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Polynomial Long Division and Synthetic Division

Polynomial Long Division and Synthetic Division. When our factoring techniques do not easily work…. Analyzing and Graphing a Function Let’s say we want to analyze this function and graph it: f(x) = x 7 - 8x 5 - 2x 4 - 21x 3 + 10x 2 + 108x + 72 We know the end behavior

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Polynomial Long Division and Synthetic Division

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  1. Polynomial Long Division and Synthetic Division

  2. When our factoring techniques do not easily work… • Analyzing and Graphing a Function • Let’s say we want to analyze this function and graph it: f(x) = x7 - 8x5 - 2x4 - 21x3 + 10x2 + 108x + 72 • We know the end behavior • We know the y- intercept • To get a good approximation of the graph, we need to know the x-intercepts or the “zeros”. To find all the real zeros of the function we must factor it completely. • Determining if one polynomial is a factor of another polynomial • Factoring a polynomial Polynomial division will help with this.

  3. Polynomial Division • Polynomial Division is very similar to long division. • Example:

  4. Polynomial Division Subtract!! Subtract!! Subtract!!

  5. Polynomial Division • Example: • Notice that there is no x term. However, we need to include it when we divide.

  6. Polynomial Division

  7. Try This • Example: • Answer:

  8. Synthetic Division • Synthetic Division is a ‘shortcut’ for polynomial division that only works when dividing by a linear factor (x + b). • It involves the coefficients of the dividend, and the zero of the divisor.

  9. Example • Divide: • Step 1: • Write the coefficients of the dividend in a upside-down division symbol. 1 5 6

  10. Example • Step 2: • Take the zero of the divisor, and write it on the left. • The divisor is x – 1, so the zero is 1. 1 1 5 6

  11. Example • Step 3: • Carry down the first coefficient. 1 1 5 6 1

  12. Example • Step 4: • Multiply the zero by this number. Write the product under the next coefficient. 1 1 5 6 1 1

  13. Example • Step 5: • Add. 1 1 5 6 1 1 6

  14. Example • Step etc.: • Repeat as necessary 1 1 5 6 1 6 12 1 6

  15. Example The numbers at the bottom represent the coefficients of the answer. The new polynomial will be one degree less than the original. 1 1 5 6 1 6 12 1 6

  16. Synthetic Division The pattern for synthetic division of a cubic polynomial is summarized as follows. (The pattern for higher-degree polynomials is similar.)

  17. Synthetic Division This algorithm for synthetic division works only for divisors of the form x – k. Remember that x + k = x – (–k).

  18. Using Synthetic Division Use synthetic division to divide x4 – 10x2 – 2x + 4 by x + 3. Solution: You should set up the array as follows. Note that a zero is included for the missing x3-term in the dividend.

  19. Example – Solution cont’d • Then, use the synthetic division pattern by adding terms in columns and multiplying the results by –3. • So, you have • .

  20. Try These • Examples: • (x4 + x3 – 11x2 – 5x + 30)  (x – 2) • (x4 – 1)  (x + 1)[Don’t forget to include the missing terms!] • Answers: • x3 + 3x2 – 5x – 15 • x3 – x2 + x – 1

  21. Application of Long Division To begin, suppose you are given the graph of f (x) = 6x3 – 19x2 + 16x – 4.

  22. Long Division of Polynomials Notice that a zero of f occurs at x = 2. Because x = 2 is a zero of f,you know that (x – 2) isa factor of f (x). This means thatthere exists a second-degree polynomial q (x) such that f (x) = (x – 2) q(x). To find q(x), you can uselong division.

  23. Example - Long Division of Polynomials Divide 6x3 – 19x2 + 16x – 4 by x – 2, and use the result to factor the polynomial completely.

  24. Think Think Think Example 1 – Solution Multiply: 6x2(x – 2). Subtract. Multiply: –7x(x – 2). Subtract. Multiply: 2(x – 2). Subtract.

  25. Example – Solution cont’d From this division, you can conclude that 6x3 – 19x2 + 16x – 4 = (x – 2)(6x2 – 7x + 2) and by factoring the quadratic 6x2 – 7x + 2, you have 6x3 – 19x2 + 16x – 4 = (x – 2)(2x – 1)(3x – 2).

  26. Long Division of Polynomials

  27. Example – Factoring a Polynomial: Repeated Division Show that (x – 2) and (x + 3) are factors of f (x) = 2x4 + 7x3 – 4x2– 27x – 18. Then find the remaining factors of f (x). Solution:Using synthetic division with the factor (x – 2), you obtain the following. 0 remainder, so f(2) = 0 and (x – 2) is a factor.

  28. Example – Solution cont’d Take the result of this division and perform synthetic division again using the factor (x + 3). Because the resulting quadratic expression factors as 2x2 + 5x + 3 = (2x + 3)(x + 1) the complete factorization of f (x) is f (x) = (x – 2)(x + 3)(2x + 3)(x + 1). 0 remainder, so f(–3) = 0 and (x + 3) is a factor.

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