Section 5.9. Applications that involve factoring

Section 5.9: Applications that involve factoring Chapter 5: Factoring A few important phrases are described below that can give us clues for how to create an equation to solve the kinds of problems we will see in this section. • The word a number often becomes the variable in the algebra. (I usually replace the word number with an x, unless a different letter makes more sense. (number = x) • The word is generally is replaced with an equal sign. (is =) y is 12 becomes y=12 • Key phrases will need to be transformed into expressions. We learned many of the phrases in section 1.1. • Plus, often represents addition A number plus 5 becomes x + 5 • Twice, often represents multiplication Twice a number becomes 2x • We will transform the key phrases into expressions and use the expressions to create equations to solve our problems. Steps to solve application problems that involve factoring Step 1: Read the problem to gain an understanding of what is known and what needs to be solved for Step 2: Identify what we are looking for and create variables Step 3: Create an equation Step 4: Solve for the first unknown using the appropriate technique, be sure to omit any solution that is not feasible. Step 5: Solve for any additional unknowns using the answer created in step 4. Step 6: Check Step 7: Write answer in a sentence Section 5.9: Applications that involve factoring For Example: The product of two consecutive even integers is 80. Find all such numbers. Step 1: Read the problem to gain an understanding of what is known and what needs to be solved for Step 2: Identify what we are looking for and create variables Step 3: Create an equation Step 4: Solve for the first unknown using the appropriate technique, be sure to omit any solution that is not feasible. Step 5: Solve for any additional unknowns using the answer created in step 4. Step 6: Check Step 7: Write answer in a sentence Chapter 5: Factoring Step 1: We need to find two integers that satisfy the conditions. Specifically, the numbers are consecutive even integers, and the product of the numbers equals 80. Step 2: We had problems that involved consecutive even integers earlier this semester. We should define the variables as follows: 𝑙𝑒𝑡 𝑥 = 𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝐿𝑒𝑡 𝑥 + 2 = 𝑙𝑎𝑟𝑔𝑒𝑟 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 (remember consecutive even integers as well as consecutive odd integers are two apart. Therefore the larger number is defined as x + 2.) Step 3: The product of the numbers (when the numbers are multiplied) the result must equal 80. 𝑥(𝑥 + 2) = 80 Step 4: Solve: Step 6: Check Check x = -10 Check x = 8 −10(−10 + 2) = 80 8(8 + 2) = 80 −10(−8) = 80 8(10) = 80 80 = 80 ✓ 80 = 80 ✓ 𝑥 2 + 2𝑥 = 80 𝑥 2 + 2𝑥 − 80 = 0 (𝑥 + 10)(𝑥 − 8) = 0 𝑥 = −10, 𝑥 = 8 Step 5: Solve for additional unknowns Determine the larger integer when x = -10 Larger number = 𝑥 + 2 = −10 + 2 = −8 First solution: Small integer 𝑥 = −10, large integer x + 2 = −8 Step 7: There are two solutions. Each solution contains a smaller number and a larger number. Answer: First solution: The smaller number is −10 and the larger number is −8. Second solution: The smaller number is 8 and the larger number is 10. Determine the larger integer when x = 8 Larger integer = 𝑥 + 2 = 8 + 2 = 10 Second Solution: small integer 8, large integer 10 Section 5.9: Applications that involve factoring Chapter 5: Factoring I generally label the 4 sides of a rectangle as depicted below. I generally label the three sides of a right triangle as depicted below. 𝑙 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 2 𝑙𝑜𝑛𝑔𝑒𝑟 𝑠𝑖𝑑𝑒𝑠 𝑤 = 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 2 𝑠ℎ𝑜𝑟𝑡𝑒𝑟 𝑠𝑖𝑑𝑒𝑠 𝑎 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠ℎ𝑜𝑟𝑡 𝑙𝑒𝑔 𝑏 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑙𝑜𝑛𝑔 𝑙𝑒𝑔 𝑐 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 Triangle formula needed for this section: Rectangle formula needed for this section: Pythagorean Theorem for right triangles 𝐴𝑟𝑒𝑎 (𝑎) = 𝑙𝑒𝑛𝑔𝑡ℎ × 𝑤𝑖𝑑𝑡ℎ 𝑎 = 𝑙𝑤 𝑎2 + 𝑏 2 = 𝑐 2 Section 5.9: Applications that involve factoring For Example: The length of a rectangular bedroom is 3 feet longer than its width. The area of the bedroom is 88 square feet. Find the dimensions of the room. Step 1: Read the problem to gain an understanding of what is known and what needs to be solved for Step 2: Identify what we are looking for and create variables Step 3: Create an equation Step 4: Solve for the first unknown using the appropriate technique, be sure to omit any solution that is not feasible. Step 5: Solve for any additional unknowns using the answer created in step 4. Step 6: Check Step 7: Write answer in a sentence Chapter 5: Factoring Step 1: We are asked to find the length and width of a rectangular bedroom. Step 2: 𝐿 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑏𝑒𝑑𝑟𝑜𝑜𝑚 𝑊 = 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑏𝑒𝑑𝑟𝑜𝑜𝑚 Step 3: 𝑙𝑒𝑛𝑔𝑡ℎ 𝑖𝑠 3 𝑙𝑜𝑛𝑔𝑒𝑟 𝑡ℎ𝑎𝑛 𝑤𝑖𝑑𝑡ℎ 𝐿 =𝑊+3 𝐴𝑟𝑒𝑎 = 88 𝑠𝑞𝑢𝑎𝑟𝑒 𝑓𝑒𝑒𝑡 𝐿𝑊 = 88 (𝑊 + 3)𝑊 = 88 Step 4: 𝑊(𝑊 + 3) = 88 𝑊 2 + 3𝑊 = 88 𝑊 2 + 3𝑊 − 88 = 0 (𝑊 + 11)(𝑊 − 8) = 0 𝑊 + 11 = 0 𝑊 − 8 = 0 𝑊 = −11 𝑊 = 8 The solution 𝑊 = −11 is not feasible and needs to be omitted. The width of a bedroom must be greater than 0. Step 5: 𝐿 =𝑊+3 𝐿 = 8 + 3 = 11 Step 6: 𝑊(𝑊 + 3) = 88 8(8 + 3) = 88 8(11) = 88 88 = 88 ✓ Step 7: The length of the bedroom is 11 feet and the width of the bedroom is 8 feet. Section 5.9: Applications that involve factoring For Example: The length of the hypotenuse of a right triangle is 8 inches more than the shortest leg. The length of the longer leg is 7 inches more than the length of the shorter leg. Find the length of each side of the triangle. Chapter 5: Factoring Step 1: I need to find the length of the three sides of the triangle. Step 2: Each sentence compares the length of one of the sides to the length of the short leg. I call the short leg “a” and this will be the variable I use. 𝑎 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠ℎ𝑜𝑟𝑡 𝑙𝑒𝑔 Step 1: Read the problem to gain an understanding of what is known and what needs to be solved for Step 2: Identify what we are looking for and create variables Step 3: Create an equation Step 4: Solve for the first unknown using the appropriate technique, be sure to omit any solution that is not feasible. Step 5: Solve for any additional unknowns using the answer created in step 4. Step 6: Check Step 7: Write answer in a sentence Step 6: 𝑎2 + (𝑎 + 7)2 = (𝑎 + 8)2 length of the hypotenuse (c)of a right triangle is 8 inches more than the shortest leg (a). 𝑐 =𝑎+8 The length of the longer leg (b) is 7 inches more than the length of the shorter leg (a) 𝑏 = 𝑎+7 Step 3: 𝑎2 + 𝑏 2 = 𝑐 2 𝑎2 + (𝑎 + 7)2 = (𝑎 + 8)2 Step 4: 𝑎2 + (𝑎 + 7)(𝑎 + 7) = (𝑎 + 8)(𝑎 + 8) 1𝑎2 + 1𝑎2 + 7𝑎 + 7𝑎 + 49 = 1𝑎2 + 8𝑎 + 8𝑎 + 64 (5)2 + (5 + 7)2 = (5 + 8)2 (5)2 + (12)2 = (13)2 25 + 144 = 169 169 = 169 ✓ 2𝑎2 + 14𝑎 + 49 = 1𝑎2 + 16𝑎 + 64 −1𝑎2 − 16𝑎 − 64 − 1𝑎2 − 16𝑎 − 64 𝑎2 − 2𝑎 − 15 = 0 (𝑎 + 3)(𝑎 − 5) = 0 𝑎 = −3 𝑎 = 5 Step 7: Answer: The short leg measures 5 inches. The long leg measures 12 inches. The hypotenuse measures 13 inches. The solution a= −3 is not feasible and needs to be omitted. The length of a side of a triangle must be greater than 0. Step 5: 𝑎= 5 𝑏 = 𝑎 + 7 = 5 + 7 = 12 𝑐 = 𝑎 + 8 = 5 + 8 = 13 𝑎 = 5, 𝑏 = 12, 𝑐 = 13