2x^5-3x^4-5x^3-15x^2-207x+108zeros: 3ifind the remaing zeros

Mathematics

1 answer:

Veseljchak [2.6K]3 years ago

4 0

The real solutions are { $x=-3, x= \frac{1}{2} , x=4$ }.

The imaginary solutions are { $x=-3i, x=3i$ }.

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