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anonymous

  • 5 years ago

Factor Completely: 8x^3+12x^2-10x-15

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  1. anonymous
    • 5 years ago
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    Everything after the \[8x^3\] I could do now, but would you factor the above like so? (2x)^3

  2. anonymous
    • 5 years ago
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    \[(2x)^3\] rather

  3. anonymous
    • 5 years ago
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    I tried going the right side by itself, but no divisors of 180 = -10. The closest was 10*-8

  4. anonymous
    • 5 years ago
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    I think you're going to have to factor into three linear equations. ie, (2x-a)(2x-b)(2x-c)

  5. anonymous
    • 5 years ago
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    Yech, I will try that

  6. anonymous
    • 5 years ago
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    That left me with \[2x(4x^2+6x-5)-15\]

  7. anonymous
    • 5 years ago
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    Trying to solve that too

  8. anonymous
    • 5 years ago
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    But the 4x^2+6-5 won't factor either.

  9. anonymous
    • 5 years ago
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    I'm sorry, 4x^2+6x-5 I multiplied 5*4 and got 20. I put in the negative sign and got -20. The only things that go into that are 10*2 and 5*4

  10. anonymous
    • 5 years ago
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    Neither of those, no matter how you switch them around, equals +6

  11. anonymous
    • 5 years ago
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    Unless the simplest factored form of the top problem is: \[(2x-15)(4x^2+6x-5)\]

  12. anonymous
    • 5 years ago
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    And that can't be right either. I just started long multiplication, and I already got -75 and that isn't 15

  13. anonymous
    • 5 years ago
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    Yeah, that's definitely not right.

  14. anonymous
    • 5 years ago
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    Maybe you just can't factor it and it is a trick question?

  15. anonymous
    • 5 years ago
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    I think you can, I just don't remember how to do 3rd power equations. Still plugging and chugging. ;)

  16. anonymous
    • 5 years ago
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    He does give us equations that have no solutions all of the time, among ones that look like they don't have solutions but they really do. Ugh, I am going to screw this final up so bad :P

  17. anonymous
    • 5 years ago
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    I found something on the net saying this is a cubic function and not a quadratic function

  18. anonymous
    • 5 years ago
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    That's annoying. Have you had other problems that factor out to three linear equations? ic, (ax+d)(bx+e)(cx+f)?

  19. anonymous
    • 5 years ago
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    right, that is true. Quadratic should have a highest power of x^2.

  20. anonymous
    • 5 years ago
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    Never heard of it before. But the problem I looked up had something similar and the answered turned into: \[\pm i \sqrt{?}\]

  21. anonymous
    • 5 years ago
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    a 3 was supposed to be in the square root box

  22. anonymous
    • 5 years ago
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    nvm, that was solving for x.

  23. anonymous
    • 5 years ago
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    \[i=\sqrt{-1}\] And I doubt that is what you are working with!

  24. anonymous
    • 5 years ago
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    No matter how I break up the problem, nothing factors anywhere other than: \[2x(4x^2+6x-5)-15\]

  25. anonymous
    • 5 years ago
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    The next 2 problems after this are easy, so it makes me feel like I should know how to do this.

  26. anonymous
    • 5 years ago
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    look 3/4 of the page down here: http://salinesports.org/mr_frederick/Algebra%202A/Unit%206/Notes_0605.pdf

  27. anonymous
    • 5 years ago
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    I'm not very familiar with that method, but it does work.

  28. anonymous
    • 5 years ago
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    I found a formula, one sec

  29. anonymous
    • 5 years ago
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    \[x=y-(b)/(3a)\]

  30. anonymous
    • 5 years ago
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    I have an answer for you.

  31. anonymous
    • 5 years ago
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    The b/3a part is a fraction

  32. anonymous
    • 5 years ago
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    what is y? Look at that link if you can. I'll type out the equation here if you want/rather.

  33. anonymous
    • 5 years ago
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    I will let you go on this, I have bothered you for way too long :P

  34. anonymous
    • 5 years ago
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    It is calling it a depressed cubic

  35. anonymous
    • 5 years ago
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  36. anonymous
    • 5 years ago
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    That is what it says to do

  37. anonymous
    • 5 years ago
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    Our objective is to find a real root of the cubic equation ax^3+bx^2+cx+d=0.

  38. anonymous
    • 5 years ago
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    is what it says

  39. anonymous
    • 5 years ago
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    But anyway, I am going to skip this, because I need to be working on logs and matricies :P

  40. anonymous
    • 5 years ago
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    It's fine. I'll type out what I did. First group: \[(8x ^{3}+12x ^{2})+(-10x - 15)\] Then factor: \[4x ^{2}(2x+3) - 5(2x+3)\] since the factored part (2x+3) is the same, you pair the multiplying factors and get this: \[(4x ^{2}-5)(2x+3)\] Which works, and is in it's simplest form (since 4x^2-5 does not factor further as an integer).

  41. anonymous
    • 5 years ago
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    Good luck. :)

  42. anonymous
    • 5 years ago
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    tyvm!

  43. anonymous
    • 5 years ago
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    Welcome - thanks for sticking it out. It was going to drive me crazy! ;)

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