Answer to Question #42524 in Calculus for shasha
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
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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