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 2−1+i3⋅r ? (Note that i=−1 )
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 2−1+i3=e2πi/3. That is because of Euler's Formula : eix=cos(x)+i⋅sin(x). 2−1+i3=−21+i⋅23=cos(120∘)+i⋅sin(120∘)=e2πi/3.
In order r to be a root of P,re2πi/3 must also be a root of P, meaning that 3 of the roots of P must be rrei32π,rei34π. 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 we2πi/3 and we4πi/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,re2πi/3, or re4πi/3. Thus we know that the polynomial P can be written in the form (x−r)m(x−re2πi/3)n(x−re4πi/3)p. Moreover, by Vieta's, we know that there is only one possible value for the magnitude of r as ∥r∥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 re2πi/3 and re4πi/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)2 .
OR
Let x1=r, then
x2=2−1+i3r x3=(2−1+i3)2r=(2−1−i3)r x4=(2−1+i3)3r=r
which means x4 is the same as x1.
Now we have 3 different roots of the polynomial, x1,x2, and x3. 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 x4=p which is different from the three roots we already know, then there must be two other roots,
x5=(2−1+i3)2p=(2−1−i3)p x6=(2−1+i3)3p=p
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 x1,x2, and x3.
The polynomial then can be written like f(x)=(x−x1)m(x−x2)n(x−x3)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)2.