Chapter 4.4, Problem 13E

To determine

To list: The possible rational roots of equation $x^3-5x^2-4x+20=0$ and then determine the rational roots.

Answer to Problem 13E

0 or 2 positive roots, 1 negative root

All roots are: $\pm 2,5$

Explanation of Solution

Given information: $x^3-5x^2-4x+20=0$

Calculation:

  $x^3-5x^2-4x+20=0$

Starting simple, you can find the first root at $x=2$

  $\frac{x^3-5x^2-4x+20}{x-2}=x^2-3x-10$

  $x^2-3x-10=\left(x+2\right)\left(x-5\right)$

Therefore all roots are: $\pm 2,5$

The possible roots are all factors of 20:

  $\pm 1,\pm 2,\pm 4\text{, }\pm 5\text{, }\pm 10\text{, }\pm 20$

Using Descartes Rule of signs there are:

0 or 2 positive roots, 1 negative root