1. LESSON 3: USING FACTORS AND MULTIPLES WITH FRACTIONS Numbers

2. Todays Objectives Students will be able to demonstrate an understanding of factors of whole numbers by determining the: prime factors, greatest common factor (GCF), least common multiple (LCM), square root, cube root, including:  Solving problems that involve perfect squares and cubes and square and cube roots

3. Using Factors and Multiples with Fractions When simplifying a fraction (reducing to its lowest terms), the numerator and denominator are divided by common factors until the only common positive factor is the number 1. This can be done by writing the numerator and denominator as a product of their prime factors, and cancelling.

4. Example

6. Lowest Common Denominator (LCD) Addition and subtraction of fractions requires that the fractions have a common denominator. The most desirable common denominator is the lowest common denominator (LCD), which is the smallest possible denominator.

7. Example Write the fractions,, and as equivalent fractions in terms of their LCD by using factoring to determine the LCM of the denominators. Solution: Prime Factorization of 8: 2 x 2 x 2 = 2 3 Prime Factorization of 12: 2 x 2 x 3 = 2 2 x 3 1 Prime Factorization of 18: 2 x 3 x 3 = 2 1 x 3 2 The LCM, which is also the LCD, is 2 3 x 3 2 = 72.

8. Example

10. Solving Problems with Factors, Multiples, and Square or Cube Roots A variety of problems involving factors, multiples, square and cube roots are possible. Look at the following examples: In a store window display, one set of lights flashes every 12 seconds while another set of lights flashes every 14 seconds. If the two sets of lights are turned on at the same time, how often do they flash simultaneously? Solution: The set of lights will flash simultaneously at a time interval that is the LCM of 12 and 14. P.F. of 12: 2 x 2 x 3 = 2 2 x 3 1 P.F. of 14: 2 x 7 = 2 1 x 7 1 The LCM is 2 2 x 3 1 x 7 1 = 84 The two sets of lights will flash simultaneously every 84 seconds.

11. Example Tom and Jeremy have ordered 450 square patio stones with side lengths of 6in. What are the dimensions of the largest square patio they can build out of the patio stones, and how many stones will they have left over, if any?

12. Example Solution: The patio will be built from the number of stones that is the largest perfect square ≤ 450. Since 20 2 = 400 and 30 2 = 900, the perfect square must be between 20 2 and 30 2, and closer to 20 2. Try 21 2 and 22 2. Since 21 2 = 441 and 22 2 = 484, the required perfect square is 21 2 = 441. Thus, the patio dimensions will be 21 stones per side, which is 21 x 6 = 126 inches/side. This is equivalent to 10 feet 6 inches per side. Since only 441 of the 450 patio stones would be used, there would be 9 stones left over.