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Warm Up Simplify each expression. Assume all variables are nonzero.

1. y 2. x 6. y 5. y 3. x 2. Warm Up Simplify each expression. Assume all variables are nonzero. 1. x 5  x 2. x 7. 2. y 3  y 3. y 6. 3. 4. x 4. Factor each expression. 5. x 2 – 2 x – 8. ( x – 4)( x + 2). 6. x 2 – 5 x. x ( x – 5). 7. x 5 – 9 x 3.

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Warm Up Simplify each expression. Assume all variables are nonzero.

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  1. 1 y2 x6 y5 y3 x2 Warm Up Simplify each expression. Assume all variables are nonzero. 1.x5x2 x7 2.y3y3 y6 3. 4. x4 Factor each expression. 5. x2 – 2x – 8 (x – 4)(x + 2) 6. x2 – 5x x(x – 5) 7. x5 – 9x3 x3(x – 3)(x + 3)

  2. Objectives Simplify rational expressions. Multiply and divide rational expressions.

  3. Vocabulary rational expression

  4. In Lesson 8-1, you worked with inverse variation functions such as y = . The expression on the right side of this equation is a rational expression. A rational expression is a quotient of two polynomials. Other examples of rational expressions include the following: 5 x

  5. Caution! When identifying values for which a rational expression is undefined, identify the values of the variable that make the original denominator equal to 0. Because rational expressions are ratios of polynomials, you can simplify them the same way as you simplify fractions. Recall that to write a fraction in simplest form, you can divide out common factors in the numerator and denominator.

  6. 510x8 – 4 5 x4 = 36 3 10x8 6x4 Example 1A: Simplifying Rational Expressions Simplify. Identify any x-values for which the expression is undefined. Quotient of Powers Property The expression is undefined at x = 0 because this value of x makes 6x4 equal 0.

  7. (x+ 2)(x – 1) x2 + x – 2 (x+ 2) (x – 1)(x + 3) x2 + 2x – 3 (x + 3) Example 1B: Simplifying Rational Expressions Simplify. Identify any x-values for which the expression is undefined. Factor; then divide out common factors. = The expression is undefined at x = 1 and x = –3 because these values of x make the factors (x – 1) and (x + 3) equal 0.

  8. 0 4 0 0 (–3)2 + (–3) – 2 (1)2 + (1) – 2 (–3)2 + 2(–3) – 3 (1)2 + 2(1) – 3 Example 1B Continued Check Substitute x = 1 and x = –3 into the original expression. = = Both values of x result in division by 0, which is undefined.

  9. 4 3 (3x + 4) 3x + 4 1 (3x + 4)(x – 1) 3x2 + x – 4 (x – 1) The expression is undefined at x = 1 and x = –because these values of x make the factors (x – 1) and (3x + 4) equal 0. Check It Out! Example 1b Simplify. Identify any x-values for which the expression is undefined. Factor; then divide out common factors. =

  10. 4 Check Substitute x = 1 and x = – into the original expression. 3 7 0 3(1) + 4 3(1)2 + (1) – 4 Check It Out! Example 1b Continued = Both values of x result in division by 0, which is undefined.

  11. (2x + 1)(3x + 2) 6x2 + 7x + 2 (2x + 1) (3x + 2)(2x – 3) 6x2 – 5x – 5 (2x – 3) 2 3 The expression is undefined at x =– and x = because these values of x make the factors (3x + 2) and (2x – 3) equal 0. 3 2 Check It Out! Example 1c Simplify. Identify any x-values for which the expression is undefined. Factor; then divide out common factors. =

  12. Check Substitute x = and x = – into the original expression. 3 2 2 3 Check It Out! Example 1c Continued Both values of x result in division by 0, which is undefined.

  13. –1(x)(x – 4) –1(x2 – 4x) 4x – x2 –x (x – 4)(x + 2) x2 – 2x – 8 x2 – 2x – 8 (x + 2 ) Example 2: Simplifying by Factoring by –1 Simplify . Identify any x values for which the expression is undefined. Factor out –1 in the numerator so that x2 is positive, and reorder the terms. Factor the numerator and denominator. Divide out common factors. Simplify. The expression is undefined at x = –2 and x = 4.

  14. 4x – x2 –x x2 – 2x – 8 (x + 2) Example 2 Continued Check The calculator screens suggest that = except when x = – 2 or x = 4.

  15. –1(2)(x – 5) –1(2x – 10) 10 – 2x –2 (x – 5) x – 5 x – 5 1 Check It Out! Example 2a Simplify . Identify any x values for which the expression is undefined. Factor out –1 in the numerator so that x is positive, and reorder the terms. Factor the numerator and denominator. Divide out common factors. Simplify. The expression is undefined at x = 5.

  16. 10 – 2x x – 5 Check It Out! Example 2a Continued Check The calculator screens suggest that = –2 except when x = 5.

  17. –1(x)(x – 3) –1(x2– 3x) –x2 + 3x –x 1 (x – 3)(2x – 1) 2x2 – 7x + 3 2x2 – 7x + 3 2x – 1 2 The expression is undefined at x = 3 and x = . Check It Out! Example 2b Simplify . Identify any x values for which the expression is undefined. Factor out –1 in the numerator so that x is positive, and reorder the terms. Factor the numerator and denominator. Divide out common factors. Simplify.

  18. Check The calculator screens suggest that = except when x = and x = 3. –x2 + 3x –x 1 2x2 – 7x + 3 2x – 1 2 Check It Out! Example 2b Continued

  19. You can multiply rational expressions the same way that you multiply fractions.

  20.    1 10x3y4 3x5y3 x – 3 3x5y3 x – 3 10x3y4 x + 5 x + 5 5x3 4(x + 3) (x – 3)(x + 3) 4x+ 20 4(x+ 5) x2 – 9 9x2y5 2x3y7 2x3y7 9x2y5 3y5 Example 3: Multiplying Rational Expressions Multiply. Assume that all expressions are defined. A. B. 3 5 3

  21. x x x7 x7      15 15 2x 2x  2 10x – 40 10(x – 4) x + 3 x + 3 2x3 20 20 (x – 2) (x – 4)(x – 2) x2 – 6x + 8 5x + 15 5(x + 3) x4 x4 3 Check It Out! Example 3 Multiply. Assume that all expressions are defined. A. B. 2 2 2 3

  22. ÷ = 1 3 1 4 2 2 4 2 3 3 You can also divide rational expressions. Recall that to divide by a fraction, you multiply by its reciprocal. 2 =

  23. 8y5 15 8y5 5x4 5x4 5x4 15 15 8y5 ÷   8x2y2 8x2y2 8x2y2 x2y3 3 Example 4A: Dividing Rational Expressions Divide. Assume that all expressions are defined. Rewrite as multiplication by the reciprocal. 2 3 3

  24. ÷   (x + 3)(x – 4) x2(x– 3)(x + 3) x4+ 2x3 – 8x2 (x + 4)(x – 4) x2(x2 – 9) x4– 9x2 x4– 9x2 x2 – 16 x2 – 16 (x – 1)(x – 2) x2(x – 2)(x + 4) x2(x2 + 2x – 8) x4+ 2x3 – 8x2 (x – 3)(x – 1) x2 – 4x + 3 x2 – 4x + 3 x2 – 4x + 3 x2 – 16  Example 4B: Dividing Rational Expressions Divide. Assume that all expressions are defined. Rewrite as multiplication by the reciprocal.

  25. ÷   4(x – 4) 4(2x – 1)(x – 3) 4(2x2 – 7x + 3) (2x + 1)(x – 4) (2x+ 1)(x – 4) 8x2 – 28x +12 2x2– 7x – 4 2x2– 7x – 4 4x2– 1 (x +3) (2x + 1)(2x – 1) (2x + 1)(2x – 1) 8x2 – 28x +12 (x + 3)(x – 3) (x + 3)(x – 3) x2 – 9 x2 – 9 4x2– 1  Check It Out! Example 4b Divide. Assume that all expressions are defined.

  26. x2 – 25 = 14 x –5 = 14 (x+ 5)(x – 5) (x – 5) Example 5A: Solving Simple Rational Equations Solve. Check your solution. Note that x ≠ 5. x + 5 = 14 x = 9

  27. (9)2 – 25 x2 – 25 = 14 14 9 –5 x –5 56 4 Example 5A Continued Check 14 14 14

  28. x2 + x – 12 = –7 x + 4 = –7 (x– 3)(x + 4) (x + 4) Check It Out! Example 5a Solve. Check your solution. Note that x ≠ –4. x – 3 = –7 x = –4 Because the left side of the original equation is undefined when x = –4, there is no solution.

  29. Check It Out! Example 5a Continued Check A graphing calculator shows that –4 is not a solution.

  30. 4x2 – 9 = 5 2x +3 Note that x ≠ – . = 5 (2x+ 3)(2x – 3) (2x + 3) 3 2 Check It Out! Example 5b Solve. Check your solution. 2x – 3 = 5 x = 4

  31. 4(4)2 – 9 4x2 – 9 = 5 5 2(4) + 3 2x +3 55 11 Check It Out! Example 5b Continued Check 5 5 5

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