Chapter 5.8, Problem 39PPS

a.

To determine

To calculate: The polynomial function that depicts the volume of swimming pool.

  

Answer to Problem 39PPS

The polynomial function is $V\left(x\right)=324x^3+54x^2-19x-2$ .

Explanation of Solution

Given information:

The dimensions of swimming pool. $9175$ cubic feet of water is hold by the pool.

  

Calculation:

Consider the dimensions of swimming pool. $9175$ cubic feet of water is hold by the pool.

  

The pool resembles like a parallelepiped.

Therefore, the volume $V\left(x\right)$ is expressed as,

  $\begin{array}{c}V\left(x\right)=4x\left(3x+1\right)\left(9x-2\right)+\frac{\left[x+\left(3x+1\right)\right]\left(12x+2\right)\left(9x-2\right)}{2}\end{array}$

     $\begin{array}{c}=4x\left(3x+1\right)\left(9x-2\right)+\frac{2\left(6x+1\right)\left(4x+1\right)\left(9x-2\right)}{2}\end{array}$

     $\begin{array}{c}=4x\left(3x+1\right)\left(9x-2\right)+\left(6x+1\right)\left(4x+1\right)\left(9x-2\right)\end{array}$

     $\begin{array}{c}=\left(9x-2\right)\left(12x^2+4x+24x^2+6x+4x+1\right)\end{array}$

Simplify it further as,

  $\begin{array}{c}V\left(x\right)=\left(9x-2\right)\left(36x^2+14x+1\right)\end{array}$

     $\begin{array}{c}=324x^3+54x^2-19x-2\end{array}$

Thus, the polynomial function is $V\left(x\right)=324x^3+54x^2-19x-2$ .

b.

To determine

To calculate: The zeros of the function $V\left(x\right)=324x^3+54x^2-19x-2$ .

Answer to Problem 39PPS

The zeros of the function $V\left(x\right)=324x^3+54x^2-19x-2$ are $x=3,x=-\frac{19}{12}+i\frac{\sqrt{8987}}{36}\text{ and }x=-\frac{19}{12}-i\frac{\sqrt{8987}}{36}$ . The reasonable value of x is $x=3$ .

Explanation of Solution

Given information:

The dimensions of swimming pool. $9175$ cubic feet of water is hold by the pool.

  

The function that represent volume of pool is $V\left(x\right)=324x^3+54x^2-19x-2$ .

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 dimensions of swimming pool. $9175$ cubic feet of water is hold by the pool.

  

The function that represent volume of pool is $V\left(x\right)=324x^3+54x^2-19x-2$ .

Therefore, the function is $9175=324x^3+54x^2-19x-2$ that is,

  $f\left(x\right)=324x^3+54x^2-19x-9177$

Observe that degree of polynomial is 3, so the functions has 3 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 $f\left(x\right)=324x^3+54x^2-19x-9177$

There is 1 positive real zeros.

Now,

  $f\left(-x\right)=-324x^3+54x^2+19x-9177$

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 $f\left(-x\right)=-324x^3+54x^2+19x-9177$ .

There are 2 or 0 negative real zero.

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

  $\begin{array}{cccc}\begin{array}{c}\text{Number of Positive}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of Negative}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of}\\ \text{Imaginary zeros}\end{array}& \begin{array}{c}\text{Total Number}\\ \text{of zeros}\end{array}\\ 1& 0& 2& 1+0+2=3\\ 1& 2& 0& 1+2+0=3\end{array}$

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.

Use the scientific calculator to solve the roots of the equation since the coefficients are too large.

The zeros of the function $V\left(x\right)=324x^3+54x^2-19x-2$ are $x=3,x=-\frac{19}{12}+i\frac{\sqrt{8987}}{36}\text{ and }x=-\frac{19}{12}-i\frac{\sqrt{8987}}{36}$ .

Since, only one real zero is obtained and other two complex that is imaginary roots so, the reasonable value of x is $x=3$ as it represent the dimension of the pool.