Chapter 2.6, Problem 13Q

Solution

To find: what are the dimensions of the patio?

Sides: $16$ in

Thickness: $0.5$ in

Given information: 

The thickness of the patio is $15.5$ feet less than each side length.

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:

Let $x$ be the length of the side of the patio. Then, its thickness is $x-15.5$ .

Thus, the patio's volume is

  $(x)(x)(x-15.5)=x^2(x-15.5)\stackrel{(1)}{=}{x}^3-15.5x^2$

In (1) used the Distributive Property.

From the text, the volume is $128$ .

Thus, with us the equation

  $x^3-15.5x^2=128$

Subtract $128$ from both sides:

  $x^3-15.5x^2-128=0$

Multiply both sides by $2$ (so you have integer coefficients!):

  $2x^3-31x^2-256=0$

All possible rational solutions are of the form

  $\frac{p}{q},$

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

The constant term is $-256$ , and its factors are

  $\pm 1,\pm 2,\pm 4,\pm 8,\pm 16,\pm 32,\pm 64,\pm 128,\pm 256$

The leading coefficient is $2$ , and its factors are

  $\pm 1,\pm 2$

All possible rational solutions are:

  $\begin{array}{l}\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{4}{1},\pm \frac{8}{1},\pm \frac{16}{1},\pm \frac{32}{1},\pm \frac{64}{1},\pm \frac{128}{1},\pm \frac{256}{1}\\ \pm \frac{1}{2},\pm \frac{2}{2},\pm \frac{4}{2},\pm \frac{8}{2},\pm \frac{16}{2},\pm \frac{32}{2},\pm \frac{64}{2},\pm \frac{128}{2},\pm \frac{256}{2}\end{array}$

Simplify fractions:

  $\begin{array}{l}\pm 1,\pm 2,\pm 4,\pm 8,\pm 16,\pm 32,\pm 64,\pm 128,\pm 256,\\ \pm \frac{1}{2},\pm 1,\pm 2,\pm 4,\pm 8,\pm 16,\pm 32,\pm 64,\pm 128\end{array}$

Remove duplicates:

  $\pm \frac{1}{2},\pm 1,\pm 2,\pm 4,\pm 8,\pm 16,\pm 32,\pm 64,\pm 128,\pm 256$

Try it $16$ is a solution (use the synthetic substitution):

  $\begin{array}{l}16\overline{)\begin{array}{llll}2\hfill & -31\hfill & 0\hfill & -256\hfill \\ \hfill & 32\hfill & 16\hfill & 256\hfill \end{array}}\\ \begin{array}{lllll}\hfill & 2\hfill & 1\hfill & 16\hfill & 0\hfill \end{array}\end{array}$

So, $16$ is indeed the solution of the equation! Also,

  $2x^3-31x^2-256=(x-16)\left(2x^2+x+16\right)$

To find other solution and solve the equation

  $2x^2+x+16=0$

The solution of the equation

  $a{x}^2+bx+c=0$

Is

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here $a=2,b=1$ and $c=16$ , so the solution is

  $\begin{array}{c}x=\frac{-1\pm \sqrt{{(1)}^2-4(2)(16)}}{2(2)}\\ =\frac{-1\pm \sqrt{1-128}}{4}\\ =\frac{-1\pm \sqrt{-127}}{4}\end{array}$

However, these numbers are imaginary!

So, the only real solution of the equation from step $3$ is $16$ .

Thus, the length of the sides of the patio is 16 inches, and the thickness is $16-15.5=0.5$ in.