1 / 37

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.

minnieg
Télécharger la présentation

Polynomial Division: Long Division and Synthetic Division

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related