Chapter 2.6, Problem 50PS

Solution

(a)

To find: a polynomial equation.

The polynomial equation $t^4-22t^3+169t^2-327t-1032=0$

Given information:

The function $F\left(t\right)=12t^4-264t^3+2028t^2-3924t+43,916$ .

Calculation:

The number of travelers (in thousands), $F\left(t\right)$ needs to be equal to $56300$ .

Simplify the resulting equation.

  $\begin{array}{c}F(t)=5630012t^4-264t^3+2028t^2-3924t+43916\\ =5630012t^4-264t^3+2028t^2-3924t-12384\\ =0t^4-22t^3+169t^2-327t-1032\\ =0\end{array}$

(b)

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

The list only the possible solutions that are whole numbers less than or equal to $10$ is

  $1,2,3,4,6,8$ .

Given information:

The function $F\left(t\right)=12t^4-264t^3+2028t^2-3924t+43,916$ .

Calculation:

The equation is

  $t^4-22t^3+169t^2-327t-1032=0$

All possible rational solutions are of the form

  $\underset{\_}{p}$

where $p$ is a factor of the constant term (-1032) and $q$ is a factor of the leading coefficient $\left(1\right)$ .

Factors of $-1032$ :

  $\begin{array}{l}\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 8,\pm 12,\pm 24,\\ \pm 43,\pm 86,\pm 129,\pm 172,\pm 258,\pm 344,\end{array}$

Factors of $1$ :

  $\pm 1$

All possible rational solutions:

  $\begin{array}{l}\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{3}{1},\pm \frac{4}{1},\pm \frac{6}{1},\pm \frac{8}{1},\pm \frac{12}{1},\pm \frac{24}{1},\pm \frac{43}{1},\pm \frac{86}{1},\pm \frac{129}{1},\\ \pm \frac{172}{1},\pm \frac{258}{1},\pm \frac{344}{1},\pm \frac{516}{1},\pm \frac{1032}{1}\end{array}$

Simplify fractions:

  $\begin{array}{l}\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 8,\pm 12,\pm 24,\pm 43,\pm 86,\pm 129,\\ \pm 172,\pm 258,\pm 344,\pm 516,\pm 1032\end{array}$

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

They are $1,2,3,4,6,8$ .

(c)

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

  $t=8$ .

Given information:

The function $F\left(t\right)=12t^4-264t^3+2028t^2-3924t+43,916$ .

Calculation:

For $t=1$ .

  $\begin{array}{l}1\overline{)\begin{array}{rrrrr}\hfill 12& \hfill -264& \hfill 2028& \hfill -3924& \hfill -12384\\ \hfill & \hfill 12& \hfill -252& \hfill 1776& \hfill -2148\end{array}}\\ \begin{array}{rrrrrr}\hfill & \hfill 12& \hfill -252& \hfill 1776& \hfill -2148& \hfill -14532\end{array}\end{array}$

For $t=2$ .

  $\begin{array}{l}2\overline{)\begin{array}{rrrrr}\hfill 12& \hfill -264& \hfill 2028& \hfill -3924& \hfill -12384\\ \hfill & \hfill 24& \hfill -480& \hfill 3096& \hfill -1656\end{array}}\\ \begin{array}{rrrrrr}\hfill & \hfill 12& \hfill -240& \hfill 1548& \hfill -828& \hfill -14040\end{array}\end{array}$

For $t=3$ .

  $\begin{array}{l}3\overline{)\begin{array}{rrrrr}\hfill 12& \hfill -264& \hfill 2028& \hfill -3924& \hfill -12384\\ \hfill & \hfill 36& \hfill -684& \hfill 4032& \hfill 324\end{array}}\\ \begin{array}{rrrrrr}\hfill & \hfill 12& \hfill -228& \hfill 1344& \hfill 108& \hfill -12060\end{array}\end{array}$

For $t=4$ .

  $\begin{array}{l}4\overline{)\begin{array}{rrrrr}\hfill 12& \hfill -264& \hfill 2028& \hfill -3924& \hfill -12384\\ \hfill & \hfill 48& \hfill -864& \hfill 4656& \hfill 2928\end{array}}\\ \begin{array}{rrrrrr}\hfill & \hfill 12& \hfill -216& \hfill 1164& \hfill 732& \hfill -9456\end{array}\end{array}$

For $t=5$ .

  $\begin{array}{l}5\overline{)\begin{array}{rrrrr}\hfill 12& \hfill -264& \hfill 2028& \hfill -3924& \hfill -12384\\ \hfill & \hfill 60& \hfill -1020& \hfill 5040& \hfill 5580\end{array}}\\ \begin{array}{rrrrrr}\hfill & \hfill 12& \hfill -204& \hfill 1008& \hfill 1116& \hfill -6804\end{array}\end{array}$

For $t=6$ .

  $\begin{array}{l}6\overline{)\begin{array}{rrrrr}\hfill 12& \hfill -264& \hfill 2028& \hfill -3924& \hfill -12384\\ \hfill & \hfill 72& \hfill -1152& \hfill 5256& \hfill 7992\end{array}}\\ \begin{array}{rrrrrr}\hfill & \hfill 12& \hfill -192& \hfill 876& \hfill 1332& \hfill -4392\end{array}\end{array}$

For $t=7$ .

  $\begin{array}{l}7\overline{)\begin{array}{rrrrr}\hfill 12& \hfill -264& \hfill 2028& \hfill -3924& \hfill -12384\\ \hfill & \hfill 84& \hfill -1260& \hfill 5376& \hfill 10164\end{array}}\\ \begin{array}{rrrrrr}\hfill & \hfill 12& \hfill -180& \hfill 768& \hfill 1452& \hfill -2220\end{array}\end{array}$

For $t=8$ .

Zero remainder means this is the solution.

  $\begin{array}{l}8\overline{)\begin{array}{rrrrr}\hfill 12& \hfill -264& \hfill 2028& \hfill -3924& \hfill -12384\\ \hfill & \hfill 96& \hfill -1344& \hfill 5472& \hfill 12384\end{array}}\\ \begin{array}{rrrrrr}\hfill & \hfill 12& \hfill -168& \hfill 684& \hfill 1548& \hfill 0\end{array}\end{array}$

For $t=9$ .

  $\begin{array}{l}9\overline{)\begin{array}{rrrrr}\hfill 12& \hfill -264& \hfill 2028& \hfill -3924& \hfill -12384\\ \hfill & \hfill 108& \hfill -1404& \hfill 5616& \hfill 15228\end{array}}\\ \begin{array}{rrrrrr}\hfill & \hfill 12& \hfill -156& \hfill 624& \hfill 1692& \hfill 2844\end{array}\end{array}$

For $t=10$ .

  $\begin{array}{l}10\overline{)\begin{array}{rrrrr}\hfill 12& \hfill -264& \hfill 2028& \hfill -3924& \hfill -12384\\ \hfill & \hfill 120& \hfill -1440& \hfill 5880& \hfill 19560\end{array}}\\ \begin{array}{rrrrrr}\hfill & \hfill 12& \hfill -144& \hfill 588& \hfill 1956& \hfill 7176\end{array}\end{array}$

The possible solutions t=8.

(d)

To find: Graph the function $F\left(t\right)$ .

The graph

  

Given information:

The function $F\left(t\right)=12t^4-264t^3+2028t^2-3924t+43,916$ .

Calculation:

The graph

  

Remember that $F$ is in thousands, so $56300000$ is at $y=56300$ .

There are no other reasonable solutions because the only other intersection point occurs at a negative $t$ value. Also, $F$ is a $4th$ degree polynomial and past $t=8$ it looks like it is done with its turning points and is taking off, not coming back.