The number of zeros does not exceed 5, because it is equation of degree 5. The closed-form formula for solutions does not exist, though some types of equations can be solved. Possible roots are divisors of 7 by Rational Zeros Theorem. By trial and error method, -7 is a zero of this equation. Divide x5+7x4+2x3+14x2+x+7 by (x+7). So,
x5+7x4+2x3+14x2+x+7=(x+7)(x2+1)2. Zeros of equation x2+1=0 are i and (-i), factor x2+1 is squared, therefore, zeros of the initial equation are i,i,-i, -i and -7. Thus, the answer is 1 real and 4 complex zeros.
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