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OregonDuck

  • one year ago

@mathmate

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  1. OregonDuck
    • one year ago
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    factoring by grouping 2x^+4x+1+4

  2. OregonDuck
    • one year ago
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    my answer is (2x+1)(x+2)(x+4)

  3. OregonDuck
    • one year ago
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    @mathmate @welshfella

  4. OregonDuck
    • one year ago
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    @DarkMoonZ

  5. Flvs.net
    • one year ago
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    Need help?

  6. OregonDuck
    • one year ago
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    yes plz

  7. Flvs.net
    • one year ago
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    I like your answer, it looks good to me.

  8. Flvs.net
    • one year ago
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    7th advanced, you?

  9. Flvs.net
    • one year ago
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    Oh really?

  10. Flvs.net
    • one year ago
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    I was in public school advanced too.

  11. mathmate
    • one year ago
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    @OregonDuck If the answer is (2x+1)(x+2)(x+4) the question should have been 2x^3+13x^2+22x+8 Do you still want to know \(how\) to solve it?

  12. OregonDuck
    • one year ago
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    yes plz

  13. OregonDuck
    • one year ago
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    @mathmate yes plz

  14. mathmate
    • one year ago
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    We note that all coefficients are positives, therefore so are the factors. The leading coeff. is 2, and the constant term is 8, so the possible factors are 1/2, 1,2,4, and 8. we can try group them assuming a factor of x+2 2x^3+13x^2+22x+8 =2x^3+4x^2 + 9x^2+18x + 4x+8 =2x^2(x+2) + 9x(x+2) + 4(x+2) and it seems to work well, so =(x+2)(2x^2+9x+4) factor the quadratic the same way, if you want: =(x+2)(2x^2+8x + x+4) =(x+2)(x+4)(2x+1)

  15. OregonDuck
    • one year ago
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    so (x+2)(x+4)(2x+1) is the answer?

  16. mathmate
    • one year ago
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    Yes, for the following question: 2x^3+13x^2+22x+8

  17. OregonDuck
    • one year ago
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    NO WHAT IS THE ANSWER FOR 2x^+4x+1+4

  18. mathmate
    • one year ago
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    you cannot factor 2x^+4x+1+4 because it is not a polynomial. exponent is missing, and I suspect there are typos. Please check the question.

  19. mathmate
    • one year ago
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    It also shows that you have not made any attempt to solve the problem, when you have posted your question for over half an hour and not realize that there are typos.

  20. OregonDuck
    • one year ago
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    no 2x^2+4x+1+4

  21. mathmate
    • one year ago
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    Are you sure about the +1+4 part? The problem could have been written 2x^2+4x+5

  22. OregonDuck
    • one year ago
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    yes i am sure

  23. mathmate
    • one year ago
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    There are no rational factors for the problem 2x^2+4x+5.

  24. OregonDuck
    • one year ago
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    k if it was written this way then the answer would be?

  25. mathmate
    • one year ago
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    There are no rational factors for the problem 2x^2+4x+5.

  26. OregonDuck
    • one year ago
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    then my answer was right then for 2x^+4x+1+4 right?

  27. OregonDuck
    • one year ago
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    or prime?

  28. mathmate
    • one year ago
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    what was your answer?

  29. OregonDuck
    • one year ago
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    (2x+1)(x+2)(x+4)

  30. pooja195
    • one year ago
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    #CAUGHT XD you got the answer and now you are presenting it here lol

  31. mathmate
    • one year ago
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    Your "answer" expands to a cubic polynomial. You're given a quadratic polynomial to factor, so your "answer" cannot be right. I already gave you the question that corresponds to your "answer".

  32. pooja195
    • one year ago
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    http://openstudy.com/study#/updates/55b25896e4b0ce10565f5149

  33. OregonDuck
    • one year ago
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    so it is prime?

  34. mathmate
    • one year ago
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    There are no rational factors, as I said earlier.

  35. mathmate
    • one year ago
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    Just to satisfy my curiosity, can you tell me how you obtained the answer (2x+1)(x+2)(x+4) to "factor 2x^+4x+1+4" ?

  36. mathmate
    • one year ago
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    The reason I said there are no rational factors is because the expression 2x^2+4x+5 can be factored into (x+1+sqrt(3/2)i)(x+1-sqrt(3/2)i) where i is the complex number where i^2=-1 by completing the square!

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