Chapter 2.6, Problem 45PS

Solution

To find: the dimensions of the mold.

Base's sides are $3$ inches long. The height is $7$ inches.

Given information: 

Each mold is a square prism with a square base and a height that is $4$ inches more than the length of each side. $63$ inch-cubic of molten glass may fit inside each mold.

Formula used:

The Rational Zero Theorem:

If $f\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ consists of integer coefficients, then the shape of each rational zero of $f$ is as follows:

  $\frac{p}{q}=\frac{\text{ factor of constant term }{a}_0}{\text{ factor of leading coefficient }{a}_n}$

Calculation:

The length and the width of the box is $x$ , and the height is $4$ inches greater, so the height is $x+4$ .

The volume is:

  $\begin{array}{c}V=(x)(x)(x+4)\\ =x^2(x+4)\\ \stackrel{(1)}{=}{x}^3+4x^2\end{array}$

In $\left(1\right)$ , the Distributive Property.

Since, from the text of the exercise, $V=63$ , the equation

  $x^3+4x^2=63$

Subtract $63$ from both sides:

  $x^3+4x^2-63=0$

So, of the function's zeros $f\left(x\right)=x^3+4x^2-63$ .

Examine potential rational zeros.

Each rational zero has the following form:

  $\underset{\_}{p}$

where $p$ is a factor of the constant term, and $q$ is a factor of the leading coefficient.

The constant term is $-63$ , and its factors are

  $\pm 1,\pm 3,\pm 7,\pm 9,\pm 21,\pm 63$

The leading coefficient is $1$ , and its factors are

  $\pm 1$

Therefore, each and every rational zero is $\pm \frac{1}{1},\pm \frac{3}{1},\pm \frac{7}{1},\pm \frac{9}{1},\pm \frac{21}{1},\pm \frac{63}{1}$

Simplify each fraction:

  $\pm 1,\pm 3,\pm 7,\pm 9,\pm 21,\pm 63$

Try $1$

Use the synthetic division

  $\begin{array}{l}1\overline{)\begin{array}{llll}1\hfill & 4\hfill & 0\hfill & -63\hfill \\ \hfill & 1\hfill & 5\hfill & 5\hfill \end{array}}\\ \begin{array}{lllll}\hfill & 1\hfill & 5\hfill & 5\hfill & -58\hfill \end{array}\end{array}$

So, $f\left(1\right)=-58$ so $1$ is not a zero of this function.

Try $3$

Use the synthetic division

  $\begin{array}{l}3\overline{)\begin{array}{llll}1\hfill & 4\hfill & 0\hfill & -63\hfill \\ \hfill & 3\hfill & 21\hfill & 63\hfill \end{array}}\\ \begin{array}{lllll}\hfill & 1\hfill & 7\hfill & 21\hfill & 0\hfill \end{array}\end{array}$

So, $f(3)=0$ , so it is indeed the zero of $f(x)$

Also, from the last row.

  $f(x)=(x-3)\left(x^2+7x+21\right)$

(The coefficients of the computed quotient are represented by the numbers in the last row, excluding the final number when dividing $f(x)$ by $x-3)$ .

So, the equation from step 4 is now

  $(x-3)\left(x^2+7x+21\right)=0$

From before, $x=3$ is the solution (because it is the zero of the function $f(x)$ .

Consider if $x^2+7x+21=0$

Let us solve the equation

  $x^2+7x+21=0$

The solution of the equation

  $a{x}^2+bx+c=0$ is

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here, $a=1,b=7$ , and $c=21$ , so

  $\begin{array}{c}x=\frac{-7\pm \sqrt{{(7)}^2-4(1)(21)}}{2(1)}\\ =\frac{-7\pm \sqrt{-35}}{2}\end{array}$

So, there are no real solutions of the equation $x^2+7x+21=0$ .

To conclude, the only real solution of the starting equation is x=3.