Chapter 4.5, Problem 29E

To determine

To find: all real solutions of the given equation.

Answer to Problem 29E

The real solutions are $-3,4$ , and $5$ .

Explanation of Solution

Given information:

Given equation,

  $x^3-6x^2-7x+60=0$

Calculation:

Consider the equation.

  $x^3-6x^2-7x+60=0$

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

Here, the leading coefficient is $1$ and constant term is $60$ .

The possible rational solutions are $\pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 10,\pm 12,\pm 15,\pm 20,\pm 30\text{and}\pm 60$ .

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

For $x=4$

  $\begin{array}{ccccc}4& 1& -6& -7& 60\\ & & 4& -8& -60\\ & 1& -2& -15& 0\end{array}$

Therefore, $x-4$ is a factor.

Factor completely using the result of above synthetic division.

  $\begin{array}{c}{x}^3-6x^2-7x+60=0\\ \left(x-4\right)\left(x^2-2x-15\right)=0\\ \left(x-4\right)\left(x-5\right)\left(x+3\right)=0\end{array}$

Solve further,

  $\begin{array}{c}\left(x-4\right)\left(x-5\right)\left(x+3\right)=0\\ x=-3,4,5\end{array}$

Therefore, all real solutions are $-3,4$ , and $5$ .