Chapter 8.7, Problem 6WE

To determine

To find the rational roots of the equation.

Answer to Problem 6WE

There is not found rational root in the equation

Explanation of Solution

Given information:

  $x^4+3x^2-8x+10=0$

Calculation:

The possible rational roots can be obtained as

  $x^4+3x^2-8x+10=0$

First find h and k

  $\begin{array}{l}h=1,2,5\text{ and 10 }\left[\begin{array}{l}h=\text{factor of the constant term of the polynomial}\\ \text{since, 10 is divisible by 1,2,5 and 10}\end{array}\right]\\ k=1\left[\text{k= factor of the cofficient of the highest-degree term}\right]\end{array}$

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

Therefore,

  $\pm \frac{1}{1}=\pm 1\text{ or }\pm \frac{2}{1}=\pm 2\text{ or }\pm \frac{5}{1}=\pm 5\text{ or }\pm \frac{10}{1}=\pm 10$

  $\begin{array}{l}\overline{)1}\text{ 1 0 3 }-8\text{ 10}\\ \underset{\_}{\text{ 1 1 4 }-4}\\ \text{ 1 1 4 }-4⋮\text{ 6}\leftarrow \text{ 1 is not a root}\end{array}$

  $\begin{array}{l}\overline{)-1}\text{ 1 0 3 }-8\text{ 10}\\ \underset{\_}{-\text{1 1 }-\text{4 12}}\\ \text{ 1}-\text{1 4 }-12⋮\text{ 22}\leftarrow -\text{1 is not a root}\end{array}$

  $\begin{array}{l}\overline{)\text{ 2}}\text{ 1 0 3 }-8\text{ 10}\\ \underset{\_}{\text{ 2 4 14 12}}\\ \text{ 1 2 7 6 }⋮\text{ 22}\leftarrow \text{ 2 is not a root}\end{array}$

  $\begin{array}{l}\overline{)-2}\text{ 1 0 3 }-8\text{ 10}\\ \underset{\_}{-2\text{ 4 }-1\text{4 44}}\\ \text{ 1}-2\text{ 7 }-22⋮\text{ 54}\leftarrow -2\text{ is not a root}\end{array}$

  $\begin{array}{l}\overline{)\text{ 5}}\text{ 1 0 3 }-8\text{ 10}\\ \underset{\_}{\text{ 5 25 140 660}}\\ \text{ 1 5 28 132 }⋮\text{ 670}\leftarrow \text{ 5 is not a root}\end{array}$

  $\begin{array}{l}\overline{)-5}\text{ 1 0 3 }-8\text{ 10}\\ \underset{\_}{-5\text{ 25 }-1\text{40 740}}\\ \text{ 1}-5\text{ 28 }-148⋮\text{ 750}\leftarrow -5\text{ is not a root}\end{array}$

  $\begin{array}{l}\overline{)\text{ 10}}\text{ 1 0 3 }-8\text{ 10}\\ \underset{\_}{\text{ 10 100 1030 10220}}\\ \text{ 1 10 103 1022 }⋮\text{ 10230}\leftarrow \text{ 10 is not a root}\end{array}$

  $\begin{array}{l}\overline{)-10}\text{ 1 0 3 }-8\text{ 10}\\ \underset{\_}{-10\text{ 100 }-1\text{030 10380}}\\ \text{ 1 }-10\text{ 103 }-1038⋮\text{ 10390}\leftarrow -10\text{ is not a root}\end{array}$

Hence,

There is not found rational root in the equation