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tien.lam12
What are the possible number of positive, negative, and complex zeros of f(x) = -3x4 - 5x3 - x2 - 8x + 4 ? A) Positive: 2 or 0; Negative: 2 or 0; Complex: 4 or 2 or 0 B) Positive: 1; Negative: 3 or 1; Complex: 2 or 0 C) Positive: 3 or 1; Negative: 1; Complex: 2 or 0 D) Positive: 4 or 2 or 0; Negative: 2 or 0; Complex: 4 or 2 or 0
the numbers after the x's are exponents, correct?
there are 2 positive and 2 complex roots
i've tried using descartes rule of signs but i got confused to be honest so i solved it
one root lie b/w 0 and 1
other will be negative as f(x) is increasing in R+
yeah 1 will be positive and one will not be possitive it can also be possilbe that roots may be irrational becuase irrational roots also exist in pairs
there is 1 sign change so by DRS there is 1 positive root Sub -x for x to get negative roots, 3 sign changes so 3 or 1 negative roots Complex roots are total roots, 4, less positive and negative roots so 0 or 2 of those. Answer is B