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Section 2.4 Dividing Polynomials; Remainder and Factor TheoremsPowerPoint Presentation

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

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Section 2.4Dividing Polynomials;Remainder and Factor Theorems

Long Division of Polynomials

Long Division of Polynomials

Long Division of Polynomials with Missing Terms

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

Example

Divide using Long Division.

(4x2 – 8x + 6) ÷ (2x – 1)

Example

Divide using Long Division.

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

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

Put in a 0 for the missing term.

Using synthetic division instead of long division.

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

Thus:

Example

Divide using synthetic division.

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

Example

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

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

Example

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

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.