Inspect polynomial
Given that , and are all real roots of the above polynomial, find .
The answer is 2.
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1 solution
Moderator note:
What is the polynomial whose roots are of the form ?
r + 1 r − 1 + s + 1 s − 1 + t + 1 t − 1 = 1 + 1 + 1 − 2 ( r + 1 1 + s + 1 1 + t + 1 1 ) = 3 − 2 ( ( r + 1 ) ( s + 1 ) ( t + 1 ) ( s + 1 ) ( t + 1 ) + ( r + 1 ) ( t + 1 ) + ( s + 1 ) ( r + 1 ) ) = 3 − 2 ( r s t + ( r s + s t + r t ) + ( r + s + t ) + 1 ( s t + s r + r t ) + 2 ( r + s + t ) + 3 ) By vieta's formula - higher degrees ,we have that: ⎩ ⎪ ⎨ ⎪ ⎧ r + s + t = 0 r s + s t + r t = − 2 0 0 7 r s t = − 2 0 0 2 Putting the respective values in the expression,we get: 3 − 2 ( − 2 0 0 2 − 2 0 0 7 + 0 + 1 − 2 0 0 7 + 2 ( 0 ) + 3 ) = 3 − − 4 0 0 8 2 × ( − 2 0 0 4 ) = 3 + − 4 0 0 8 4 0 0 8 = 3 − 1 = 2