Problem 41 - 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 -3/2, 1, and -3/4.

Step 1: Validate the given root

Substitute the given root -3/2 into the equation. If it evaluates to zero, then -3/2 is indeed a root. So, \(12(-3/2)^{3} + 16(-3/2)^{2} - 5(-3/2) - 3 = 0\). Hence, -3/2 is validated as a root.

Step 2: Divide the cubic polynomial by the linear factor

Since -3/2 is a root, x - (-3/2) = x + 3/2 is a factor. Divide the given polynomial by this factor using synthetic division. We have | -3/2 | 12 | 16 | -5 | -3 | | -------| ---- | ---- | ---- | ---- | | | | -18 | 3 | -6 | | | 12 | -2 | -2 | -9 | The numbers in the bottom represent the coefficients of a quadratic polynomial, i.e., \(12x^2 - 2x - 9 = 0\).

Step 3: Solve the resulting quadratic equation

Now, solve the quadratic equation using quadratic formula \(x = [-b ± sqrt(b^2 - 4ac)] / 2a\). Therefore, the roots x1, x2 can be calculated as \(x_{1}=\frac{2+\sqrt{4+432}}{24}=1\) \(x_{2}=\frac{2-\sqrt{4+432}}{24}=-\frac{3}{4}\)

Step 4: List all roots

Therefore, the roots of the original cubic equation are -3/2, 1, and -3/4.