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Function F has exactly 2 complex solutions
(2x+5)(2x-5)(2x+5i)(2x-5i)
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x=(4+3i)/5, x=(4=3i)/5
=?
did you mean -?
yes sorry
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The first one is: x^2+2x+2 The second one is: x^2+4 The third one is: x^2-4x+8
The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. This includes polynomials with real coefficients, since every real number is a complex number with an imaginary part equal to zero. So, B
failed again lol
i hate my life
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I'm sorry if I gave you the wrong answers, I did the best I could
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