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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats easy , first find where derivative is zero

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you know where the function turns around ,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then compare the values of the function at the turning points

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you go from positive to negative , or negative to positive function value between two turning points then you must have an intersection with the x axis ( ie a zero )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if the sign of the function value doesnt change then there are no roots in the interval .

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0find the number of the real roots , then to find the number of complex root you subtract the number of real roots from the degree of the function

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0Or you can use Descartes theorem and count the sign changes in f(x) for positive roots and f(x) for neg roots positive: 2 or 4 negative: 1 imaginary: 0 or 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0By the fundamental theorem of algebra, this polynomial will have five roots. Descartes rule of signs has that r has either two or zero positive real roots, and one negative real root. The rational root theorem says that 1 is a root of r, and since this is real and positive, r clearly does not have zero real positive roots  it must have two. So, we have so far: Two real positive roots One real negative root Since r has rational coefficients, the remaining roots are complex conjugates by the complex conjugate root theorem. So for r: Two real, positive roots One real, negative root Two complex conjugate roots
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