Chapter 6.8, Problem 3BCYP

To determine

To calculate: The simplified form of the equation $k\left(x\right)=2x^4-5x^3+20x^2-45x+18$ ,

Find all the possible zeros of rational function.

Answer to Problem 3BCYP

The simplified form of the equation $h\left(x\right)=x^3+11x^2+24$ has no real roots.

Explanation of Solution

Given information:

The given equation:

  $k\left(x\right)=2x^4-5x^3+20x^2-45x+18$

Calculation:

Consider the expression as:

  $k\left(x\right)=2x^4-5x^3+20x^2-45x+18$

Now take $f\left(x\right)=0$ ,

Here,

  $\begin{array}{l}f\left(x\right)=0\\ 2x^4-5x^3+20x^2-45x+18=0\end{array}$

The leading coefficient is $2$ and the trilling coefficient is $18$ .

Now factor of leading coefficient is: $1,2$

Now to find the list of positive rational solutions for the expression look at the factors of the constants.

That are:

  $1,2,3,6,9,18$

Here the last two terms subtracting values from the cube of the first and want to get zero,

So as have to take the positive number, and mostly take the less than mid-way.

So take a startup with $1,1$ and check the result, then again tried to get zero.

  $\begin{array}{c}2x^4-5x^3+20x^2-45x+18=0\\ 2{\left(1\right)}^4-5{\left(1\right)}^3+20{\left(1\right)}^2-45\left(1\right)+18=0\\ 2-5+20-45+18=0\\ -10=0\end{array}$

So the value is not zero its too grater than zero.

So check with the value $2$ .

  $\begin{array}{c}2x^4-5x^3+20x^2-45x+18=0\\ 2{\left(2\right)}^4-5{\left(2\right)}^3+20{\left(2\right)}^2-45\left(2\right)+18=0\\ 2\left(16\right)-5\left(8\right)+20\left(4\right)-90+18=0\\ 32-40+80-90+18=0\end{array}$

To get zero:

  $\begin{array}{c}32-40+80-90+18=0\\ 0=0\end{array}$

Here got the real root, it means $\left(x-2\right)$ is a factor.

Now divide the expression $2x^4-5x^3+20x^2-45x+18=0$ to the factor $\left(x-2\right)$ , to check for the other possible solutions.

  $\begin{array}{l}x-2\begin{array}{c}\hfill 2x^3-x^2+18x-9\\ \hfill \overline{)\begin{array}{ccccccccc}2x^4& -& 5x^3& +& 20x^2& -& 45x& +& 18\end{array}}\end{array}\\ \begin{array}{ccc}2x^4& -& 4x^3\end{array}\\ \begin{array}{ccc}{x}^3& -& 20x^2\end{array}\\ \begin{array}{ccc}{x}^3& -& 2x^2\end{array}\\ \begin{array}{ccc}18x^2& -& 45x\end{array}\\ \begin{array}{ccc}18x^2& -& 36x\end{array}\\ \begin{array}{ccc}9x& +& 18\end{array}\\ \begin{array}{ccc}9x& +& 18\end{array}\\ \begin{array}{c}0\end{array}\end{array}$

So the other factor is:

  $2x^3-x^2+18x-9$

Now find the factors of the expression $2x^3-x^2+18x-9$:

The leading coefficient is $2$ and the trilling coefficient is $-9$ .

Now factor of leading coefficient is: $1,2$

Now to find the list of positive rational solutions for the expression look at the factors of the constants.

That are:

  $1,3,9$

Here the last two terms subtracting values from the cube of the first and want to get zero,

So as have to take the positive number, and mostly take the less than mid-way.

So take a startup with $1,1$ and check the result, then again tried to get zero.

  $\begin{array}{c}2x^3-x^2+18x-9=0\\ 2{\left(1\right)}^3-{\left(1\right)}^2+18\left(1\right)-9=0\\ 2-1+18-9=0\\ 10=0\end{array}$

So the value is not zero its too grater than zero.

So check with the value $1,2$ that means $p/q=0.5$ .

  $\begin{array}{c}2x^3-x^2+18x-9=0\\ 2{\left(0.5\right)}^3-{\left(0.5\right)}^2+18\left(0.5\right)-9=0\\ 2\left(0.125\right)-\left(0.25\right)+9-9=0\\ 0.25-0.25+9-9=0\end{array}$

To get zero:

  $\begin{array}{c}0.25-0.25+9-9=0\\ 0=0\end{array}$

As for the factor $1,2$ its become zero.

Here got the real root, it means $\left(2x-1\right)$ is a factor.

Now divide the expression $2x^3-x^2+18x-9=0$ to the factor $\left(2x-1\right)$ , to check for the other possible solutions.

  $\begin{array}{l}2x-1\begin{array}{c}\hfill x^2+9\\ \hfill \overline{)\begin{array}{ccccccc}2x^3& -& x^2& +& 18x& -& 9\end{array}}\end{array}\\ \begin{array}{ccccccc}2x^3& -& x^2& & & & \end{array}\\ \begin{array}{ccc}18x& -& 9\end{array}\\ \begin{array}{ccc}18x& -& 9\end{array}\\ \begin{array}{c}0\end{array}\end{array}$

So the other factor is:

  $x^2+9$

Now find the roots of $x^2+9$ .

  $x^2+9=0$

The leading coefficient is: $1$

The trilling coefficient is: $1,3,9$

Here the last two terms subtracting values from the cube of the first and want to get zero,

So as have to take the positive number, and mostly take the less than mid-way.

So take a startup with $1$ and check the result, then again tried to get zero.

  $\begin{array}{c}{x}^2+9=0\\ {\left(1\right)}^2+9=0\\ 10=0\end{array}$

So the value is not zero its too grater than zero.

So check with the value $3$ .

  $\begin{array}{c}{x}^2+9=0\\ {\left(3\right)}^2+9=0\\ \left(9\right)+9=0\\ 18=0\end{array}$

Here no polynomial rational roots are found.

Thus, the simplified form of the equation $k\left(x\right)=2x^4-5x^3+20x^2-45x+18$ has no rational roots.