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Unformatted text preview: Math 461 F Spring 2011 Homework 6 Solutions Drew Armstrong Book Problems. Problem 6.4.8. Let r,s,t be the zeros of the real polynomial x 3 + 23 x + 1. Find the real cubic (monic) polynomial whose zeros are r 2 ,s 2 ,t 2 . By assumption we have x 3 + 23 x + 1 = ( x r )( x s )( x t ) = x 3 ( r + s + t ) x 2 + ( rs + rt + st ) x rst. Then equating coefficients on the left and right gives 0 = e 1 = r + s + t, 23 = e 2 = rs + rt + st, 1 = e 3 = rst. Now we are looking for the coefficients of the polynomial f ( x ) = ( x r 2 )( x s 2 )( x t 2 ) = x 3 ( r 2 + s 2 + t 2 ) x 2 + ( r 2 s 2 + r 2 t 2 + s 2 t 2 ) x r 2 s 2 t 2 . First note that r 2 s 2 t 2 = e 2 3 = ( 1) 2 = 1. Next ( Problem 6.4.1 ) we wish to express r 2 + s 2 + t 2 in terms of e 1 ,e 2 ,e 3 . Using Gauss’ algorithm, or just trial and error, we have ( r 2 + s 2 + t 2 ) = ( r + s + t ) 2 2( r + s + t ) = e 2 1 2 e 2 = ( 0) 2 2 · 23 = 46 . Next ( Problem 6.4.4 ) we wish to express X = r 2 s 2 + r 2 t 2 + s 2 t 2 in terms of e 1 ,e 2 ,e 3 . We will use Gauss’ algorithm explicitly this time. First note that the leading term is r 2 s 2 . To eliminate this we will subtract e 2 2 which has the same leading term. This gives X e 2 2 = 2 r 2 st 2 rs 2 t 2 rst 2 = 2( r + s + t )( rst ) = 2...
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 Spring '11
 Armstrong
 Math, Algebra, Euler, PQR, real quadratics, r2 s2 t2

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