Finding Real Roots of Polynomial Equations

Question

A tank's length is 3 feet longer than its width. The height is 1 foot less than the width. The volume of the tank is 160 cubic feet. What is the width of the tank?

A. 6 ft.
B. 3 ft.
C. 1 ft.
D 5 ft.

Answer 1

Length times width times height is the volume.

 

Let the width = x

Let the length = x + 3

Let the height = x - 1

Volume = 160 ft³

 

Use the variables, we can set an equation that represents the volume.

 

x(x + 3)(x - 1) = 160

 

Solve for x from this equation.

 

x(x² + 2x - 3) = 160

 

x³ + 2x² - 3x = 160

 

 

Subtract 160 on both sides of the equation.

 

x³ + 2x² - 3x - 160 = 0

 

 

According to the ration root theorem, the possible roots are

 

c = ±1, ±2, ±4, ±5, ±8, ±10, ±16, ±20, ±32, ±40, ±80, ±160 

 

 

Using synthetic division, 5 is a root.  So we can factor this as

 

(x - 5)(x² + 7x + 32) = 0

 

 

Next, we can factor the second factor using FOIL, if possible

 

x² + 7x + 32 = 0

 

 

Factoring using FOIL is not possible.  Therefore, we would have to use the quadratic formula, and the solutions will not be an integer.

 

The only solution that is an integer is 5.

 

 

The width is 5 feet.

Answer 2

L = W + 3

H = W - 1

V = W * H * L = 160

 

V = W *(W-1) * (3 + W) = (W² -W)*(W + 3) = W³ - W² +3W² - 3W = W³ + 2W² - 3W = 160

 

W³ + 2W² - 3W - 160 = 0  

Factoring this is really annoying, but if you plug it in here http://akiti.ca/Quad3Deg.html

 

You'll get 5 ft, so choice D.

 

 

Answer 3

width = w

length = w + 3

height = w - 1

 

w(w+3)(w-1) = 160

w(w² + 2w - 3) = 160

w³ + 2w² - 3w -160 = 0

 

(w-5)(w² + 7w +32) = 0

 

Setting w -5 = 0, we get w = 5ft.  D.