Applications Using Factoring Polynomials This section will discuss applications involving the area of a rectangle, consecutive integers, and right triangles. Recall the steps that will help to translate and solve the problem: 1. Read through the entire problem 2. Organize the information (a drawing may be useful) 3. Write the equation 4. Solve the equation 5. Check the answer Area of a Rectangle Area of a rectangle is the number of squared units it takes to completely fill a rectangle. To solve this type of application, we need to use the formula for finding the area of a rectangle. Length Width = Area Since we are multiplying units of measurement by units of measurement, an area will always have squared units of measurement: ft 2, m 2, in 2, etc. Whenever an application involves a shape, it is useful to draw and label the shape. This will help to translate the information and write the equation.
Example 1: Jose purchased carpet to cover the floor of a rectangular room that has an area of 96 ft 2. The width of the room he is carpeting is 4 feet less than the length. Find the length and width of the room. length = x width = x 4 x(x 4) = 96 Use the formula for area of a rectangle x 2 4x = 96 Distribute x x 2 4x 96 = 0 Subtract 96 (x + 8)(x 12) = 0 Factor the polynomial x + 8 = 0 x 12 = 0 Set factor each equal to zero - 8 8 + 12 + 12 Solve the resulting equations x = -8 x = 12 Since a measurement cannot be negative, the length is 12 ft. To find the width, substitute 12 for x: x 4 = 12 4 = 8 The width is 8 ft. To check the answer: 12 ft 8 ft (12 8)(ft ft) 96 ft 2 True
Example 2: The length of a rectangle is three times the width. If the area of the rectangle is 432 m 2, find the length and width. length = 3x width = x 3x(x) = 432 Use the formula for area of a rectangle 3x 2 = 432 Multiply 3x and x 3x 2 432 = 0 Subtract 432 3(x 2 144) = 0 Factor out the GCF 3 3(x + 12)(x 12) = 0 Factor the polynomial 3 = 0 x + 12 = 0 x 12 = 0 Set each factor equal to zero 3 0-12 12 + 12 + 12 Solve the resulting equations x = -12 x = 12 Since a measurement cannot be negative, the width is 12 m. To find the length, substitute 12 for x: 3x = 3(12) = 36 The length is 36 m. To check the answer: 12 m 36 m (12 36)(m m) 432 m 2 True
Consecutive Integers Consecutive integers are numbers that come one after the other, such as 3, 4, 5. To get from one number to the next, we only have to add 1, so if the first number is x, the next number is x+1. Consecutive odd or even integers are numbers that are spaced apart by two, such as 2, 4, 6 or 3, 5, 7. If the first number is x, the next number is x+2. Example 3: The product of two consecutive integers is 210. Find all such pairs of integers. The first number is x The next number is x+1 x(x + 1) = 210 x 2 + x = 210 Distribute x x 2 + x 210 = 0 Subtract 210 (x + 15)(x 14) = 0 Factor the polynomial x + 15 = 0 x 14 = 0 Set each factor equal to zero - 15 15 + 14 + 14 Solve the resulting equations x = -15 x = 14 The first numbers are -15 and 14 The second numbers are: -15 + 1 = -14 and 14 + 1 = 15 The pairs of integers are: -15, -14 and 14, 15 To check the answer: -15(-14) = 210 True 14(15) = 210 True
Example 4: The product of two consecutive even integers is 120. Find all such pairs of integers. The first number is x The second number is x+2 x(x + 2) = 120 x 2 + 2x = 120 Distribute x x 2 + 2x 120 = 0 Subtract 120 (x + 12)(x 10) = 0 Factor the polynomial x + 12 = 0 x 10 = 0 Set each factor equal to zero - 12 12 + 10 + 10 Solve the resulting equations x = -12 x = 10 The first numbers are -12 and 10 The second numbers are: -12 + 2 = -10 and 10 + 2 = 12 The pairs of integers are: -12, -10 and 10, 12 To check the answer: -12(-10) = 120 True 10(12) = 120 True Example 5: The product of two consecutive odd integers is 63. Find all such pairs of integers. The first number is x The second number is x+2 x(x + 2) = 63 x 2 + 2x = 63 Distribute x x 2 + 2x 63 = 0 Subtract 63 (x + 9)(x 7) = 0 Factor the polynomial x + 9 = 0 x 7 = 0 Set each factor equal to zero - 9 9 + 7 + 7 Solve the resulting equations x = -9 x = 7
The first numbers are -9 and 7 The second numbers are: -9 + 2 = -7 and 7 + 2 = 9 The pairs of integers are: -9, -7 and 7, 9 To check the answer: -9(-7) = 63 True 7(9) = 63 True Right Triangle If we are given the measurements of two sides of a right triangle, we can easily find the measurement of the third side by using the Pythagorean Theorem. The Pythagorean Theorem states that the square of hypotenuse is equal to the sum of the squares of the other two sides. The formula is: a 2 + b 2 = c 2 where c is the hypotenuse; a and b are the legs
Example 6: The hypotenuse of a right triangle is 13 yards long. If one leg is seven yards longer than the other leg, how long are the two legs? a 13 a + 7 a 2 + b 2 = c 2 Use the Pythagorean Theorem a 2 + (a + 7) 2 = 13 2 Substitute the values into the formula a 2 + (a + 7) 2 = 169 Square 13 a 2 + a 2 + 14a + 49 = 169 Multiply (a + 7) 2 or (a + 7)(a + 7) 2a 2 + 14a + 49 = 169 Add a 2 +a 2 2a 2 + 14a 120 = 0 Subtract 169 2(a 2 + 7a 60) = 0 Factor out the GCF 2 2(a + 12)(a 5) = 0 Factor the polynomial 2 = 0 a + 12 = 0 a 5 = 0 Set each factor equal to zero 2 0-12 12 + 5 + 5 Solve the resulting equations a = -12 a = 5 Since a measurement cannot be negative, leg a is 5 yd long. To find the measurement of side b, substitute 5 for a: a + 7 = a + 5 = 12 Leg b is 12 yd long. To check the answer: 5 2 + 12 2 = 13 2 25 + 144 = 169 169 = 169 True