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pinkdollbaby1234

  • 2 years ago

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

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

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

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

  4. richyw
    • 2 years ago
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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.

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

  6. richyw
    • 2 years ago
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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?

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

  8. richyw
    • 2 years ago
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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.

  9. pinkdollbaby1234
    • 2 years ago
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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

  10. richyw
    • 2 years ago
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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...

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