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Dividing Polynomials

Dividing Polynomials. ALGEBRA 2 LESSON 6-3. (For help, go to Lessons 5-1 and 6-1.). Simplify each expression. 1. ( x + 3)( x – 4) + 2 2. (2 x + 1)( x – 3) 3. ( x + 2)( x + 1) – 11 4. –3(2 – x )( x + 5) Write each polynomial in standard form. Then list the coefficients.

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Dividing Polynomials

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  1. Dividing Polynomials ALGEBRA 2 LESSON 6-3 (For help, go to Lessons 5-1 and 6-1.) Simplify each expression. 1. (x + 3)(x – 4) + 2 2. (2x + 1)(x – 3) 3. (x + 2)(x + 1) – 11 4. –3(2 – x)(x + 5) Write each polynomial in standard form. Then list the coefficients. 5. 5x – 2x2 + 9 + 4x36. 10 + 5x3 – 9x2 7. 3x + x2 – x + 7 – 2x28. –4x4 – 7x2 + x3 + x4 9. divide 5x3 – 6x2 + 4x – 1 by x – 3. 6-3

  2. Dividing Polynomials ALGEBRA 2 LESSON 6-3 Solutions • 1. (x + 3)(x – 4) + 2 = x2 – 4x + 3x – 12 + 2 • = x2 – x – 10 • 2. (2x + 1)(x – 3) = 2x2 – 6x + x – 3 = 2x2 – 5x – 3 • 3. (x + 2)(x + 1) – 11 = x2 + x + 2x + 2 – 11 • = x2 + 3x – 9 • 4. –3(2 – x)(x + 5) = –3(2x + 10 – x2 – 5x) = –3(–x2 – 3x + 10) = 3x2 + 9x – 30 • 5. 5x – 2x2 + 9 + 4x3 = 4x3 – 2x2 + 5x + 9 coefficients: 4, –2, 5, 9 • 6. 10 + 5x3 – 9x2 = 5x3 – 9x2 + 10 coefficients: 5, –9, 0, 10 • 7. 3x + x2 – x + 7 – 2x2 = –x2 + 2x + 7 coefficients: –1, 2, 7 • –4x4 – 7x2 + x3 + x4 = –3x4 + x3 – 7x2 coefficients: –3, 1, –7, 0, 0 • The quotient is 5x2 + 9x + 31, R 92. 6-3

  3. Assignment 54 • Page 318 1-12, 37-40

  4. Dividing Polynomials ALGEBRA 2 LESSON 6-3 pages 318–320  Exercises 1.x – 8 2. 3x – 5 3.x2 + 4x + 3, R 5 4. 2x2 + 5x + 2 5. 3x2 – 7x + 2 6. 9x – 12, R –32 7.x – 10, R 40 8.x2 + 4x + 3 9. no 10. yes 11. yes 12. no 13.x2 + 4x + 3 14.x2 – 2x + 2 15. x2 – 11x + 37, R –128 16. x2 + 2x + 5 17. x2 – x – 6 18. –2x2 + 9x – 19, R 40 19.x + 1, R 4 20. 3x2 + 8x – 3 21.x2 – 3x + 9 22. 6x – 2, R –4 23.y = (x + 1)(x + 3)(x – 2) 6-3

  5. Dividing Polynomials ALGEBRA 2 LESSON 6-3 24.y = (x + 3)(x – 4)(x – 3) 25. = x + 3 and h = x 26. 18 27. 0 28. 0 29. 12 30. 168 31. 10 32. 51 33. 0 34. P(a) = 0; x – a is a factor of P(x). 35.x – 1 is not a factor of x3 – x2 – 2x because it does not divide into x3 – x2 – 2x evenly. 36. Answers may vary. Sample: (x2 + x – 4) ÷ (x – 2) 37.x2 + 4x + 5 38. x3 – 3x2 + 12x – 35, R 109 39. x4 – x3 + x2 – x + 1 40. x + 4 41.x3 – x2 + 1 42. no 43. yes 44. yes 6-3

  6. Dividing Polynomials • Synthetic division Synthetic division is a method of dividing polynomials in which you omit all variables and exponents and perform the division on the list of coefficients. You also reverse the sign of the divisor so that you can add throughout the process, rather than subtract.

  7. Write x – 3 5x3 – 6x2 + 4x – 1 as 3 5 –6 4 –1 Bring down the 5. 3 5 –6 4 –1 This begins the quotient. 5 Dividing Polynomials ALGEBRA 2 LESSON 6-3 Use synthetic division to divide 5x3 – 6x2 + 4x – 1 by x – 3. Step 1: Reverse the sign of the constant term in the divisor. Write the coefficients of the polynomial in standard form. Step 2: Bring down the coefficient. 6-3

  8. Multiply 3 by 5. Write 35 –6 4 –1 the result under –6. x15 5 9 Add –6 and 15. 3 5 –6 4 –1 15 27 93 5 9 31 92 5x2 + 9x + 31  Remainder Dividing Polynomials ALGEBRA 2 LESSON 6-3 (continued) Step 3:  Multiply the first coefficient by the new divisor. Write the result under the next coefficient. Add. Step 4:  Repeat the steps of multiplying and adding until the remainder is found. The quotient is 5x2 + 9x + 31, R 92. 6-3

  9. Dividing Polynomials • Use synthetic division to divide 3x3 − 4x2 + 2x − 1 by x + 1. • Step 1 Reverse the sign of the constant term in the divisor. Write the coefficients of the polynomial in standard form • Step 2 Bring down the first coefficient

  10. Dividing Polynomials • Step 3Multiply the first coefficient by the new divisor. Write the result under the next coefficient. Add. • Step 4Repeat the steps of multiplying and adding until the remainder is found. • The quotient is 3x2 − 7x + 9, R −10.

  11. 5 1 –6 3 10 Divide. 5 –5 –10 1 –1 –2 0 x2 – x – 2  Remainder Dividing Polynomials ALGEBRA 2 LESSON 6-3 The volume in cubic feet of a shipping carton is V(x) = x3 – 6x2 + 3x + 10. The height is x – 5 feet. a. Find linear expressions for the other dimensions. Assume that the length is greater than the width. x2 – x – 2 = (x – 2)(x + 1) Factor the quotient. The length and the width are x + 1 and x – 2, respectively. b. If the width of the carton is 4 feet, what are the other two dimensions? x – 2 = 4  Substitute 4 into the expression for width. Find x. x = 6 Since the height equals x – 5 and the length equals x + 1, the height is 1 ft. and the length is 7 ft. 6-3

  12. Dividing Polynomials • Remainder Theorem If a polynomial P(x) of degree n ≥ 1 is divided by (x − a), where a is a constant, then the remainder is P(a).

  13. 3 1 –2 0 1 –9 3 3 9 30 1 1 3 10 21 Dividing Polynomials ALGEBRA 2 LESSON 6-3 Use synthetic division to find P(3) for P(x) = x4 – 2x3 + x – 9. By the Remainder Theorem, P(3) equals the remainder when P(x) is divided by x – 3. The remainder is 21, so P(3) = 21. 6-3

  14. Assignment 45 • Page 318 14-24 even, 25, 49, 51

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