Chapter 8.7, Problem 12WE

To determine

To find the possible rational roots of the equation.

Answer to Problem 12WE

The rational roots are $\frac{1}{2},\frac{2}{3},\pm i$

Explanation of Solution

Given information:

  $6x^4-7x^3+8x^2-7x+2=0$

Calculation:

The possible rational roots can be obtained as

  $6x^4-7x^3+8x^2-7x+2=0$

First find h and k

  $\begin{array}{l}h=\pm 1\text{ or}\pm \text{ 2 }\left[\begin{array}{l}h=\text{factor of the constant term of the polynomial}\\ \text{since, 2 is divisible by 1 or 2}\end{array}\right]\\ k=\pm 1,\pm 2,\pm 3\text{ or }\pm \text{6 }\left[\begin{array}{l}k=\text{ factor of the cofficient of the highest-degree term}\\ \text{since, 6 is divisible by 1,2,3 and 4}\end{array}\right]\end{array}$

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

Therefore,

  $\begin{array}{l}\pm \frac{1}{1}=\pm 1\text{ or }\pm \frac{2}{1}=\pm 2\\ \pm \frac{1}{2}=\pm \frac{1}{2}\text{ or }\pm \frac{2}{2}=\pm 1\\ \pm \frac{1}{3}=\pm \frac{1}{3}\text{ or }\pm \frac{2}{3}=\pm \frac{2}{3}\\ \pm \frac{1}{4}=\pm \frac{1}{4}\text{ or }\pm \frac{2}{4}=\pm \frac{1}{2}\end{array}$

  $\begin{array}{l}\overline{)1}\text{ 6 }-7\text{ 8 }-7\text{ 2}\\ \underset{\_}{\text{ 6 }-1\text{ 7 0}}\\ \text{ 6 }-1\text{ 7 0 }⋮\text{ 2}\leftarrow \text{ 1 is not a root}\\ \overline{)2}\text{ 6 }-7\text{ 8 }-7\text{ 2}\\ \underset{\_}{\text{ 12 10 36 58}}\\ \text{ 6 5 18 29 }⋮\text{ 60}\leftarrow \text{ 2 is not a root}\\ \overline{)\frac{1}{2}}\text{ 6 }-7\text{ 8 }-7\text{ 2}\\ \underset{\_}{\text{ 3 }-2\text{ 3 }-2}\\ \text{ 6 }-4\text{ 6 }-4⋮\text{ 0}\leftarrow \frac{1}{2}\text{ is a root}\end{array}$

The depressed equation is $6x^3-4x^2+6x-4=0$

  $\begin{array}{l}\overline{)\frac{1}{3}}\text{ 6 }-4\text{ 6 }-4\\ \underset{\_}{\text{ 2 }-\frac{2}{3}\frac{16}{9}}\\ \text{ 6 }-2\frac{16}{3}⋮-\frac{20}{9}\leftarrow \frac{1}{3}\text{ is not a root}\\ \overline{)\frac{1}{6}}\text{ 6 }-4\text{ 6 }-4\\ \underset{\_}{\text{ 1 }-\frac{1}{2}\frac{11}{12}}\\ \text{ 6 }-3\frac{11}{2}⋮-\frac{37}{12}\leftarrow \frac{1}{6}\text{ is not a root}\\ \overline{)\frac{2}{3}}\text{ 6 }-4\text{ 6 }-4\\ \underset{\_}{\text{ 4 0 4}}\\ \text{ 6 0 6 }⋮\text{ 0}\leftarrow \frac{2}{3}\text{ is a root}\end{array}$

Again, the depressed equation is $6x^2+6=0$

  $\begin{array}{l}6x^2+6=0\\ 6x^2=-6\\ x^2=-1\left[\text{divide both side by 6}\right]\\ x=\pm i\left[\text{since, }{i}^2=-1\right]\end{array}$

Hence,

The rational roots are $\frac{1}{2},\frac{2}{3},\pm i$