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2.3 Polynomial and 2.3 Polynomial and Synthetic DivisionSynthetic Division
Why teach long division in Why teach long division in grade school?grade school?
Long Division
Find 2359 ÷ 51 by hand
Which one goes inside
√
Long Division
Find 2359 ÷ 51 by hand
51 √ 2359
Long Division
Find 2359 ÷ 51 by hand
4
51 √ 2359
204
What do you do now ?
Long Division
Find 2359 ÷ 51 by hand
4
51 √ 2359
- 204
319
Long Division
Find 2359 ÷ 51 by hand
46
51 √ 2359
- 204
319
- 306
13
Long Division
Find 2359 ÷ 51 by hand46 13/51
51 √ 2359
- 204
319
- 306
13
Lets do the same with a Polynomial
Divide 6x3 + 4x2 – 10x – 5 by 2x2 + 1
2x2 + 1 √ 6x3 + 4x2 – 10x - 5
Lets do the same with a Polynomial
Divide 6x3 + 4x2 – 10x – 5 by 2x2 + 1
3x
2x2 + 1 √ 6x3 + 4x2 – 10x – 5
6x3 + 3x
Lets do the same with a Polynomial
Divide 6x3 + 4x2 – 10x – 5 by 2x2 + 1
3x
2x2 + 1 √ 6x3 + 4x2 – 10x – 5
6x3 + 3x
4x2 – 13x - 5
Lets do the same with a Polynomial
Divide 6x3 + 4x2 – 10x – 5 by 2x2 + 1
3x + 2
2x2 + 1 √ 6x3 + 4x2 – 10x – 5
6x3 + 3x
4x2 – 13x - 5
4x2 + 2
- 13x - 7
Lets do the same with a Polynomial
Divide 6x3 + 4x2 – 10x – 5 by 2x2 + 1
3x + 2 +
2x2 + 1 √ 6x3 + 4x2 – 10x – 5
6x3 + 3x
4x2 – 13x - 5
4x2 + 2
- 13x - 7
12x
7 -13x -2
The Division Algorithm
f(x) = d(x)g(x) + r(x)
(6x3 + 4x2 – 10x – 5) = (3x + 2)(2x2 + 1) +(-13x – 7)
= 6x3 + 4x2 + 3x + 2 – 13x – 7
= 6x3 + 4x2 - 10x – 5
WE can use the division Algorithm to find G.C.D.(greatest common divisors )
What is the G.C.D. of 3461, 4879
4879 = 3461(1) + 14183461 = 1418(2) + 6251418 = 625(2) + 168625 = 168(3) + 121168 = 121(1) + 47121 = 47(2) + 2747 = 27(1) + 2027 = 20(1) + 720 = 7(2) + 67 = 6(1) + 1 ← G.C.D.6 = 1(6) + 0
Ruffini’s rule3 or Synthetic Division
• Paolo Ruffini (September 22, 1765 – May 10, 1822) was an Italian mathematician and 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.
Synthetic Division
Can be used when dividing by x – r term, where r is a number.
(4x3 + 5x2 + 8)÷(x – 2) What is x; x – 2 = 0
x = 2
This will go in the little box in the
first line.
Synthetic Division
Can be used when dividing by x – r term, where r is a number.
(4x3 + 5x2 + 8)÷(x – 2) What is x; x – 2 = 0 2 | 4 5 0 8
The coefficients are written out in descending exponential order.
(even leaving a zero for the 1st degree term)
Synthetic Division
Can be used when dividing by x – r term, where r is a number.
(4x3 + 5x2 + 8)÷(x – 2) What is x; x – 2 = 0 2 | 4 5 0 8
4 The first number is dropped,
then multiply by 2 and add to 5
Synthetic Division
Can be used when dividing by x – r term, where r is a number.
(4x3 + 5x2 + 8)÷(x – 2) What is x; x – 2 = 0
2 | 4 5 0 8
8
4 13
Then the steps are repeated added and multiply by 2.
Synthetic Division
Can be used when dividing by x – r term, where r is a number.
(4x3 + 5x2 + 8)÷(x – 2) What is x; x – 2 = 0
2 | 4 5 0 8
8 26 52
4 13 26 60
Then the steps are repeated added and multiply by 2.
Synthetic Division
Can be used when dividing by x – r term, where r is a number.
(4x3 + 5x2 + 8)÷(x – 2) What is x; x – 2 = 0
2 | 4 5 0 8
8 26 52
4 13 26 60
60 is the reminder; 26 is the constant, 13 the 1st degree term, 4 the 2nd degree term
Synthetic Division
Can be used when dividing by x – r term, where r is a number.
(4x3 + 5x2 + 8)÷(x – 2) What is x; x – 2 = 0 2 | 4 5 0 8
8 26 524 13 26 60
4x2 + 13x + 26 + 2
60
x
The Remainder Theorem
The remainder is the answer!
So in f(x) = 4x3 + 5x2 + 8
f(2) = 60
The Remainder Theorem
The remainder is the answer!
So in f(x) = 4x3 + 5x2 + 8
f(2) = 60
Check it out: 4(2)3 + 5(2)2 + 8
4(8) + 5(4) + 8
32 + 20 + 8 = 60
(x2 + 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?
HomeworkHomework
Page 140 – 142 Page 140 – 142
##2, 8, 14, 17, 2, 8, 14, 17,
21, 24, 28, 36,21, 24, 28, 36,
42, 44, 51, 55,42, 44, 51, 55,
63, 74, 82, 9263, 74, 82, 92
HomeworkHomework
Page 140-142 Page 140-142
# # 7, 13, 15, 22,7, 13, 15, 22,
27, 35, 40, 43,27, 35, 40, 43,
50, 53, 62, 70,50, 53, 62, 70,
81, 8681, 86