Problem 80 - Algebra and Trigonometry Real Mathematics, Real People

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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.

Step 3: Apply the Rational Root Theorem

List all factors of the trailing constant (-2) which are ±1, ±2. Then, list all factors of the leading coefficient (6) which are ±1, ±2, ±3, ±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 ±1/1, ±2/1, ±1/2, ±1/3, ±2/3, ±1/6, ±2/6.

Step 4: 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.

Step 5: 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.