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# Answer to Question #42524 in Calculus for shasha

Question #42524
State how many imaginary and real zeros the function has.
f(x) = x5 + 7x4 + 2x3 + 14x2 + x + 7

3 imaginary; 2 real

4 imaginary; 1 real

0 imaginary; 5 real

2 imaginary; 3 real

show work and explain it to me please
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 &nbsp;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.&nbsp;&nbsp; &nbsp; &nbsp;&nbsp;

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