Chapter 8.7, Problem 2WE

To determine

To find the possible rational roots of the equation.

Answer to Problem 2WE

There is no rational roots found in this equation

Explanation of Solution

Given information:

  $x^3+x^2-4x+4=0$

Calculation:

The possible rational roots can be obtained as

  $x^3+x^2-4x+4=0$

First find h and k

  $\begin{array}{l}h=1,2,\text{and 4 }\left[\begin{array}{l}h=\text{factor of the constant term of the polynomial}\\ \text{since, 4 is divisible by 1,2,and 4}\end{array}\right]\\ k=1\left[\text{k= factor of the cofficient of the highest-degree term}\right]\end{array}$

  $$ All possible rational roots are in the form of $\pm \frac{h}{k}$

Therefore,

  $\pm \frac{1}{1}=\pm 1\text{ or }\pm \frac{2}{1}=\pm 2\text{ or }\pm \frac{4}{1}=\pm 4$

  $\begin{array}{l}\overline{)1}\text{ 1 1 }-4\text{ 4}\\ \underset{\_}{\text{ 1 2 }-2}\\ \text{ 1 2}-2⋮\text{ 2}\leftarrow \text{ 1 is not a root}\\ -\overline{)1}\text{ 1 1 }-4\text{ 4}\\ \underset{\_}{-1\text{ 0 4}}\\ \text{ 1 0 }-4⋮\text{ 8}\leftarrow -\text{1 is not a root}\\ \overline{)2}\text{ 1 1 }-4\text{ 4}\\ \underset{\_}{\text{ 2 6 4}}\\ \text{ 1 3 }2⋮\text{ 8}\leftarrow \text{ 2 is not a root}\\ -\overline{)2}\text{ 1 1 }-4\text{ 4}\\ \underset{\_}{-2\text{ 2 4}}\\ \text{ 1}-1-2⋮\text{ 8}\leftarrow -2\text{ is not a root}\\ \overline{)4}\text{ 1 1 }-4\text{ 4}\\ \underset{\_}{\text{ 4 20 64}}\\ \text{ 1 5 16 }⋮\text{ 68}\leftarrow \text{ 4 is not a root}\\ \overline{)-4}\text{ 1 1 }-4\text{ 4}\\ \underset{\_}{-\text{4 12 32}}\\ \text{ 1}-3\text{ 8 }⋮\text{ 8}\leftarrow -\text{4 is not a root}\end{array}$

Hence,

There is no rational roots found in this equation