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The polynomial f(x) = x^3 + x^2 - 2x - 3  has three distinct roots. Let g(x) = x^3 + bx^2 + cx + d  be a cubic polynomial with leading coefficient 1 such that the roots of g(x) are the squares of the roots of f(x). Find the ordered triple (b,c,d).

 Jul 6, 2021
 #1
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Let  the roots  of  the  first  be  p, q  and   r

 

Sum of the  roots  =  -1

Sum of the products of the roots, taken two at a time = - 2

Product of all three roots =  3

 

p + q + r =   -1   

pq + pr  + qr =  -2   ⇒  (pq + pr + qr)^2  =  4  

pqr  = 3   ⇒ (pqr)^2  =  9    ⇒  p^2q^2r^2  =   9

 

Square  both sides  of  the first equation

(p + q + r)^2   =    1

p^2 + 2 p q + 2 p r + q^2 + 2 q r + r^2  =   1

p^2  + q^2  + r^2  +  2 (pq + pr + qr)  =  1

p^2  + q^2  + r^2  +  2 ( -2)  = 1

p^2  + q^2  + r^2  =   5 

So  b =  -5  and  d  = -9 

 

And

(pq + pr  +  qr)^2  =   4

p^2 q^2 + 2 p^2 q r + p^2 r^2 + 2 p q^2 r + 2 p q r^2 + q^2 r^2  = 4

(pr)^2  + (pq)^2  + (qr)^2  +  2pqr ( p + q + r)   =   4

(pr)^2  + (pq)^2  + (qr)^2  +  2 (3)  (-1)  =  4

(pr)^2  + (pq)^2  + (qr)^2  -  6   =  4

(pr)^2  + (pq)^2  + (qr)^2  = 10    =    c

 

(b , c, d)  =  ( -5 , 10 , -9)

 

cool cool cool

 Jul 6, 2021

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