M1410203 - 2.25 SECTION 2.3: LONG AND SYNTHETIC POLYNOMIAL...

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2.25 SECTION 2.3: LONG AND SYNTHETIC POLYNOMIAL DIVISION PART A: LONG DIVISION Ancient Example with Integers 4 9 2 8 1 We can say: 9 4 = 2 + 1 4 By multiplying both sides by 4, this can be rewritten as: 9 = 4 2 + 1 In general: dividend, f divisor, d = quotient, q ( ) + remainder, r ( ) d where either: r = 0 (in which case d divides evenly into f ), or r d is a positive proper fraction: i.e., 0 < r < d Technical Note: We assume f and d are positive integers, and q and r are nonnegative integers. Technical Note: We typically assume f d is improper : i.e., f d . Otherwise, there is no point in dividing this way. Technical Note: Given f and d , q and r are unique by the Division Algorithm (really, it’s a theorem). By multiplying both sides by d , f d = q + r d can be rewritten as: f = d q + r
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2.26 Now, we will perform polynomial division on f x ( ) d x ( ) so that we get: f x ( ) d x ( ) = q x ( ) + r x ( ) d x ( ) where either: r x ( ) = 0 , in which case d x ( ) divides evenly into f x ( ) , or r x ( ) d x ( ) is a proper rational expression: i.e., deg r x ( ) ( ) < deg d x ( ) ( ) Technical Note: We assume f x ( ) and d x ( ) are nonzero polynomials, and q x ( ) and r x ( ) are polynomials. Technical Note: We assume f x ( ) d x ( ) is improper ; i.e., deg f x ( ) ( ) deg d x ( ) ( ) . Otherwise, there is no point in dividing. Technical Note: Given f x ( ) and d x ( ) , q x ( ) and r x ( ) are unique by the Division Algorithm (really, it’s a theorem). By multiplying both sides by d x ( ) , f x ( ) d x ( ) = q x ( ) + r x ( ) d x ( ) can be rewritten as: f x ( ) = d x ( ) q x ( ) + r x ( )
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2.27 Example Use Long Division to divide: 5 + 3 x 2 + 6 x 3 1 + 3 x 2 Solution Warning: First, write the N and the D in descending powers of
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M1410203 - 2.25 SECTION 2.3: LONG AND SYNTHETIC POLYNOMIAL...

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