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pinkdollbaby1234 Group Title

i dont understand how to do this use descartes rule of signs to describe the root of h(x)=4x^4-5x^3+2x^2-x+5

  • 2 years ago
  • 2 years ago

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  1. richyw Group Title
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    \[h(x)=4x^4-5x^3+2x^2-x+5\]how many times does it change signs?

    • 2 years ago
  2. richyw Group Title
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    i see 4 now do it for \(h(-x)\)...

    • 2 years ago
  3. pinkdollbaby1234 Group Title
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    okay so what do you mean by that solve for h?

    • 2 years ago
  4. richyw Group Title
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    no \[h(-x)=4(-x)^4-5(-x)^3+2(-x)^2-(-x)+5\]Now simplify that, and then count the sign changes.

    • 2 years ago
  5. pinkdollbaby1234 Group Title
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    oh okay so i would plug in 4 for -x since it changes four times or is im wrong again?

    • 2 years ago
  6. richyw Group Title
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    yeah you are wrong. this is what you should get.\[h(-x)=4x^4+5x^3+2x^2+x+5\]Now how many times does the sign change places there?

    • 2 years ago
  7. pinkdollbaby1234 Group Title
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    the sign doesnt change anymore cuz all are equal?

    • 2 years ago
  8. richyw Group Title
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    yup so you can conclude that there are no negative roots, and either 4, 2 or 0 positive roots. You also know that there are 4 roots total. so you can either have 4 positive real and 0 complex, 2 and 2, or 0 real positive roots and 4 complex roots.

    • 2 years ago
  9. pinkdollbaby1234 Group Title
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    omg thay seem soo simple i have to practice on this more but thankyou so much u help me understand it a bit

    • 2 years ago
  10. richyw Group Title
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    yup no worries! And in this case that is all it can tell you. Sometimes it can tell you exactly how many of each there are, but sometimes it gives you a few options. It is still useful though because you can use it before you use other methods to get an idea what you are looking for...

    • 2 years ago
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