FACTORING POLYNOMIAL TO DETERMINE THE DIMENSION OF RECTANGLE

To find the area of the rectangle, we can use the formula

Write two binomials that could represent the length and width of the rectangle.

Problem 1 :

4x² - 7x – 2

Solution :

Area of rectangle = length × width

Area = 4x² - 7x – 2

= 4x² - 8x + x – 2

= 4x(x - 2) + 1(x - 2)

= (4x + 1) (x - 2)

So, the length of rectangle is (4x + 1), and width is (x - 2).

Problem 2 :

16x² - 25

Solution :

Area of rectangle = length × width

Area = 16x² - 25

(a² - b²) = (a + b) (a - b)

= (4x + 5) (4x - 5)

So, the length of rectangle is (4x + 5), and width is (4x - 5).

Problem 3 :

9x² - 6x + 1

Solution :

Area of rectangle = length × width

Area = 9x² - 6x + 1

= 9x² - 3x – 3x + 1

= 3x(3x - 1) – 1(3x - 1)

= (3x - 1) (3x - 1)

So, the length of rectangle is (3x - 1) and width is (3x - 1).

Problem 4 :

3x² + 5x – 2

Solution :

Area of rectangle = length × width

Area = 3x² + 5x – 2

= 3x² + 6x – x – 2

= 3x(x + 2) – 1(x + 2)

= (3x - 1) (x + 2)

So, the length of rectangle is (3x - 1) and width is (x + 2).

Problem 5 :

A projector displays an image on a wall. The area (in square feet) of the rectangular projection can be represented by x² − 8x + 15.

a. Write a binomial that represents the height of the projection.

b. Find the perimeter of the projection when the height of the wall is 8 feet.

Solution :

a)

Area of rectangular projection = x² − 8x + 15

Factoring,

= x² − 5x - 3x + 15

= x(x - 5) - 3(x - 5)

= (x - 3)(x - 5)

Length of rectangular image = x - 3

Height of rectangular image = x - 5

b)  When x = 8

Length = 8 - 3 ==> 5

Width = 8 - 5 ==> 3

Perimeter = 2(5 + 3)

= 2(8)

= 16 feet

Problem 6 :

A company’s profit (in millions of dollars) can be represented by x² − 6x + 8, where x is the number of years since the company started. When did the company have a profit of $3 million?

Solution :

Profit = x² − 6x + 8

x² − 6x + 8 = 3

x² − 6x + 8 - 3 = 0

x² − 6x + 5 = 0

x² − 5x - 1x + 5 = 0

x(x - 5) - 1(x - 5) = 0

(x - 1)(x - 5) = 0

x = 1 and x = 5

It yeilds the profit in 1 year or in 5 years.

Problem 7 :

A web browser is open on your computer screen.

a. The area of the browser is 24 square inches. Find the value of x.

b. The browser covers 3/13 of the screen. What are the dimensions of the screen?

Solution :

a)

Area of the browser = 24 square inches

(x - 2)x = 24

x² - 2x = 24

x² - 2x - 24 = 0

x² - 6x + 4x - 24 = 0

x(x - 6) + 4(x - 6) = 0

(x - 6)(x + 4) = 0

x = 6 and x = -4 (is not possible)

b)

3/13 of area of screen = area of browser

(3/13) ⋅ length ⋅ width = 24

(3/13) ⋅ length ⋅ (x + 2) = 24

Applying x = 6

(3/13) ⋅ length ⋅ 8 = 24

length = (24/8) ⋅ (13/3)

length = 13

So, the length of the rectangle is 13 inches

Problem 8 :

You enlarge a photograph on a computer. The area (in square inches) of the enlarged photograph can be represented by x² + 17x + 70.

a. Write binomials that represent the length and width of the enlarged photograph.

b. How many inches greater is the length of the enlarged photograph than the width? Explain.

c. The area of the enlarged photograph is 154 square inches. Find the dimensions of each photograph.

Solution :

a) 

Area of enlarged photo = x² + 17x + 70

Length of the enlarged figure = x + 2 + 8

= (x + 10) inches

x² + 17x + 70 = x² + 10x + 7x + 70

= x(x + 10) + 7(x + 10)

= (x + 7)(x + 10)

Widht of the enlarged picture = x + 10

b) Comparing x + 10 and x + 7, the width of the enlarged rectangular figure is 3 inches more.

c)  x² + 17x + 70 = 154

x² + 17x + 70 - 154 = 0

x² + 17x - 84 = 0

x² + 21x - 4x - 84 = 0

x(x + 21) - 4(x + 21) = 0

(x - 4)(x + 21) = 0

x = 4 and x = -21

Length of the enlarged picture = x + 10 ==> 4 + 10 ==> 14 inches

Widht of the enlarged picture = x + 7 ==> 4 + 7 ==> 11 inches