Chapter 2.6, Problem 48PS

Solution

To find: a polynomial equation.

Equation: $2x^3-5x^2-9=0$

Possible rational solutions of the problem: $3,9,\frac{9}{2}$ .

Given information: 

The base of a pyramid is square, sides that are $x$ in length, a height of $2x-5$ and a volume of $3$ .

Formula used:

The pyramid's volume is

  $V=\frac{1}{3}Bh$

where $B$ is the area of its base, and $h$ is its height.

Calculation:

The base is a square with sides of length $x$ , so the area of the base is the square of the length of its sides, or

  $B=x^2$

Now, the volume is

  $V=\frac{1}{3}{x}^2(2x-5)$

Also, from the exercise,

  $V=3$

Now, obtain the equation

  $\frac{1}{3}{x}^2\left(2x-5\right)=3$

Multiply both sides by $3:$

  $x^2\left(2x-5\right)=9$

Using the Distributive Property,

  $x^2(2x-5)=\left(x^2\right)(2x)+\left(x^2\right)(-5)=2x^3-5x^2$

To get the equation, take the equality from the previous step and deduct $9$ from both sides of it.

  $2x^3-5x^2-9=0$

All rational solutions of the equation are of the form

  $\frac{p}{q},$

where $p$ is a constant term factor $-9$ , and a component of the leading coefficient is $q$ $\left(2\right)$ .

Factors of $-9$ :

  $\pm 1,\pm 3,\pm 9$

Factors of 2:

  $\pm 1,\pm 2$

So, all possible rational zeros are

  $\pm \frac{1}{1},\pm \frac{3}{1},\pm \frac{9}{1},\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{9}{2}$

Simplify fractions:

  $\pm 1,\pm 3,\pm 9,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{9}{2}$

However, not all of these values can be the solution of the starting problem! Since $x$ is the length of the side of the base, it must be positive. Also, $2x-5$ must be positive. Using these, the possible solutions of the starting problem are $3,9,\frac{9}{2}$ .