2. Martin-Gay, Beginning Algebra, 5ed 22

3. 33

4. 44

5. 55

6. 66

7. 77

8. 88

9. 99

10. 10 Example For each pair of polynomials, find the least common multiple. a) 16a and 24b b) 24x 4 y 4 and 6x 6 y 2 c) x 2  4 and x 2  2x  8 Solution a) 16a = 2  2  2  2  a 24b = 2  2  2  3  b The LCM = 2  2  2  2  a  3  b The LCM is 2 4  3  a  b, or 48ab 16a is a factor of the LCM 24b is a factor of the LCM

11. Martin-Gay, Beginning Algebra, 5ed 11 Example continued b) 24x 4 y 4 = 2  2  2  3  x  x  x  x  y  y  y  y 6x 6 y 2 = 2  3  x  x  x  x  x  x  y  y LCM = 2  2  2  3  x  x  x  x  y  y  y  y  x  x Note that we used the highest power of each factor. The LCM is 24x 6 y 4 c) x 2  4 = (x  2)(x + 2) x 2  2x  8 = (x + 2)(x  4) LCM = (x  2)(x + 2)(x  4) x 2  4 is a factor of the LCM x 2  2x  8 is a factor of the LCM

12. Martin-Gay, Beginning Algebra, 5ed 12 a) b) c) d) e)