section 2.4 dividing polynomials; remainder and factor theorems

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Section 2.4 Dividing Polynomials; Remainder and Factor Theorems

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Section 2.4 Dividing Polynomials; Remainder and Factor Theorems. Long Division of Polynomials. Long Division of Polynomials. Long Division of Polynomials with Missing Terms. - PowerPoint PPT Presentation

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Page 1: Section 2.4 Dividing Polynomials; Remainder and Factor Theorems

Section 2.4Dividing Polynomials;

Remainder and Factor Theorems

Page 2: Section 2.4 Dividing Polynomials; Remainder and Factor Theorems
Page 3: Section 2.4 Dividing Polynomials; Remainder and Factor Theorems

Long Division of Polynomials

Page 4: Section 2.4 Dividing Polynomials; Remainder and Factor Theorems

Long Division of Polynomials

2

3 2 9 6 5x x x

12 5x

4

13

3x

29 6x x

12 8x

13

3 2x

Page 5: Section 2.4 Dividing Polynomials; Remainder and Factor Theorems

Long Division of Polynomials with Missing Terms

2 3

3 2

2

2

x +5x -3 x 3x 2

x +5x 3x

-5x 6x 2

-5x 25x 15

31x- 17

5x 2

31 17

5 3

x

x x

You need to leave a hole when you have missing terms. This technique will help you line up like terms. See the dividend above.

Page 6: Section 2.4 Dividing Polynomials; Remainder and Factor Theorems
Page 7: Section 2.4 Dividing Polynomials; Remainder and Factor Theorems

Example

Divide using Long Division.

(4x2 – 8x + 6) ÷ (2x – 1)

Page 8: Section 2.4 Dividing Polynomials; Remainder and Factor Theorems

Example

Divide using Long Division.

2 4 32 1 8 3 +5 1x x x x x

Page 9: Section 2.4 Dividing Polynomials; Remainder and Factor Theorems
Page 10: Section 2.4 Dividing Polynomials; Remainder and Factor Theorems

Comparison of Long Division and Synthetic Division of X3 +4x2-5x+5 divided by x-3

Page 11: Section 2.4 Dividing Polynomials; Remainder and Factor Theorems

Steps of Synthetic Division dividing 5x3+6x+8 by x+2

Put in a 0 for the missing term.

Page 12: Section 2.4 Dividing Polynomials; Remainder and Factor Theorems

2 5 + 7 - 1

Using synthetic division instead of long division.

Notice that the divisor has to be a binomial of degree 1 with no coefficients.

5

103

65

2

55 3

22 5 7 1

xx

x x x

Thus:

Page 13: Section 2.4 Dividing Polynomials; Remainder and Factor Theorems

Example

Divide using synthetic division.

3 23 5 7 8

4

x x x

x

Page 14: Section 2.4 Dividing Polynomials; Remainder and Factor Theorems

If you are given the function f(x)=x3- 4x2+5x+3 and you want to find f(2), then the remainder of this function when divided by x-2 will give you f(2)

f(2)=5

Page 15: Section 2.4 Dividing Polynomials; Remainder and Factor Theorems

2(1) for f(x)=6x 2 5 is

1 6 -2 5

6 4

6 4 9

f(1)=9

f x

Page 16: Section 2.4 Dividing Polynomials; Remainder and Factor Theorems

Example

Use synthetic division and the remainder theorem to find the indicated function value.

3 2( ) 3 5 1; f(2)f x x x

Page 17: Section 2.4 Dividing Polynomials; Remainder and Factor Theorems

Solve the equation 2x3-3x2-11x+6=0 given that 3 is a zero of f(x)=2x3-3x2-11x+6. The factor theorem tells us that x-3 is a factor of f(x). So we will use both synthetic division and long division to show this and to find another factor.

Another factor

Page 18: Section 2.4 Dividing Polynomials; Remainder and Factor Theorems

Example

Solve the equation 5x2 + 9x – 2=0 given that -2 is a zero of f(x)= 5x2 + 9x - 2

Page 19: Section 2.4 Dividing Polynomials; Remainder and Factor Theorems

Example

Solve the equation x3- 5x2 + 9x - 45 = 0 given that 5 is a zero of f(x)= x3- 5x2 + 9x – 45. Consider all complex number solutions.

Page 20: Section 2.4 Dividing Polynomials; Remainder and Factor Theorems

3 2Divide 2 8 3x x x x

Page 21: Section 2.4 Dividing Polynomials; Remainder and Factor Theorems

3 2

Use Synthetic Division and the Remainder

Theorem to find the value of f(2) for the function

f(x)=x +x - 11x+10