1. Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Chapter 5 Polynomials and Factoring

2. 5-2 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Solving Applications Applications The Pythagorean Theorem 5.8

3. 5-3 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example The Mitchell’s are designing a garden. The garden will be in the shape of a rectangle and have an area of 270 square feet. The width of the garden is 3 feet less than the length. Find the length and width. Solution 1. Familiarize. We first make a drawing. Recall that the area of a rectangle is Length  Width. We let x = the length, in feet. The width is then x  3. x x  3

4. 5-4 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example 2. Translate. Rewording: The area of the rectangle is 270 ft 2. Translating: x(x  3) = 270 3. Carry out. We solve the equation. x(x  3) = 270 x 2  3x = 270 x 2  3x  270 = 0 (x  18)(x + 15) = 0 x  18 = 0 or x + 15 = 0 x = 18 or x =  15

5. 5-5 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example 4. Check. The solutions of the equation are 18 and  15. Since the length must be positive,  15 cannot be a solution. To check 18, we note that if the length is 18, then the width is x  3 or 15 and the area is 18 ft  15 ft = 270 ft 2. Thus the solution checks. 5. State. The garden is 18 feet long and 15 feet wide.

6. 5-6 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example h h  6 The math club is designing a brochure. The design calls for a triangle to be placed on the front. The triangle has a base that is 6 centimeters less than the height. If the area of the triangle is 216 cm 2. Find the height and base. Solution 1. Familiarize. We first make a drawing. The formula for the area of a triangle is A = ½ (base)(height). We let h = the height, in cm, and the base = h  6, in cm.

7. 5-7 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example 2. Translate. Rewording: The area of the triangle is 216 cm 2. Translating: = 216 3. Carry out. We solve the equation.

8. 5-8 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example 3. Carry out. h  24 = 0 or h + 18 = 0 h = 24 or h =  18 4. Check. The height must be positive, so  18 cannot be a solution. Suppose the height is 24 cm. The base would be 24  6, or 18 cm, and the area ½(24)(18), or 216 cm 2. These numbers check in the original problem. 5. State. The height of the triangle would be 24 cm and the base would be 18 cm.

9. Copyright © 2014, 2010, and 2006 Pearson Education, Inc. The Pythagorean Theorem In any right triangle, if a and b are the lengths of the legs and c is the length of the hypotenuse, then a 2 + b 2 = c 2 or (Leg) 2 + (Other leg) 2 = (Hypotenuse) 2. The equation a 2 + b 2 = c 2 is called the Pythagorean equation.* *The converse of the Pythagorean theorem is also true. That is, if a 2 + b 2 = c 2, then the triangle is a right triangle. c b a

10. 5-10 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example A 13-ft ladder is leaning against a house. The distance from the bottom of the ladder to the house is 7 ft less than the distance from the top of the ladder to the ground. How far is the bottom of the ladder from the house? Solution 1. Familiarize. We first make a drawing. The ladder and the missing dimensions form a right triangle. x = distance from top of the ladder to the ground x  7 = distance from bottom ladder to house. The hypotenuse has length 13 ft. 13 ft x x  7

11. 5-11 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example 2. Translate. Since a right triangle is formed, we can use the Pythagorean theorem: 3. Carry out. We solve the equation.

12. 5-12 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Example 3. Carry out. x  12 = 0 or x + 5 = 0 x = 12 or x =  5 4. Check. The integer  5 cannot be a length of a side because it is negative. When x = 12, x  7 = 5, and 12 2 + 5 2 = 13 2. So 12 checks. 5. State. The distance from the bottom of the ladder to the house is 5 ft. The distance from the top of the ladder to the ground is 12 ft.