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Lesson 2.3 Real Zeros of Polynomials

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The Division Algorithm

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Dividing by a polynomial Set up in long division 2 terms in divisor (x + 1). How does this go into 1 st two terms in order to eliminate the 1 st term of the dividend. 2x 2x 2 + 2x Multiply by the divisor Write product under dividend Subtract Carry down next term Repeat process -- x + 5 + 1 x + 1 -- 4 Answer:

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HINTS: If a term is missing in the dividend – add a “0” term. If there is a remainder, put it over the divisor and add it to the quotient (answer) Example 1 (x 4 – x 2 + x) ÷ (x 2 - x + 1)

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Synthetic Division Less writing Uses addition Setting Up Divisor must be of the form: x – a Use only “a” and coefficients of dividend Write in “zero terms” 450-25 x – 2: a = 2 x + 3: a = -3

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450-25

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Steps Bring down Multiply diagonally Add Remainder = last addition Answer Numbers at bottom are coefficients Start with 1 degree less than dividend REPEAT

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Example 2: (2x 3 – 7x 2 – 11x – 20) (x – 5)

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Example 3: (2x 4 – 30x 2 – 2x – 1) (x – 4) Problem Set 2.3 (1 – 21 EOO)

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The Factor Theorem If f(x) has a factor (x – a) then f(a) = 0 The Remainder Theorem If f(x) is divided by x – a, the remainder is r = f(a)

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Example 4 Show that (x – 2) and (x + 3) are factors of

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Rational Zero Test Every rational zero = Factors of constant term Factors of leading coefficient =

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Descartes’ Rule Number of positive real roots is: ► the number of variations in the signs, or ► less than that by a positive even integer 5x 4 – 3x 3 + 2x 2 – 7x + 1 variations: possible positive real roots:

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Example 5 List possible zeros, verify with your calculator which are zeros, and check results with Descartes’ Rule Problems Set 2.3

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