{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

461hw6sol

# 461hw6sol - Math 461 F Spring 2011 Homework 6 Solutions...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

461hw6sol - Math 461 F Spring 2011 Homework 6 Solutions...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online