Chapter 4.5, Problem 31E

To determine

To find: all real solutions of the given equation.

Answer to Problem 31E

The real solutions are $-4,6$ , and $-\frac{1}{2}$ .

Explanation of Solution

Given information:

Given equation,

  $2x^3-3x^2-50x-24=0$

Calculation:

Consider the equation.

  $2x^3-3x^2-50x-24=0$

The above equation is of degree 3, so it has 3 solutions.

Here, the leading coefficient is $2$ and constant term is $-24$ .

The possible rational solutions are $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 8,\pm 12,\pm 24,\pm \frac{1}{2}\text{and}\pm \frac{3}{2}$ .

Test possible solution using synthetic division until a solution is found.

For $x=6$

  $\begin{array}{ccccc}6& 2& -3& -50& -24\\ & & 12& 54& 24\\ & 2& 9& 4& 0\end{array}$

Therefore, $x-6$ is a factor.

Factor completely using the result of above synthetic division.

  $\begin{array}{c}2x^3-3x^2-50x-24=0\\ \left(x+4\right)\left(2x^2+9x+4\right)=0\\ \left(x+4\right)\left(x-6\right)\left(x+\frac{1}{2}\right)=0\end{array}$

Solve further,

  $\begin{array}{c}\left(x+4\right)\left(x-6\right)\left(x+\frac{1}{2}\right)=0\\ x=-4,6,-\frac{1}{2}\end{array}$

Therefore, all real solutions are $-4,6$ , and $-\frac{1}{2}$ .