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MelindaR Group Title

What can we afirm about p(x)=x^5 - 5x^3 + 4x^2 - 3x - 2? a) p(x) has only 3 real roots (1 integer and 2 rationals) b)p(x) has only one real root, which is also integer c) x=2 isn't a root of p(x) d)p(x) has only real roots (1 integer, 2 rationals and 2 irrationals) So, c is wrong because I divided p(x) by x - 2 and I got x^4 + 2x^3 - x^2 + 2x + 1. Problem is I have no idea how to find the other roots! Help!

  • 2 years ago
  • 2 years ago

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  1. amitlpu91 Group Title
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    jst check for those value whose value for thoce given equation gives "0".& this shows that root is real .

    • 2 years ago
  2. MelindaR Group Title
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    I've already tried some

    • 2 years ago
  3. MelindaR Group Title
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    @ash2326 @Mertsj Any ideas?

    • 2 years ago
  4. Mertsj Group Title
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    2 is one rational zero

    • 2 years ago
  5. MelindaR Group Title
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    So it's either a or d.

    • 2 years ago
  6. Mertsj Group Title
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    It has three real zeros

    • 2 years ago
  7. Mertsj Group Title
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    One integer and two rationals.

    • 2 years ago
  8. Mertsj Group Title
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    A

    • 2 years ago
  9. MelindaR Group Title
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    how did you get that?

    • 2 years ago
  10. Mertsj Group Title
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    i graphed it on Wolfram

    • 2 years ago
  11. MelindaR Group Title
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    Clever. Thx

    • 2 years ago
  12. Mertsj Group Title
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    yw

    • 2 years ago
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