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).
+129850 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)
