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anonymous

  • 5 years ago

how do you simplify 2a^2+5a-3??

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  1. amistre64
    • 5 years ago
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    may be it means to put it into its counterpart... y = M(x-h)^2 +k ?

  2. anonymous
    • 5 years ago
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    well the whole equation is 16a^4-1/2a^2+5a-3 * 2a^2+9a+9/4a^2+1

  3. anonymous
    • 5 years ago
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    You have to factor it: \[2a^2+5a-3\] You get \[(2a^2)(-a-3)\] Then: \[2a(a+3) -1(a+3)\] Finally: \[(2a-1)(a+3)\]

  4. amistre64
    • 5 years ago
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    \[\frac{16x^4 -1}{2a^2 +5a -3} * \frac{2a^2 +9a +9}{4a^2 +1}\]

  5. anonymous
    • 5 years ago
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    you people are a bit confusing- i mean that you're making it more complicated

  6. anonymous
    • 5 years ago
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    Yeah, amistre, that's totally irrelevant. It's a quadratic equation; all you have to do is factor it.

  7. anonymous
    • 5 years ago
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    so far I have (4a^2+1)(2a+1)(2a-1)/(2a+?) * (2a+9)(a+1)/(4a^2+a)

  8. amistre64
    • 5 years ago
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    lol....well it helps to know what the actual question is :)

  9. amistre64
    • 5 years ago
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    how do YOU simplify 2x^2 +5a -3 ?? lol

  10. anonymous
    • 5 years ago
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    yeah lol that was my question

  11. anonymous
    • 5 years ago
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    yes, but not is you're making it more complicated you;re just adding mroe steps

  12. anonymous
    • 5 years ago
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    She/he asked for \[2a2+5a−3\] to be simplified. That means factoring it into binomials.

  13. anonymous
    • 5 years ago
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    she

  14. amistre64
    • 5 years ago
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    i didnt add any steps... I simply was wondering what it was talking about to begin with. right?

  15. anonymous
    • 5 years ago
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    \[2a^2+5a−3\] should be the equation; my bad.

  16. anonymous
    • 5 years ago
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    I just need that to be factored so I can simplify the rest of the problem

  17. anonymous
    • 5 years ago
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    Please read the steps that I posted previously, and tell me if it helps at all.

  18. anonymous
    • 5 years ago
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    not really. It only makes it more confusing.

  19. amistre64
    • 5 years ago
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    in the context of the rest of the "question", then yes; factoring it is appropriate.

  20. amistre64
    • 5 years ago
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    i dont see it factoring to anything useful; so factor the difference of cubes and leave that part alone ;)

  21. anonymous
    • 5 years ago
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    Ok thanks :D

  22. amistre64
    • 5 years ago
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    you get square roots involved if you factor that thing there, and the other one is clean of them....so they dont cancel anyways... make sense?

  23. anonymous
    • 5 years ago
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    A little more than it did. This is what I got: (2a-1)(2a+1)(2a+9)(a+1)/2a^2+5a-3

  24. amistre64
    • 5 years ago
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    without doing the math myself, that looks about right :)

  25. anonymous
    • 5 years ago
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    Awesome lol thank you

  26. amistre64
    • 5 years ago
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    x = -5/4 +- 7/4 x = 12/4 = 3 x = -2/4 = -1/2 (2x-1)(x+3) .... or this :)

  27. amistre64
    • 5 years ago
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    (x+6/2)(x-1/2) (x+3)(2x-1) is right... now that I have a moment to see it correctly ;)

  28. anonymous
    • 5 years ago
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    That's what I thought it was. Except I had the signs switched.

  29. amistre64
    • 5 years ago
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    (2a-1)(2a+1)(2a+9)(a+1) --------------------- (2a-1)(a+3)

  30. amistre64
    • 5 years ago
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    2a^2 +9a +9 (a+6/2)(a+3/2) (a+3)(2a+3) is what I get on the top over there

  31. anonymous
    • 5 years ago
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    so the answer would be: \[(2a+1)(2a+9)(a+1)\div (a+3)\]

  32. amistre64
    • 5 years ago
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    (a+3)(2a+3) (4x^2+1) (4x^2-1) ----------- ---------------- (2a-1)(a+3) (4x^2+1) (2a+3) (4a^2-1) --------------- (2a-1) (2a+3) (2a-1)(2a+1) --------------- (2a-1) (2a+3)(2a+1)

  33. anonymous
    • 5 years ago
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    how did you get a+3 on the top?

  34. amistre64
    • 5 years ago
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    4a^2 +8a +3 is what I get if I did it right ;) (a+3)(2a+3) = 2a^2 +9a +9

  35. amistre64
    • 5 years ago
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    (a+6/2)(a+3/2) (a+3) (a+3/2) (a+3)(2a+3)

  36. anonymous
    • 5 years ago
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    but when you factor it back to 2a^2+9a+9...the 3's don't add up to 9

  37. amistre64
    • 5 years ago
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    2aa + 3a +2(3)a + 9 2a^2 +3a +6a +9 2a^2 +9a +9

  38. anonymous
    • 5 years ago
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    Ooooh ok I think I got

  39. amistre64
    • 5 years ago
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    am I right :)

  40. anonymous
    • 5 years ago
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    Yup!! Thanks! Now I have 4 more to go....jeez

  41. amistre64
    • 5 years ago
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    lol good luck ;)

  42. anonymous
    • 5 years ago
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    thanks

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