anonymous
  • anonymous
find zeros..will give out medals!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
anonymous
  • anonymous
descartes rule of sign for this one
anonymous
  • anonymous
can you help me find the answer?

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More answers

anonymous
  • anonymous
\[f(x)=-2x^3-5x^2+6x+4\] coefficients are \(-2,-5,6,4\)
anonymous
  • anonymous
there is one "change of sign" for \(f(x)\) from \(-5\) to \(+6\)
anonymous
  • anonymous
ok what is next?
anonymous
  • anonymous
so there is one positive zero you count down by 2s and there cannot be -1 positive zero
anonymous
  • anonymous
ok
anonymous
  • anonymous
oh btw there are 3 zeros all together because the degree is 3, although some may be complex or repeated
anonymous
  • anonymous
now we compute \(f(-x)\)
anonymous
  • anonymous
ok
anonymous
  • anonymous
\[f(-x)=2x^3-5x^2-6x+4\]
anonymous
  • anonymous
ok
anonymous
  • anonymous
the sign of the coefficients of the terms of odd degree change, and the even degree stays the same the coefficients are \(2,-5,-6,4\)
anonymous
  • anonymous
now there are two "changes in sign" from 2 to -5 and from -6 to 4 is that more or less clear?
anonymous
  • anonymous
ok is that the final answer?
anonymous
  • anonymous
can you help me with one more? it is easier this time
anonymous
  • anonymous
oh no two changes in sign of \(f(-x)\) means either 2 or 0 negative zeros
anonymous
  • anonymous
anonymous
  • anonymous
oh ok sorry..please finish the problem
anonymous
  • anonymous
so final answer is this: 1 positive zero for sure either 2 negative zeros and therefore no complex zzeros or no negative zeros and two complex zeros
anonymous
  • anonymous
ok thanks..can you do one more? please?
anonymous
  • anonymous
this one is easier \[\sum_3^53n+2\]
anonymous
  • anonymous
put \(n=3\) get \(3\times 3+2=11\) then put \(n=4\) get \(3\times 4+2=14\) finally put \(n=5\) get \(17\) the "sigma" \(\sum\) means add them up
anonymous
  • anonymous
\[11+14+17\]is what you need to compute, and you are done
anonymous
  • anonymous
thats it?
anonymous
  • anonymous
thanks

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