2020 AMC 12B Problems/Problem 17
How many polynomials of the form , where , , , and are real numbers, have the property that whenever is a root, so is ? (Note that )
Let . We first notice that . That is because of Euler's Formula : . = = = = . In order to be a root of , must also be a root of P, meaning that 3 of the roots of must be , , . However, since is degree 5, there must be two additional roots. Let one of these roots be , if is a root, then and must also be roots. However, is a fifth degree polynomial, and can therefore only have roots. This implies that is either , , or . Thus we know that the polynomial can be written in the form . Moreover, by Vieta's, we know that there is only one possible value for the magnitude of as , meaning that the amount of possible polynomials is equivalent to the possible sets . In order for the coefficients of the polynomial to all be real, due to and being conjugates and since , (as the polynomial is 5th degree) we have two possible solutions for which are and yielding two possible polynomials. The answer is thus .
~Murtagh (edited by totoro.selsel)
Let , then , x_3=〖((-1+i√3)/2)〗^2 r=(-1-i√3)/2 r, x_4=〖((-1+i√3)/2)〗^3 r=r which means x_4 will be back to x_1 Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2, and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of the different root is wrong. The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
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