anonymous
  • anonymous
http://i1142.photobucket.com/albums/n608/nancylam67/Page3664.jpg do 4 only
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
thats easy , first find where derivative is zero
anonymous
  • anonymous
so you know where the function turns around ,
anonymous
  • anonymous
then compare the values of the function at the turning points

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anonymous
  • anonymous
if 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
  • anonymous
if the sign of the function value doesnt change then there are no roots in the interval .
anonymous
  • anonymous
*no real roots
anonymous
  • anonymous
find 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
  • dumbcow
Or 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
  • anonymous
no
anonymous
  • anonymous
By 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
anonymous
  • anonymous
thank all

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