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7.4 The Remainder and Factor Theorems

7.4 The Remainder and Factor Theorems. Use Synthetic Substitution to find Remainders. A review of Synthetic Substitution. Synthetic Division a method of dividing polynomials using the coefficients of the divided polynomial and its divisor. Synthetic Division definition.

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7.4 The Remainder and Factor Theorems

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  1. 7.4 The Remainder and Factor Theorems Use Synthetic Substitution to find Remainders

  2. A review of Synthetic Substitution Synthetic Division a method of dividing polynomials using the coefficients of the divided polynomial and its divisor.

  3. 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.

  4. 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

  5. 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.

  6. 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.

  7. Synthetic Division definition 2 | 1 -4 6 -4 Inside the box is 2 -4 4 the zero of x - 2 1 - 2 2 0 Since the Remainder is Zero, (x – 2) is a factor of

  8. A new way to Evaluate polynomials Find Use Synthetic Substitution

  9. A new way to Evaluate polynomials Find Use Synthetic Substitution 4 | 3 -2 1 0 - 2 12 3

  10. A new way to Evaluate polynomials Find Use Synthetic Substitution 4 | 3 -2 1 0 - 2 12 40 3 10

  11. A new way to Evaluate polynomials Find Use Synthetic Substitution 4 | 3 -2 1 0 - 2 12 40 164 3 10 41

  12. A new way to Evaluate polynomials Find Use Synthetic Substitution 4 | 3 -2 1 0 - 2 12 40 164 656 3 10 41 164

  13. A new way to Evaluate polynomials Find Use Synthetic Substitution 4 | 3 -2 1 0 - 2 12 40 164 656 3 10 41 164 654

  14. A new way to Evaluate polynomials Find Use Synthetic Substitution 4 | 3 -2 1 0 - 2 12 40 164 656 3 10 41 164 654 Remainder is 654, so f(4) = 654

  15. To show a binomial is a Factor the remainder equals zero Show (x – 3) is a factor of x3 + 4x2 – 15x – 18.

  16. To show a binomial is a Factor the remainder equals zero Show (x – 3) is a factor of x3 + 4x2 – 15x – 18. 3 | 1 4 -15 -18 3 21 18 1 7 6 0 Since the remainder is zero, ( x – 3) is a factor

  17. To show a binomial is a Factor the remainder equals zero Show (x – 3) is a factor of x3 + 4x2 – 15x – 18. 3 | 1 4 -15 -18 3 21 18 1 7 6 0 If you were going to find the other factors, you would use the depressed equation. x2 + 7x + 6 = (x + 6)(x + 1)

  18. x3 + 4x2 – 15x – 18 factor Using the synthetic substitution and factoring the depressed equation gives us the factors of the polynomial. x3 + 4x2 – 15x – 18 = (x – 3)(x + 1)(x + 6)

  19. Homework Page 368 – 369 # 13 – 29 odd

  20. Homework Page 368 – 369 # 14 – 30 even

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