1. 12.4 Warm Up Warm Up Lesson Quiz Lesson Quiz Lesson Presentation Lesson Presentation Simplify Rational Expressions

2. 12.4 Warm-Up 1.x 2 + 8x + 15 2.2x 2 + 15x – 8 ANSWER (x + 3)(x + 5) ANSWER (x + 8)(2x – 1) Factor the polynomial. 3. You pay $20 to join an aerobics center and pay $4 per class session. Write an equation that gives the average cost C per session as a function of the number a of aerobics sessions that you take. ANSWER C = + 4 20 a

3. 12.4 Example 1 Find the excluded values, if any, of the expression. a. x + 8 10x SOLUTION a. The expression x + 8 is undefined when 10x = 0, or x = 0. 10x ANSWER The excluded value is 0. b. 2y + 14 5 c. v 2 – 9 4v4v d. 7w + 2 8w 2 + w + 5

4. 12.4 Example 1 SOLUTION The expression 5 is undefined when 2y + 14 = 0, or x = –7. 2y + 14 ANSWER The excluded value is –7. b.

5. 12.4 Example 1 SOLUTION c.c. The expression 4v is undefined when v 2 – 9 = 0, or (v + 3)(v – 3) = 0. The solutions of the equation are –3 and 3. v2 – 9v2 – 9 The excluded values are –3 and 3. ANSWER

6. 12.4 Example 1 SOLUTION d. The expression 7w + 2 8w 2 + w + 5 is undefined when 8w 2 + w + 5 = 0. The discriminant is b 2 – 4ac = 1 2 – 4(8)(5) < 0. So, the quadratic equation has no real roots. ANSWER There are no excluded values.

7. 12.4 Guided Practice Find the excluded values, if any, of the expression. x + 2 3x – 5 ANSWER The excluded value is. 5 3 2. 2 5y 2 + 2y +3 ANSWER There are no excluded values. 1.

8. 12.4 Guided Practice Find the excluded values, if any, of the expression. 3. n – 6 2n 2 – 5n – 12 ANSWER The excluded value is and 4. 3 2 – 4. 2m m 2 – 4 ANSWER The excluded value is 2, and 2. –

9. 12.4 Example 2 Simplify the rational expression, if possible. State the excluded values. a. r 2r SOLUTION Divide out common factor. 2r2r a. r = r 2r2r = 1 2 Simplify. ANSWER The excluded value is 0. b. 5x5x 5(x + 2) c. 6m 3 – 12m 3 18m 2 d. y 7 – y

10. 12.4 Example 2 SOLUTION b. 5x5x 5(x + 2) = 5 x Divide out common factor. Simplify. = x (x + 2) ANSWER The excluded value is – 2.

11. 12.4 Example 2 SOLUTION c. 18m 2 6m 3 – 12m 3 = 6m 2 (m – 2) 63m2m2 Factor numerator and denominator. = 6m 2 (m – 2) 6 3 m2m2 Divide out common factors. = m – 2 3 Simplify. ANSWER The excluded value is 0.

12. 12.4 Example 2 SOLUTION d. The expression y 7 – y is already in simplest form. ANSWER The excluded value is 7.

13. 12.4 Guided Practice 5. 4 a 3 22a 6 2 11a 3 ANSWERThe excluded value is 0. 6. 2c2c c + 5 2c2c ANSWER The excluded value is – 5. 7. 2s 2 + 8s 3s +12 ANSWER 2s2s 3 The excluded value is – 4. 8. 8x8x 8x 3 + 16x 2 ANSWER 1 x 2 + 2x The excluded values are 0 and – 2. Simplify the rational expression, if possible. State the excluded values.

14. 12.4 Example 3 Simplify x 2 – 3x – 10 x 2 + 6x + 8. State the excluded values. SOLUTION x 2 – 3x – 10 x 2 + 6x + 8 = (x – 5)(x + 2) (x + 4)(x + 2) Factor numerator and denominator. (x – 5)(x + 2) (x + 4)(x + 2) = = x – 5 x + 4 Divide out common factor. Simplify. ANSWER The excluded values are – 4 and – 2.

15. 12.4 Example 3 CHECK In the graphing calculator activity on page 560, you saw how to use a graph to check a sum or difference of polynomials. Check your simplification using a graphing calculator. Graph y 1 x 2 – 3x – 10 x 2 + 6x + 8 and y 2 = = x – 5 x + 4 The graphs coincide. So, the expressions are equivalent for all values of x other than the excluded values (–4 and –2 ).

16. 12.4 Example 4 Simplify x 2 – 7x + 12 16 – x 2. State the excluded values. SOLUTION x 2 – 7x + 12 16 – x 2 Factor numerator and denominator. (x – 3)(x – 4) – (x – 4)(4 + x) = Rewrite 4 – x as –( x – 4). Simplify. –(x – 4)(4+ x) (x – 3)(x – 4) = Divide out common factor. –(4 + x) (x – 3) = (x – 3)(x – 4) (x – 4)(4 + x) = ANSWER The excluded values are –4 and 4. (x + 4) (x – 3) = –

17. 12.4 Guided Practice Simplify the rational expression. State the excluded values. 9. x 2 + 3x + 2 x 2 + 7x + 10 (x + 1) (x + 5) ANSWER The excluded values are – 2 and – 5. 10. y 2 – 64 y 2 – 16y + 64 ANSWER (y + 8) (y – 8) The excluded value is 8 11. 5 + 4z – z 2 z 2 – 3z – 10 ANSWER (z + 1) (z + 2) – The excluded values are 5 and – 2.

18. 12.4 Example 5 46 – 2.2x C = 100 – 18x + 2.2x 2 where x is the number of years since 1991. Rewrite the model so that it has only whole number coefficients.Then simplify the model. The average cost C (in dollars per minute) for cell phone service in the United States during the period 1991–2000 can be modeled by CELL PHONE COSTS

19. 12.4 Example 5 C = 46 – 2.2x 100 – 18x + 2.2x 2 Write model. = 460 – 22x 1000 – 180x + 22x 2 Multiply numerator and denominator by 10. Factor numerator and denominator. = 2(230 – 11x) 2(500 – 90x + 11x 2 ) Divide out common factor. = 2(230 – 11x) 2(500 – 90x + 11x 2 ) SOLUTION = 230 – 11x 500 – 90x + 11x 2 Simplify.

20. 12.4 Guided Practice In Example 5, approximate the average cost per minute in 2000. 12. ANSWER The average cost per minute in 2000 is $.23/min.

21. 12.4 Lesson Quiz Simplify the rational expression, if possible. Find the excluded values. 1. 12x 3 18x 5 ANSWER 2 3 x 2 ; 0 2. 2x – 14 x 2 – 9x + 14 ANSWER 2 x – 2 ; 2, 7

22. 12.4 Lesson Quiz The average cost C (in dollars per hour ) for a cleaning service during the period 1994 – 2003 can be modeled by,where x is the numbers of years since 1994. Rewrite the model so that it has only whole number coefficients. Then simplify the model, if necessary. 3. 52 – 1.4x 10 – 14x + 2.1x 2 C =C = ANSWER 520 – 14x 100 – 140x + 2x C =