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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 2.4Dividing Polynomials;Remainder and Factor Theorems

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.

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:

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

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

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

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.

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