3. Example Add. Simplify the result, if possible. a)b) Solution a) b) Combining like terms Factoring Combining like terms in the numerator

4. Example Subtract and, if possible, simplify: a)b) Solution a) The parentheses are needed to make sure that we practice safe math. Removing the parentheses and changing the signs (using the distributive law) Combining like terms

5. Example continued b) Removing the parentheses (using the distributive law) Factoring, in hopes of simplifying Removing the clever form of 1

7. 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

8. 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

9. Example For each group of polynomials, find the least common multiple. a) 15x, 30y, 25xyzb) x 2 + 3, x + 2, 7 Solution a) 15x = 3  5  x 30y = 2  3  5  y 25xyz = 5  5  x  y  z LCM = 2  3  5  5  x  y  z The LCM is 2  3  5 2  x  y  z or 150xyz b) Since x 2 + 3, x + 2, and 7 are not factorable, the LCM is their product: 7(x 2 + 3)(x + 2).

11. Solution 1. First, we find the LCD: 9 = 3  3 12 = 2  2  3 2. Multiply each expression by the appropriate number to get the LCD. Example Add: LCD = 2  2  3  3 = 36 

12. Solution First, we find the LCD: a 2  4 = (a  2)(a + 2) a 2  2a = a(a  2) We multiply by a form of 1 to get the LCD in each expression: Example Add: LCD = a(a  2)(a + 2). 3a 2 + 2a + 4 will not factor so we are done.

13. Solution First, we find the LCD. It is just the product of the denominators: LCD = (x + 4)(x + 6). We multiply by a form of 1 to get the LCD in each expression. Then we subtract and try to simplify. Example Subtract: Multiplying out numerators When subtracting a numerator with more than one term, parentheses are important, practice safe math.

14. Solution Example Add: Adding numerators

16. Continued