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whatevs

  • 3 years ago

When asked to find the zeros of the function r(x) = 5t3 – 8t2 + 7t + 2 which task do you complete first (Rational Root Theorem or Descartes’ Rule of Signs? Use complete sentences to explain your reasoning

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  1. Dalvoron
    • 3 years ago
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    I've never heard of these before, but just looked them up! At first glance, I'd argue that the Rational Root Theorem should come first. It's more specific than Decartes'. By this I mean, RRT will get you some finite amount of numbers, the factors of the coefficients of the highest power, and lowest power of x respectively. For example, if you have \[3x^3-5x^2+5x-2=0\], RRT will return 8 possible roots. That's all of the numbers in existence, narrowed down to 8 particular numbers. On the other hand Decartes will give you a maximum number for the amount of positive and negative roots. If I understand it right, it will return "at most 3 positive roots" for the example given above. Well big whoop, right? All of infinfity narrowed down to half, \(maybe\). On its own, it doesn't really seem to be of any help whatsoever. If you do it after the other one, you can narrow 8 down to 3, so it's moderately useful there.

  2. Dalvoron
    • 3 years ago
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    Then again, it's possible I don't understand these theorems at all.

  3. whatevs
    • 3 years ago
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    hmm..its ok... can you help me with this one... What are the possible number of positive, negative, and complex zeros of f(x) = –2x3 + 5x2 + 6x – 4 ? Answer a.) Positive: 2 or 0; Negative: 1; Complex: 2 or 1 b.) Positive: 2 or 0; Negative: 1; Complex: 2 or 0 c.) Positive: 1; Negative: 2 or 0; Complex 2 or 0 d.) Positive: 1; Negative: 2 or 0; Complex: 0

  4. Dalvoron
    • 3 years ago
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    Well I see 2 sign changes there, which I think means there are at most 2 positive roots, meaning a) or b). After multiplying the odd coefficients by -1, there is 1 sign change, so that means there is 1 negative root I think. Then the number of complex roots is 3-(2+1)=0 OR 3-(0+1)=2. So I think b) is the answer.

  5. whatevs
    • 3 years ago
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    i figured the first one out...=) one last question...i promise!! =) and thanks alot for the help What are the possible number of negative zeros of f(x) = –2x7 + 2x6 + 7x5 + 7x4 + 4x3 + 4x2 ? Answer a.) 4, 2, or 0 b.) 2 or 0 c.) 3 or 1 d.) 7, 5, 3, or 1

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