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Math Help - need help factoring a problem

  1. #1
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    need help factoring a problem

    9x^2y^2+5xy-4

    I dont really even know where to begin

    My book doesnt explain what to do with this kind of problem at all. I mean it does talk about factoring trinomials, but when its setup this way its a bit confusing, and different

    any help is appreciated.
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  2. #2
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    It is a tricky one.

    Maybe before trying this one, have a go at 9a^2+5a-4

    Can you do it?
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  3. #3
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    yeah I would rewrite the problem as

    (9a2 + 9a)(-4a - 4)
    9a(a + 1) -4(a+1)
    (9a-4)(a+1)
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  4. #4
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    Good; do same with yours, after letting y=1 ; get it?
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  5. #5
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    Quote Originally Posted by Wilmer View Post
    Good; do same with yours, after letting y=1 ; get it?
    unfortunately im a little confused by what you mean lol
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  6. #6
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    Quote Originally Posted by NecroWinter View Post
    unfortunately im a little confused by what you mean lol
    9a^2 + 5a - 4

    9(xy)^2 + 5(xy) - 4
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  7. #7
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    Ok so heres what I think I should do with that, if someone can fix my train of thought i think that would help me correct the errors.

    9x^2y^2+9x-4x-4
    add ()'s
    (9x^2y^2+9x)(-4x-4)

    (9x^2y^2+9x)
    9x(y^2 + 1)-4(x+1)

    this would be so much better if my book actually talked about factoring trinomials when you have exponents set up this way, outside the norm

    i also tried something like this but it just didnt look right at all:
    9(xy)^2+5(xy)-4

    9xy(1^2 +1) -4(1(xy) +1)
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  8. #8
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    you're really making this much more difficult than it needs to be ...

    pickslides had you factor 9a^2 + 5a - 4 for a reason.

    9a^2 + 5a - 4 = (9a-4)(a+1)

    compare the factorization with a above to the factorization with xy below ...

    9(xy)^2 + 5(xy) - 4 = (9xy - 4)(xy + 1)

    see the similarity?
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  9. #9
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    Okay I gotcha. Thanks

    Is this typically how you guys tackle these kinds of problems? Will simply substituting(for a lack of a better word) make me run into a problem later on?
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  10. #10
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    can't say ... depends on the problem.
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