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# Section 2.4 Dividing Polynomials; Remainder and Factor Theorems - PowerPoint PPT Presentation

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

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## PowerPoint Slideshow about ' Section 2.4 Dividing Polynomials; Remainder and Factor Theorems' - jarrod-eaton

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### Section 2.4Dividing Polynomials;Remainder and Factor Theorems

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.

(4x2 – 8x + 6) ÷ (2x – 1)

Divide using Long Division.

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

Put in a 0 for the missing term.

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.