Problem:
How many polynomials of the form x5+ax4+bx3+cx2+dx+2020, where a,b,c, and d are real numbers, have the property that whenever r is a root, so is 2β1+i3βββ r ? (Note that i=β1β )
Answer Choices:
A. 0
B. 1
C. 2
D. 3
E. 4 Solution:
Let P(x)=x5+ax4+bx3+cx2+dx+2020. We first notice that 2β1+i3ββ=e2Οi/3. That is because of Euler's Formula : eix=cos(x)+iβ sin(x). 2β1+i3ββ=β21β+iβ 23ββ=cos(120β)+iβ sin(120β)=e2Οi/3.
In order r to be a root of P,re2Οi/3 must also be a root of P, meaning that 3 of the roots of P must be rrei32Οβ,rei34Οβ. However, since P is degree 5, there must be two additional roots. Let one of these roots be w, if w is a root, then we2Οi/3 and we4Οi/3 must also be roots. However, P is a fifth degree polynomial, and can therefore only have 5 roots. This implies that w is either r,re2Οi/3, or re4Οi/3. Thus we know that the polynomial P can be written in the form (xβr)m(xβre2Οi/3)n(xβre4Οi/3)p. Moreover, by Vieta's, we know that there is only one possible value for the magnitude of r as β₯rβ₯5=2020, meaning that the amount of possible polynomials P is equivalent to the possible sets (m,n,p). In order for the coefficients of the polynomial to all be real, n=p due to re2Οi/3 and re4Οi/3 being conjugates and since m+n+p=5, (as the polynomial is 5th degree) we have two possible solutions for (m,n,p) which are (1,2,2) and (3,1,1) yielding two possible polynomials. The answer is thus (C)2β .
OR
Let x1β=r, then
x2β=2β1+i3ββr x3β=(2β1+i3ββ)2r=(2β1βi3ββ)r x4β=(2β1+i3ββ)3r=r
which means x4β is the same as x1β.
Now we have 3 different roots of the polynomial, x1β,x2β, and x3β. Next, we will prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there is one root x4β=p which is different from the three roots we already know, then there must be two other roots,
different from all known roots. We get 6 different roots for the polynomial, which contradicts the limit of 5 roots. Therefore the assumption of a different root is wrong, thus the roots must be chosen from x1β,x2β, and x3β.
The polynomial then can be written like f(x)=(xβx1β)m(xβx2β)n(xβx3β)q, where m,n, and q are nonnegative integers and m+n+q=5. Since a,b,c and d are real numbers, then 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)2β.