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Lesson 2.3 Real Zeros of Polynomials. The Division Algorithm. 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. + 1. Multiply by the divisor
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Dividing by a polynomial Set up in long division 2 terms in divisor (x + 1). How does this go into 1st two terms in order to eliminate the 1st term of the dividend. 2x + 1 • Multiply by the divisor • Write product under dividend • Subtract • Carry down next term • Repeat process - 2x2 + 2x - x + 5 - x + 1 - 4 Answer:
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 (x4 – x2 + x) ÷ (x2 - x + 1)
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” x – 2: a = 2 x + 3: a = -3 4 5 0 -2 5
Steps • Bring down • Multiply diagonally • Add • Remainder = last addition • Answer • Numbers at bottom are coefficients • Start with 1 degree less than dividend REPEAT
Example 3: (2x4 – 30x2 – 2x – 1) (x – 4) Problem Set 2.3 (1 – 21 EOO)
The Remainder Theorem If f(x) is divided by x – a , the remainder is r = f(a) The Factor Theorem If f(x) has a factor (x – a) then f(a) = 0
Rational Zero Test Every rational zero = Factors of constant term Factors of leading coefficient =
Descartes’ Rule Number of positivereal roots is: ► the number of variations in the signs, or ► less than that by a positive even integer 5x4 – 3x3 + 2x2 – 7x + 1 variations: possible positive real roots:
Example 5 List possible zeros, verify with your calculator which are zeros, and check results with Descartes’ Rule Problems Set 2.3