5.3 Part 2 Polynomial Division

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Presentation transcript:

5.3 Part 2 Polynomial Division Synthetic Division

Synthetic Division definition Synthetic Division a method of dividing polynomials using the coefficients of the divided polynomial and its divisor.

Synthetic Division definition Synthetic Division a method of dividing polynomials using the coefficients of the divided polynomial and its divisor. Since the divided polynomial is in descending order, we can just use the coefficient as they are written.

Synthetic Division definition -2 | 1 -4 6 -4 Inside the box is the zero of x + 2 Synthetic can only be used when the divisor has a degree of 1

Synthetic Division definition 2 | 1 -4 6 -4 Inside the box is 2 -4 4 the zero of x - 2 1 - 2 2 0 We drop the first number the then multiply by 2 and add it to the next number. Repeat till done.

Synthetic Division definition 2 | 1 -4 6 -4 Inside the box is 2 -4 4 the zero of x - 2 1 - 2 2 0 We drop the first number the then multiply by 2 and add it to the next number. Repeat till done.

Synthetic Division definition 2 | 1 -4 6 -4 Inside the box is 2 -4 4 the zero of x - 2 1 - 2 2 0 Almost done. Need to fill in the variables

Synthetic Division definition 2 | 1 -4 6 -4 Inside the box is 2 -4 4 the zero of x - 2 1 - 2 2 0 The answer would be. 1x2 – 2x + 2 with a remainder of 0

Solve

Something different When dividing by a divisor whose lead coefficient is not 1. To make the divisor 1, we multiply by ½ Give us y - ½

Something different ½ | ½(4 0 – 5 2 4) Multiply by ½, before using Synthetic division ½ | 2 0 -5/2 1 2 1 ½ - 1 0 2 1 -2 0 2 Answer: 2y3 + 1y2 – 2y + 0 + 2 you can y – ½ not have a fraction in the dominator.

Something different ½ | ½(4 0 – 5 2 4) Multiply by ½, before using Synthetic division ½ | 2 0 -5/2 1 2 1 ½ - 1 0 2 1 -2 0 2 Answer: 2y3 + 1y2 – 2y + 4 Multiply 2y – 1 by 2

Something different ½ | ½(4 0 – 5 2 4) Multiply by ½, before using Synthetic division ½ | 2 0 -5/2 1 2 1 ½ - 1 0 2 1 -2 0 2 Answer: 2y3 + 1y2 – 2y + 4 Multiply 2y – 1 by 2

Something different ½ | ½(4 0 – 5 2 4) Multiply by ½, before using Synthetic division ½ | 2 0 -5/2 1 2 1 ½ - 1 0 2 1 -2 0 2 Answer: 2y3 + 1y2 – 2y + 4 Multiply 2y – 1 by 2

Something different ½ | ½(4 0 – 5 2 4) Multiply by ½, before using Synthetic division ½ | 2 0 -5/2 1 2 1 ½ - 1 0 2 1 -2 0 2 Answer: 2y3 + 1y2 – 2y + 4 Multiply 2y – 1 by 2

Something different ½ | ½(4 0 – 5 2 4) Multiply by ½, before using Synthetic division ½ | 2 0 -5/2 1 2 1 ½ - 1 0 2 1 -2 0 2 Answer: 2y3 + 1y2 – 2y + 0 + 2 you can y – ½ not have a fraction in the dominator.

Something different ½ | ½(4 0 – 5 2 4) Multiply by ½, before using Synthetic division ½ | 2 0 -5/2 1 2 1 ½ - 1 0 2 1 -2 0 2 Answer: 2y3 + 1y2 – 2y + 4 Multiply 2y – 1 by 2 2

Homework Page 236 -237 Using Synthetic division # 21, 26, 29, 30, # 21, 26, 29, 30, 32, 33, 36, 43, 44