Problem 22 - Algebra and Trigonometry

Question

In Exercises \(17-24\) a. List all possible rational roots. b. Use synthetic division to test the possible rational roots and find an actual root. c. Use the quotient from part (b) to find the remaining roots and solve the equation. $$ 2 x^{3}-5 x^{2}-6 x+4=0 $$

Step-by-Step Solution

Answer

The roots of the given equation are \(1\), \(\frac{1+\sqrt{17}}{4}\), and \(\frac{1-\sqrt{17}}{4}\).

Step 1: Identify Rational Roots

First, determine all possible rational roots using the Rational Root Theorem. This theorem suggests that any possible rational root, say \(p/q\), must be such that \(p\) is a factor of the constant term (4 in this case) and \(q\) is a factor of the leading coefficient (2 in this case). So, the list of all possible rational roots will be: \(\pm1, \pm2, \pm4\).

Step 2: Use Synthetic Division to Test Roots

Now, test these potential roots using the synthetic division to see which one is an actual root. After testing with synthetic division, it is found that 1 is an actual root of the equation.

Step 3: Transform the Cubic Equation

Use synthetic division again with root 1 to transform the original cubic equation into a quadratic equation. The quotient obtained would be \(2x^{2} - 3x - 4\).

Step 4: Find the Remaining Roots

Solve the quadratic equation \(2x^{2} - 3x - 4 = 0\). A quadratic formula should be applied here as this equation is non-factorizable. Applying the formula and simplifying, you get two roots as \(\frac{1+\sqrt{17}}{4}\) and \(\frac{1-\sqrt{17}}{4}\).