Example 6 from Algebra and Trigonometry, 5.4 Dividing Polynomials

Using Polynomial Division in an Application Problem

The volume of a rectangular solid is given by the polynomial $3 x^{4} - 3 x^{3} - 33 x^{2} + 54 x .$ The length of the solid is given by $3 x$ and the width is given by $x - 2.$ Find the height, $h,$ of the solid.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
There are a few ways to approach this problem. We need to divide the expression for the volume of the solid by the expressions for the length and width. Let us create a sketch as in Figure 3.

Figure 3

We can now write an equation by substituting the known values into the formula for the volume of a rectangular solid.

$\begin{matrix} V & = & l \cdot w \cdot h \\ 3 x^{4} - 3 x^{3} - 33 x^{2} + 54 x & = & 3 x \cdot (x - 2) \cdot h \end{matrix}$

To solve for $h,$ first divide both sides by $3 x .$

$\begin{matrix} \frac{3 x \cdot (x - 2) \cdot h}{3 x} & = & \frac{3 x^{4} - 3 x^{3} - 33 x^{2} + 54 x}{3 x} \\ (x - 2) h & = & x^{3} - x^{2} - 11 x + 18 \end{matrix}$

Now solve for $h$ using synthetic division.

$h = \frac{x^{3} - x^{2} - 11 x + 18}{x - 2}$

The quotient is $x^{2} + x - 9$ and the remainder is 0. The height of the solid is $x^{2} + x - 9.$

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

The area of a rectangle is given by $3 x^{3} + 14 x^{2} - 23 x + 6.$ The width of the rectangle is given by $x + 6.$ Find an expression for the length of the rectangle.

How did it go?

Solution

Your turn

Solution

Your turn