Chapter 6.8, Problem 11PPS

To determine

To calculate: The simplified form of the equation $f\left(x\right)=x^3+x^2-x-56$ ,

Find all the possible zeros of rational function.

Answer to Problem 11PPS

The simplified form of the equation $f\left(x\right)=x^3+x^2-x-56$ has no real roots.

Explanation of Solution

Given information:

The given equation:

  $f\left(x\right)=x^3+x^2-x-56$

Calculation:

Consider the expression as:

  $f\left(x\right)=x^3+x^2-x-56$

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

Here,

  $\begin{array}{l}f\left(x\right)=0\\ x^3+x^2-x-56=0\end{array}$

This function is not a perfect cube.

So in grouping solve the equation.

As,

  $f\left(x\right)=x^3+x^2-x-56$

Now find the rational possible roots:

  $x^3+x^2-x-56=0$

Here is no factors so make groups:

  $\begin{array}{c}\left(x+56\right)\cdot \left(-1\right)\\ \left(x+1\right)\cdot \left(x^2\right)\end{array}$

So find the leading coefficient and trilling coefficient.

The leading coefficient is $1$ and the trilling coefficient is $-56$ .

Now factor of leading coefficient is: $1$

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

That are:

  $1,2,4,7,8,14,28,56$

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.

  $\begin{array}{c}{x}^3+x^2-x-56=0\\ {\left(1\right)}^3+{\left(1\right)}^2-\left(1\right)-56=0\\ 1+1+1-56=0\\ -55=0\end{array}$

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

So check with the value $2,1$ .

  $\begin{array}{c}{x}^3+x^2-x-56=0\\ {\left(2\right)}^3+{\left(2\right)}^2-\left(2\right)-56=0\\ 8+4-2-56=0\\ -46=0\end{array}$

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

So check with the value $4$ .

  $\begin{array}{c}{x}^3+x^2-x-56=0\\ {\left(4\right)}^3+{\left(4\right)}^2-4-56=0\\ 64+16-4-56=0\\ 20=0\end{array}$

Here is no rational root.

Thus, the simplified form of the equation $f\left(x\right)=x^3+x^2-x-56$ has no rational roots.