Chapter 6.8, Problem 40PPS

To determine

To calculate: The zeros of the function $f\left(t\right)=t^4-31t^3+308t^2-1100t+1200$ for the roller coaster is at ground level.

Answer to Problem 40PPS

Whentime t is $2,4,10\text{ and }15$ the roller coaster is at ground level.

Explanation of Solution

Given information:

The function $f\left(t\right)=t^4-31t^3+308t^2-1100t+1200$ represents the height of roller coaster where t denotes the time in seconds.

Formula used:

A polynomial of n degree has n zeros, which can be either real or imaginary.

Descartes’ rule of signs states that consider a polynomial $P\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ with real coefficients then, the number of times the sign between coefficients changes is equal to count of positive real zeroes of $P\left(x\right)$ or is less than this by an even number, the number of times the sign between coefficients changes is equal to count of negative real zeroes of $P\left(-x\right)$ or is less than this by an even number.

Calculation:

Consider the function $f\left(t\right)=t^4-31t^3+308t^2-1100t+1200$ .

Observe that degree of polynomial is 4, so the functions has 4 zeros which can be either real or imaginary.

Descartes’ rule of signs states that consider a polynomial $P\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ with real coefficients then, the number of times the sign between coefficients changes is equal to count of positive real zeroes of $P\left(x\right)$ or is less than this by an even number.

So, count the number of times the sign changes between the coefficients of $f\left(t\right)=t^4-31t^3+308t^2-1100t+1200$

There are 4, 2 or 0 positive real zeros.

Now,

  $f\left(-t\right)=t^4+31t^3+308t^2+1100t+1200$

Descartes’ rule of signs states that consider a polynomial $P\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ with real coefficients then, the number of times the sign between coefficients changes is equal to count of negative real zeroes of $P\left(-x\right)$ or is less than this by an even number.

So, count the number of times the sign changes between the coefficients of $f\left(-t\right)=t^4+31t^3+308t^2+1100t+1200$ .

There are no negative real zero.

Next, construct a table with possible combinations of real and imaginary zeros.

  $\begin{array}{cccc}\begin{array}{c}\text{Number of Positive}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of Negative}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of}\\ \text{Imaginary zeros}\end{array}& \begin{array}{c}\text{Total Number}\\ \text{of zeros}\end{array}\\ 4& 0& 0& 4+0+0=4\\ 2& 0& 2& 2+0+2=4\\ 0& 0& 4& 0+0+4=4\end{array}$

Recall that the Rational zero theorem states that provided a polynomial $P\left(x\right)$ with integral coefficients then every rational zero is of the form $\frac{p}{q}$ that is a rational number in simplest form. Here, p is a factor of the constant term and q is a factor of leading coefficient.

For the provided function leading coefficient is 1 and constant term is 1200. Therefore, p is a factor of 1200 and q is a factor of 1.

The possible combinations of $\frac{p}{q}$ in simplest form are,

  $\frac{p}{q}=\pm 1,\pm 2,\pm 3,\pm 4,\pm 5,\pm 6,\pm 8,\pm 10,\pm 12,\pm \dots$

Next, construct a table with help of synthetic substitution to compute the value of $f\left(t\right)$ for real values of $t$ .

  $\begin{array}{cc}t& f\left(t\right)\\ 1& 378\\ -1& 2640\\ 2& 0\\ -2& 4896\\ 3& -84\\ -3& 8190\\ 4& 0\end{array}$

As observed two zeros are resulted at $t=2\text{ and }t=4$ .

So, factor of the polynomial is $\left(t-2\right)\left(t-4\right)=t^2-6t+8$ .

Now, divide the polynomial $f\left(t\right)=t^4-31t^3+308t^2-1100t+1200$ by $t^2-6t+8$ .

  $t^2-6t+8\begin{array}{c}\hfill t^2-25t+150\\ \hfill \overline{)\begin{array}{l}\begin{array}{ccccccccc}{t}^4& -& 31t^3& +& 308t^2& -& 1100t& +& 1200\end{array}\\ \begin{array}{ccccccccc}{t}^4& -& 6t^3& +& 8t^2& & & & \end{array}\\ \begin{array}{ccccccccc}& -& 25t^3& +& 300t^2& -& 1100t& & \end{array}\\ \begin{array}{ccccccccc}& -& 25t^3& +& 150t^2& -& 200t& & \end{array}\\ \begin{array}{ccccccccc}& & & & & 150t^2& +& 900t& +1200\end{array}\\ \begin{array}{ccccccccc}& & & & 150t^2& & +& 900t& +1200\end{array}\\ \begin{array}{ccccccccc}& & & & & & & 0& \end{array}\end{array}}\end{array}$

Now, the depressed polynomial is obtained is $t^2-25t+150$ .

Factor the polynomial.

  $\begin{array}{c}{t}^2-25t+150=0\\ t^2-15t-10t+150=0\\ t\left(t-15\right)-10\left(t-15\right)=0\\ \left(t-10\right)\left(t-15\right)=0\end{array}$

Either $\left(t-10\right)=0\text{ or }\left(t-15\right)=0$

That is, $t=10\text{ or }t=15$ .

Thus, the zeros of the function $f\left(t\right)=t^4-31t^3+308t^2-1100t+1200$ are $2,4,10\text{ and }15$ .

When time t is $2,4,10\text{ and }15$ the roller coaster is at ground level.