Problem 78 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the rational zeros of the polynomial function. $$f(x)=x^{3}-\frac{3}{2} x^{2}-\frac{23}{2} x+6=\frac{1}{2}\left(2 x^{3}-3 x^{2}-23 x+12\right)$$

Step-by-Step Solution

Answer

The rational zeros of the given polynomial function are -2 and 3.

Step 1: Rewrite the Polynomial

First, rewrite the polynomial in its simplified form: \(f(x) = \frac{1}{2}(2x^3 - 3x^2 - 23x + 12)\) which equals \(f(x) = x^3 - \frac{3}{2}x^2 - \frac{23}{2}x + 6\)

Step 2: Apply the Rational Root Theorem

Apply the Rational Root Theorem. Identify the leading term and the constant term of the polynomial. For the polynomial \(x^3 - \frac{3}{2}x^2 - \frac{23}{2}x + 6\), the leading coefficient is 1 and the constant term is 6. According to the theorem, any rational roots of the polynomial would be factors of 6 divided by factors of 1. The factors of 6 are ±1, ±2, ±3, ±6, hence the possible rational roots of the polynomial are also ±1, ±2, ±3, ±6.

Step 3: Test the Possible Rational Roots

Substitute each possible rational root into the polynomial and check if the result is zero. When trying ±1, ±2, ±3, and ±6, it can be found that -2 and 3 are roots of the polynomial since \(f(-2)=0\) and \(f(3)=0\).