exactly one common root;
b) no common root;
c) exactly three common roots;
d) exactly two common roots;
Kindly explain...
ans:
x^3 + 2x^2 + 2x + 1 = x(x+1)^2 + (x+1) = (x+1)(x^2 + x + 1) has zeros -1, w and w^2 where w is cube root of 1
-1 is not a zero of x^200 + x^130 + 1 since the expression equals 3 when x = -1.
So we are down to 2 possible roots in common.
let f(x) = x^200 + x^130 + 1
f(w) = w^200 + w^130 + 1= w^2+ w + 1 = 0 so w is a root
f(w^2) = w^400 + w^260 + 1 = w+w^2+1 = 0 so w^2 is a root
so there are exactly 2 common roots and hence d)
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