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2.3 Polynomial and Synthetic Division Why teach long division in grade school?

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Presentation on theme: "2.3 Polynomial and Synthetic Division Why teach long division in grade school?"— Presentation transcript:

1 2.3 Polynomial and Synthetic Division Why teach long division in grade school?

2 Long Division Find 2359 ÷ 51 by hand Which one goes inside √

3 Long Division Find 2359 ÷ 51 by hand 51√ 2359

4 Long Division Find 2359 ÷ 51 by hand 4 51√ 2359 204 What do you do now ?

5 Long Division Find 2359 ÷ 51 by hand 4 51√ 2359 - 204 319

6 Long Division Find 2359 ÷ 51 by hand 46 51√ 2359 - 204 319 - 306 13

7 Long Division Find 2359 ÷ 51 by hand 46 13/51 51√ 2359 - 204 319 - 306 13

8 Lets do the same with a Polynomial Divide6x 3 + 4x 2 – 10x – 5 by 2x 2 + 1 2x 2 + 1 √ 6x 3 + 4x 2 – 10x - 5

9 Lets do the same with a Polynomial Divide6x 3 + 4x 2 – 10x – 5 by 2x 2 + 1 3x 2x 2 + 1 √ 6x 3 + 4x 2 – 10x – 5 6x 3 + 3x

10 Lets do the same with a Polynomial Divide6x 3 + 4x 2 – 10x – 5 by 2x 2 + 1 3x 2x 2 + 1 √ 6x 3 + 4x 2 – 10x – 5 6x 3 + 3x 4x 2 – 13x - 5

11 Lets do the same with a Polynomial Divide6x 3 + 4x 2 – 10x – 5 by 2x 2 + 1 3x + 2 2x 2 + 1 √ 6x 3 + 4x 2 – 10x – 5 6x 3 + 3x 4x 2 – 13x - 5 4x 2 + 2 - 13x - 7

12 Lets do the same with a Polynomial Divide6x 3 + 4x 2 – 10x – 5 by 2x 2 + 1 3x + 2 + 2x 2 + 1 √ 6x 3 + 4x 2 – 10x – 5 6x 3 + 3x 4x 2 – 13x - 5 4x 2 + 2 - 13x - 7

13 The Division Algorithm f(x) = d(x)g(x) + r(x) (6x 3 + 4x 2 – 10x – 5) = (3x + 2)(2x 2 + 1) +(-13x – 7) = 6x 3 + 4x 2 + 3x + 2 – 13x – 7 = 6x 3 + 4x 2 - 10x – 5 WE can use the division Algorithm to find G.C.D. (greatest common divisors )

14 What is the G.C.D. of 3461, 4879 4879 = 3461(1) + 1418 3461 = 1418(2) + 625 1418 = 625(2) + 168 625 = 168(3) + 121 168 = 121(1) + 47 121 = 47(2) + 27 47 = 27(1) + 20 27 = 20(1) + 7 20 = 7(2) + 6 7 = 6(1) + 1← G.C.D. 6 = 1(6) + 0

15 Ruffini’s rule3 or Synthetic Division Paolo Ruffini (September 22, 1765 – May 10, 1822) was an Italian mathematician and philosopher. Italianmathematician philosopher By 1788 he had earned university degrees in philosophy, medicine/surgery, and mathematics. Among his work was an incomplete proof (Abel–Ruffini theorem1) that quintic (and higher-order) equations cannot be solved by radicals (1799), and Ruffini's rule3 which is a quick method for polynomial division.Abel–Ruffini theorem1 quintic (and higher-order) equations radicalsRuffini's rule3polynomial division

16 Synthetic Division Can be used when dividing by x – r term, where r is a number. (4x 3 + 5x 2 + 8)÷(x – 2)What is x; x – 2 = 0 x = 2 This will go in the little box in the first line.

17 Synthetic Division Can be used when dividing by x – r term, where r is a number. (4x 3 + 5x 2 + 8)÷(x – 2)What is x; x – 2 = 0 2 |4508 The coefficients are written out in descending exponential order. (even leaving a zero for the 1 st degree term)

18 Synthetic Division Can be used when dividing by x – r term, where r is a number. (4x 3 + 5x 2 + 8)÷(x – 2)What is x; x – 2 = 0 2 |4508 4The first number is dropped, then multiply by 2 and add to 5

19 Synthetic Division Can be used when dividing by x – r term, where r is a number. (4x 3 + 5x 2 + 8)÷(x – 2)What is x; x – 2 = 0 2 |4508 8 413 Then the steps are repeated added and multiply by 2.

20 Synthetic Division Can be used when dividing by x – r term, where r is a number. (4x 3 + 5x 2 + 8)÷(x – 2)What is x; x – 2 = 0 2 |4508 82652 4132660 Then the steps are repeated added and multiply by 2.

21 Synthetic Division Can be used when dividing by x – r term, where r is a number. (4x 3 + 5x 2 + 8)÷(x – 2)What is x; x – 2 = 0 2 |4508 82652 4132660 60 is the reminder; 26 is the constant, 13 the 1 st degree term, 4 the 2 nd degree term

22 Synthetic Division Can be used when dividing by x – r term, where r is a number. (4x 3 + 5x 2 + 8)÷(x – 2)What is x; x – 2 = 0 2 |4508 82652 4132660 4x 2 + 13x + 26 +

23 The Remainder Theorem The remainder is the answer! So in f(x) = 4x 3 + 5x 2 + 8 f(2) = 60

24 The Remainder Theorem The remainder is the answer! So in f(x) = 4x 3 + 5x 2 + 8 f(2) = 60 Check it out: 4(2) 3 + 5(2) 2 + 8 4(8) + 5(4) + 8 32 + 20 + 8 = 60

25 (x 2 + 3x – 40) ÷ (x - 5) 5| 1 3 - 40 5 40 1 8 0 Since the reminder is 0, 5 is a root or zero of the equation. What is the other root?

26 Homework Page 140 – 142 #2, 8, 14, 17, 21, 24, 28, 36, 42, 44, 51, 55, 42, 44, 51, 55, 63, 74, 82, 92 63, 74, 82, 92

27 Homework Page 140-142 # 7, 13, 15, 22, 27, 35, 40, 43, 50, 53, 62, 70, 50, 53, 62, 70, 81, 86 81, 86

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