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MissKelly~

  • one year ago

Find all values of k so that each polynomial can be factored using integers. 1.) x^2+kx-19 2.) x^2 - 8x+k, k>0

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  1. klimenkov
    • one year ago
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    what means using integers?

  2. klimenkov
    • one year ago
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    Integer roots or integer value of k?

  3. MissKelly~
    • one year ago
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    I don't know...I just wrote word for word for what it says in the book...All I know is that for #1 the answers will be in integers...

  4. klimenkov
    • one year ago
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    For the first one: -18, 18.

  5. MissKelly~
    • one year ago
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    How do you solve it? Distributive...factor?

  6. klimenkov
    • one year ago
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    Do you know Vieta's Theorem?

  7. MissKelly~
    • one year ago
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    Not yet, The questions I got was from the book on the chapter of Quadratic equations, and it said it required Higher-order thinking skills...

  8. MissKelly~
    • one year ago
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    So you factor it out into 2 binomials?

  9. klimenkov
    • one year ago
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    Vieta's Theorem is not so hard to get. I advice you to get it and try to use for your task.

  10. MissKelly~
    • one year ago
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    I'm not so sure my teacher wants me to apply Vieta's Theorem yet, because she hates it when we move on ahead of her teachings...btw she can't teach =_=

  11. MissKelly~
    • one year ago
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    Well I have to leave for school in a little while, there's a chemical equation quiz waiting for me...

  12. MissKelly~
    • one year ago
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    Thanks for helping me~

  13. MissKelly~
    • one year ago
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    Hehe why is that? Well I'm sorry I have to go to school now..

  14. MissKelly~
    • one year ago
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    See you guys around^^ Bye

  15. klimenkov
    • one year ago
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    If you want to factor \(x^2+kx-19\), I hope you will think it will be \((x-a)(x-b)\). So, when expanding: \(x^2-(a+b)x+ab\). Now just look at the coefficients. \(k=-(a+b), -19=ab\). You want only integers. From the second equality it will be only \(a=1,b=-19\) or \(a=19,b=-1\), because 19 is a prime number. Now try to find \(k\) in both cases.

  16. klimenkov
    • one year ago
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    Just try to explain. Not so sure he got it.

  17. klimenkov
    • one year ago
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    Or she.

  18. klimenkov
    • one year ago
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    If \(-(a+b)=-8\), so \(a+b=8\). And \(ab=k\), we get \(a(8-a)=k>0\). It will be if \(-a^2+8a>0\).|dw:1357824979524:dw| Answer: 1, 2, 3, 4, 5, 6, 7.

  19. klimenkov
    • one year ago
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    Oops. Sorry, it's not right.

  20. klimenkov
    • one year ago
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    The answer will be if you put all this numbers: 1, 2, 3, 4, 5, 6, 7 in the expression for \(k=a(8-a)\) instead of \(a\).

  21. klimenkov
    • one year ago
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    How did you get that this will be all the possible values for \(k\), that you've written k = {0, 7, 12, 15, 16}? \(k=0\) doesn't suit.

  22. klimenkov
    • one year ago
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    \(k>0\) in the statement.

  23. klimenkov
    • one year ago
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    I can't be absolutely sure that this will be ALL the values for \(k\) by just guessing its value.

  24. klimenkov
    • one year ago
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    The source of our misunderstanding is my bad English. Sometimes I can't get what is spoken about.

  25. klimenkov
    • one year ago
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    You say that it is no need in writing \(k=a(8-a), a=1,2,\ldots ,7\). Because it will have the same value for \(a=1\) and \(a=7\). I just showed the way I had solved this task. Sorry.

  26. MissKelly~
    • one year ago
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    Okay I'm back...whoooooa. And yeah I got it now^^

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