The number of minimum possible complex roots of the equation $\displaystyle x^{6}-3x^{5}+4x^{3}+3x^{2}+4=0 $ is $2$

Using the given equation x^6 - 3x^5 + 4x^3 + 3x^2 + 4 = 0, we substitute x = 1 to get 1 - 3 + 4 + 3 + 4 = 9, and x = -1 to get 1 + 3 - 4 + 3 + 4 = 7. Because neither value results in zero, the polynomial has no real roots at 1 or -1, so it cannot have four or six complex roots. The minimum number of complex roots it can have is therefore 2, confirming the assertion and its underlying reason are correct.

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