Answer to Question #42524 in Calculus for shasha

Question

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

Answer 1

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.