Chapter 2.6, Problem 49PS

Solution

(a)

To find: a polynomial equation.

The polynomial equation $t^3-14t^2+2t+215=0$

Given information:

The function $E\left(t\right)=-10t^3+140t^2-20t+18,150$ .

Calculation:

The function $t$ such that

  $E\left(t\right)=20,300$

This yields the equation

  $-10t^3+140t^2-20t+18,150=20,300$

Subtract $20,300$ from both sides:

  $-10t^3+140t^2-20t-2,150=0$

Divide both sides by $-10\text{ : }$

Then

  $t^3-14t^2+2t+215=0$ .

(b)

To find: the possible whole-number solutions of the equation in part (a) that are less than $10.$

The list only the possible solutions that are whole numbers less than $10$ is $1,5$ .

Given information:

The function $E\left(t\right)=-10t^3+140t^2-20t+18,150$ .

Calculation:

All possible rational solutions are of the form

  $\frac{p}{q}$

where $p$ is a factor of the constant term $\left(215\right)$ and $q$ is a factor of the leading coefficient $\left(1\right)$ .

Step 3

Factors of $215$ :

  $\pm 1,\pm 5,\pm 43,\pm 215$

Factors of $1$ :

  $\pm 1$

All possible rational solutions:

  $\pm \frac{1}{1},\pm \frac{5}{1},\pm \frac{43}{1},\pm \frac{215}{1}$

Simplify fractions:

  $\pm 1,\pm 5,\pm 43,\pm 215$

The list only the possible solutions that are whole numbers less than $10$ .

They are $1,5$ .

(c)

To determine: the possible solutions in part $\left(b\right)$ is an actual solution. Next, determine the year that follows the solution.

  $1994,t=5$

Given information:

The function $E\left(t\right)=-10t^3+140t^2-20t+18,150$ .

Calculation:

Try $1$ :

  $\begin{array}{l}1\overline{)\begin{array}{llll}1\hfill & -14\hfill & 2\hfill & 215\hfill \\ \hfill & 1\hfill & -13\hfill & -11\hfill \end{array}}\\ \begin{array}{lllll}\hfill & 1\hfill & -13\hfill & -11\hfill & 204\hfill \end{array}\end{array}$

So, $1$ is not a solution.

Try $5$ :

  $\begin{array}{l}5\overline{)\begin{array}{llll}1\hfill & -14\hfill & 2\hfill & 215\hfill \\ \hfill & 5\hfill & -45\hfill & -215\hfill \end{array}}\\ \begin{array}{lllll}\hfill & 1\hfill & -9\hfill & -45\hfill & 0\hfill \end{array}\end{array}$

So, $5$ is not a solution.

Since $t$ is the number of years after $1994,t=5$ is the year $1999$ .