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anonymous
 5 years ago
find a polynomial equation with integer coefficients, for which 5 is a double root and 1+ square root of 2 and 1 square root of 3i are solutions.
I dont know how to solve this or write it out.. help?
anonymous
 5 years ago
find a polynomial equation with integer coefficients, for which 5 is a double root and 1+ square root of 2 and 1 square root of 3i are solutions. I dont know how to solve this or write it out.. help?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if A, B, C , D are roots of a polynomial, the the polynomial is the product of (x A)(xB)(xC)(xD) You have the four roots as 5, 5, 1 + root2 and 1root 3i

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In the above equation, replace A, B, C and D by the given roots and take their product to get the polynomial

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so do (x5)(x5)(x1+square root of 2)(x1square root of 3i)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It will be (x5)(x5)[x(1+root2)][x (1 root of 3i)} Which becomes (x5)(x5)[x1  root2][x1+ root of 3i}

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0see if the coefficients are integer and real then complex conjugate roots must exist. and if \[1+\sqrt{2}\] exists then \[1\sqrt{2}\]also exits. let \[\sqrt{3i}=ki\] whre k is a no. \[(x5)^{2}(x1\sqrt{2})(x1+\sqrt{2})(x1ki)(x1+ki)\]this gives \[(x5)^{2}((x1)^{2}2)((x1)^{2}+3)\] if you don't consider the conjugate part then you can't get integral coefficient....
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