Chapter 6.8, Problem 1BCYP

To determine

To calculate: The simplified form of the equation $h\left(x\right)=x^3+11x^2+24$ ,

Answer to Problem 1BCYP

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:

  $h\left(x\right)=x^3+11x^2+24$

Calculation:

Consider theexpression as:

  $h\left(x\right)=x^3+11x^2+24$

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

Here,

  $\begin{array}{c}f\left(x\right)=0\\ x^3+11x^2+24=0\end{array}$

The leading coefficient is $1$ and the trilling coefficient is $24$ .

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

That are:

  $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 8,\pm 12,\pm 24$

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 $12$ and check the result, then again tried to get zero.

  $\begin{array}{c}{x}^3+11x^2+24=0\\ {\left(12\right)}^3+11{\left(12\right)}^2+24=0\\ 1728+11\left(144\right)+24=0\\ 1728+1584+24=0\end{array}$

To get zero:

  $\begin{array}{c}1728+1584+24=0\\ 3336=0\end{array}$

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

So check with the value $2$ .

  $\begin{array}{c}{x}^3+11x^2+24=0\\ {\left(2\right)}^3+11{\left(2\right)}^2+24=0\\ 8+11\left(4\right)+24=0\\ 8+44+24=0\end{array}$

To get zero:

  $\begin{array}{c}8+44+24=0\\ 76=0\end{array}$

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

So check with the value $-2$ .

  $\begin{array}{c}{x}^3+11x^2+24=0\\ {\left(-2\right)}^3+11{\left(-2\right)}^2+24=0\\ -8+11\left(4\right)+24=0\\ -8+44+24=0\end{array}$

To get zero:

  $\begin{array}{c}-8+44+24=0\\ 60=0\end{array}$

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

So check with the value $-1$ .

  $\begin{array}{c}{x}^3+11x^2+24=0\\ {\left(-1\right)}^3+11{\left(-1\right)}^2+24=0\\ -1+11\left(1\right)+24=0\\ -1+11+24=0\end{array}$

To get zero:

  $\begin{array}{c}-1+11+24=0\\ 34=0\end{array}$

Here no polynomial roots are found.

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