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Polynomial Division: Long Division and Synthetic Division

Learn how to divide polynomials using long division and synthetic division. Understand the Remainder and Factor Theorems. Examples provided.

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Polynomial Division: Long Division and Synthetic Division

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  1. Splash Screen

  2. Five-Minute Check (over Lesson 2-2) Then/Now New Vocabulary Example 1: Use Long Division to Factor Polynomials Key Concept: Polynomial Division Example 2: Long Division with Nonzero Remainder Example 3: Division by Polynomial of Degree 2 or Higher Key Concept: Synthetic Division Algorithm Example 4: Synthetic Division Key Concept: Remainder Theorem Example 5: Real-World Example: Use the Remainder Theorem Key Concept: Factor Theorem Example 6: Use the Factor Theorem Concept Summary: Synthetic Division and Remainders Lesson Menu

  3. A. B. C. D. Graph f(x) = (x – 2)3 + 3. 5–Minute Check 1

  4. A.The degree is 3 and the leading coefficient 2. Because the degree is odd and the leading coefficient is positive, . B.The degree is 3 and the leading coefficient 2. Because the degree is odd and the leading coefficient is positive, . C.The degree is 3, and the leading coefficient is 2. Because the degree is odd and the leading coefficient is positive, . D.The degree is 3 and the leading coefficient is 2. Because the degree is odd and the leading coefficient is positive, . Describe the end behavior of the graph of f (x) = 2x3 – 4x + 1 using limits. Explain your reasoning using the leading term test. 5–Minute Check 2

  5. Determine all of the real zeros of f(x) = 4x6 – 16x4. A.0, −2, 2 B.0, −2, 2, 4 C.−2, 2 D.−4, 4 5–Minute Check 3

  6. You factored quadratic expressions to solve equations. (Lesson 0–3) • Divide polynomials using long division and synthetic division. • Use the Remainder and Factor Theorems. Then/Now

  7. synthetic division • depressed polynomial • synthetic substitution Vocabulary

  8. ←Multiply divisor by 3x 2 because = 3x 2. ←Multiply divisor by 16x because = 16x. ←Multiply divisor by –12 because = –12 Use Long Division to Factor Polynomials Factor 6x3+ 17x2– 104x + 60 completely using long division if (2x – 5) is a factor. (–)6x3 – 15x2 32x2 – 104x ←Subtract and bring down next term. (–)32x2 – 80x ←Subtract and bring down next term. –24x + 60 (–)–24x + 60 ←Subtract. Notice that the remainder is 0. 0 Example 1

  9. Use Long Division to Factor Polynomials From this division, you can write 6x3 + 17x2 – 104x + 60 = (2x – 5)(3x2 + 16x – 12). Factoring the quadratic expression yields 6x3 + 17x2 – 104x + 60 = (2x – 5)(3x – 2)(x + 6). Answer:(2x – 5)(3x – 2)(x + 6) Example 1

  10. Factor 6x3 + x2 – 117x + 140 completely using long division if (3x – 4) is a factor. A. (3x – 4)(x – 5)(2x + 7) B. (3x – 4)(x + 5)(2x – 7) C. (3x – 4)(2x2 + 3x – 35) D. (3x – 4)(2x + 5)(x – 7) Example 1

  11. Key Concept 1

  12. Long Division with Nonzero Remainder Divide 6x3– 5x2+ 9x + 6 by 2x – 1. (–)6x3 – 3x2 –2x2 + 9x (–)–2x2 + x 8x + 6 (–)8x – 4 10 Example 2

  13. You can write the result as . Answer: Long Division with Nonzero Remainder CheckMultiply to check this result. (2x – 1)(3x2 – x + 4) + 10 = 6x3 – 5x2 + 9x + 6 6x3 – 2x2 + 8x – 3x2 + x – 4 + 10 = 6x3 – 5x2 + 9x + 6 6x3 – 5x2 + 9x + 6 = 6x3 – 5x2 + 9x + 6  Example 2

  14. A. B. 4x3 + 2x2 + 2x + 10 C. D. Divide 4x4 – 2x3 + 8x – 10 by x + 1. Example 2

  15. Division by Polynomial of Degree 2 or Higher Divide x3–x2– 14x + 4 by x2 – 5x + 6. (–)x3 – 5x2 + 6x 4x2 – 20x + 4 (–)4x2 – 20x + 24 –20 Example 3

  16. You can write this result as . Answer: Division by Polynomial of Degree 2 or Higher Example 3

  17. A. B. C. D. Divide 2x4 + 9x3 + x2 – x + 26 by x2 + 6x + 9. Example 3

  18. Key Concept 2

  19. 3 2 –4 –3 –6 –5 –8 Synthetic Division A. Find (2x5– 4x4– 3x3 – 6x2– 5x – 8) ÷ (x – 3) using synthetic division. Because x – 3 is x – (3), c = 3. Set up the synthetic division as follows. Then follow the synthetic division procedure. = add terms. 12 6 6 9 9 2 2 3 3 4 4 = Multiply by c, and write the product. coefficients of depressed quotient remainder Example 4

  20. The quotient has degree one less than that of its dividend, so Answer: Synthetic Division Example 4

  21. Synthetic Division B. Find (8x4 + 38x3 + 5x2 + 3x + 3) ÷ (4x + 1) using synthetic division. Rewrite the division expression so that the divisor is of the form x – c. Example 4

  22. So, . Perform the synthetic division. Synthetic Division Example 4

  23. So, . Answer: Synthetic Division Example 4

  24. A. B. C. 6x3 – 8x2 + 3 D. 6x3 + 4x2 + 12x + 3 Find (6x4 – 2x3 + 8x2 – 9x – 3) ÷ (x – 1) using synthetic division. Example 4

  25. Key Concept 3

  26. Use the Remainder Theorem REAL ESTATE Suppose 800 units of beachfront property have tenants paying $600 per week. Research indicates that for each $10 decrease in rent, 15 more units would be rented. The weekly revenue from the rentals is given by R(x) = –150x2+ 1000x + 480,000, where x is the number of $10 decreases the property manager is willing to take. Use the Remainder Theorem to find the revenue from the properties if the property manager decreases the rent by $50. To find the revenue from the properties, use synthetic substitution to evaluate f(x) for x = 5 since $50 is 5 times $10. Example 5

  27. 5 –150 1000 480,000 –750 1250 – 150 250 481,250 Use the Remainder Theorem The remainder is 481,250, so f(5) = 481,250. Therefore, the revenue will be $481,250 when the rent is decreased by $50. Example 5

  28. Use the Remainder Theorem Answer:$481,250 Check You can check your answer using direct substitution. R(x) = –150x2+ 1000x + 480,000 Original function R(5) = –150(5)2 + 1000(5) + 480,000 Substitute 5 for x. R(5) = –3750 + 5000 + 480,000 or 481,250 Simplify. Example 5

  29. REAL ESTATE Use the equation for R(x) from Example 5 and the Remainder Theorem to find the revenue from the properties if the property manager decreases the rent by $100. A. $380,000 B. $450,000 C. $475,000 D. $479,900 Example 5

  30. Key Concept 4

  31. 5 1 –18 60 25 5 –65 –25 1 –13 –5 0 Use the Factor Theorem A. Use the Factor Theorem to determine if (x –5) and (x + 5) are factors of f(x) = x3– 18x2 + 60x + 25. Use the binomials that are factors to write a factored form of f(x). Use synthetic division to test each factor, (x – 5) and (x + 5). Example 6

  32. –5 1 –18 60 25 –5 115 –875 1 –23 175 –850 Use the Factor Theorem Because the remainder when f(x) is divided by (x – 5) is 0, f(5) = 0, and (x – 5) is a factor. Because the remainder when f(x) is divided by (x + 5) is –850, f(–5) = –850 and (x + 5) is not a factor. Because (x – 5) is a factor of f(x), we can use the quotient of f(x) ÷ (x – 5) to write a factored form of f(x). Answer:f(x) = (x – 5)(x2 – 13x – 5) Example 6

  33. Use the Factor Theorem B. Use the Factor Theorem to determine if (x – 5) and (x + 2) are factors of f(x) = x3 – 2x2 – 13x – 10. Use the binomials that are factors to write a factored form of f(x). Use synthetic division to test the factor (x – 5). 5 1 –2 –13 –10 5 15 10 1 3 2 0 Because the remainder when f(x) is divided by (x – 5) is 0, f(5) = 0 and (x – 5) is a factor of f(x). Example 6

  34. –2 1 3 2 –2 –2 1 1 0 Use the Factor Theorem Next, test the second factor (x + 2), with the depressed polynomial x2 + 3x + 2. Because the remainder when the quotient of f(x) ÷ (x – 5) is divided by (x + 2) is 0, f(–2) = 0 and (x + 2) is a factor of f(x). Because (x – 5) and (x + 2) are factors of f(x), we can use the final quotient to write a factored form of f(x). Answer:f(x) = (x – 5)(x + 2)(x + 1) Example 6

  35. Use the Factor Theorem to determine if the binomials (x + 2) and (x – 3) are factors off(x) = 4x3 – 9x2 – 19x + 30. Use the binomials that are factors to write a factored form of f(x). A. yes, yes; f(x) = (x + 2)(x – 3)(–14x + 5) B. yes, yes; f(x) = (x + 2)(x – 3)(4x – 5) C. yes, no; f(x) = (x + 2)(4x2 – 17x – 15) D. no, yes; f(x) = (x – 3)(4x2 + 3x + 10) Example 6

  36. Key Concept 5

  37. End of the Lesson

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