Problem 45 - College Algebra

Question

Solve the equation \(12 x^{3}+16 x^{2}-5 x-3=0\) given that \(-\frac{3}{2}\) is a root.

Step-by-Step Solution

Answer

The roots of the equation are \(x = -\frac{3}{2}\), \(x = \frac{1 + \sqrt{7}}{6}\), and \(x = \frac{1 - \sqrt{7}}{6}\).

Step 1: Verify Given Root

Substitute the given root -3/2 into the equation to verify that it is indeed a root: \(12(-\frac{3}{2})^{3}+16(-\frac{3}{2})^{2}-5(-\frac{3}{2})-3 = 0\). Since this simplifies to zero, it confirms that -3/2 is a root.

Step 2: Carry Out Polynomial Division

Next we need to divide the original polynomial by the binomial corresponding to the given root to get a quadratic polynomial. The binomial is \(x-(-\frac{3}{2}) = x + \frac{3}{2}\). The process of polynomial division yields a quadratic equation \(12x^{2} - 2x -2 =0\).

Step 3: Solve the Quadratic Equation

Next, we solve the quadratic equation \(12x^{2} - 2x -2 =0\) for its roots. By using the quadratic formula \(x=\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\), we find that the roots are \(x = \frac{1 \pm \sqrt{7}}{6}\).