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anonymous
 3 years ago
Algebra II
What are the possible number of positive, negative, and complex zeros of f(x) = 3x4  5x3  x2  8x + 4 ?
Positive: 2 or 0; Negative: 2 or 0; Complex: 4 or 2 or 0
Positive: 1; Negative: 3 or 1; Complex: 2 or 0
Positive: 3 or 1; Negative: 1; Complex: 2 or 0
Positive: 4 or 2 or 0; Negative: 2 or 0; Complex: 4 or 2 or 0
anonymous
 3 years ago
Algebra II What are the possible number of positive, negative, and complex zeros of f(x) = 3x4  5x3  x2  8x + 4 ? Positive: 2 or 0; Negative: 2 or 0; Complex: 4 or 2 or 0 Positive: 1; Negative: 3 or 1; Complex: 2 or 0 Positive: 3 or 1; Negative: 1; Complex: 2 or 0 Positive: 4 or 2 or 0; Negative: 2 or 0; Complex: 4 or 2 or 0

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0PLEEEEEAAAAAAAAASSSSEEEEE HEEEELLLLPPPPP ! & EXPLAIN :C

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0f(x)=3x^(4)5x^(3)x^(2)8x+4 If a polynomial function has integer coefficients, then every rational zero will have the form (p)/(q) where p is a factor of the constant and q is a factor of the leading coefficient. p=4_q=3 Find every combination of \(p)/(q). These are the possible roots of the polynomial function. \1,\(1)/(3),\2,\(2)/(3),\4,\(4)/(3)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I would say; Positive: 1; Negative: 3 or 1; Complex: 2 or 0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0f(x)=3x^(4)5x^(3)x^(2)8x+4 Since there are 1 sign change from the highest order term to the lowest, there is at most 1 positive root (Descartes' Rule of Signs). There is 1 positive root. To find the possible number of negative roots, replace x with x and repeat the sign comparison. f(x)=3(x)^(4)5(x)^(3)(x)^(2)8(x)+4 Simplify the polynomial. f(x)=3x^(4)+5x^(3)x^(2)+8x+4 Since there are 3 sign changes from the highest order term to the lowest, there are at most 3 negative roots (Descartes' Rule of Signs). There are 3 or 1 possible negative roots.
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