must be factors of -2 over factors of 6. Doing so, the possible rational roots include 卤1/1, 卤2/1, 卤1/2, 卤1/3, 卤2/3, 卤1/6, 卤2/6.
04

Test the possible rational roots using synthetic division

Now that we have all the possible rational roots, these can be tested using synthetic division to check which will result in a remainder of 0. Starting with -1, we see it is a rational root because the remainder is 0. So, \( f(z) = 6z^{3} + 11z^{2} - 3z - 2 \) can be factored as \( (z + 1)(6z^{2} + 5z - 2) \). We continue to test the remaining possible rational roots on the quadratic \( 6z^{2} + 5z - 2 \), and find that none of the remaining possible roots are roots of the quadratic.
05

Solve the quadratic equation

Since none of the remaining possible roots are roots of the quadratic, we can solve \( 6z^{2} + 5z - 2 = 0 \) using the quadratic formula. Which leads to the irrational roots.

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Most popular questions from this chapter

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