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Math Help - AC Method question

  1. #1
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    AC Method question

    I am asked to factor 2x^2+13x-24

    I multipled -24 and 2 and came to -48 and found that 16 and -3 were the solutions because they equal the X coifficient when added.

    My question is this - When I rewrite the equation where do I stick the 16 and the -3

    should it be 2x^2+16x-3x-24

    or

    should it be 2x^2-3x+16x-24

    Is this a contstant? Does the largest/smallet number always get plugged in first?

    Thanks!!!!!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathfailure View Post
    I am asked to factor 2x^2+13x-24

    I multipled -24 and 2 and came to -48 and found that 16 and -3 were the solutions because they equal the X coifficient when added.

    My question is this - When I rewrite the equation where do I stick the 16 and the -3

    should it be 2x^2+16x-3x-24

    or

    should it be 2x^2-3x+16x-24

    Is this a contstant? Does the largest/smallet number always get plugged in first?

    Thanks!!!!!
    no, its not always the largest or smallest first.
    you want to plug in in such a way that the first two terms and the last two terms factor so that you can get a common factor among them. in this case it doesn't matter:

    let's say you use
    2x^2+16x-3x-24
    = 2x(x + 8) -3(x + 8)
    = (x + 8)(2x - 3)

    if you used
    2x^2-3x+16x-24
    = x(2x - 3) + 8(2x - 3)
    = (2x - 3)(x + 8)...............same thing
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    is up to his old tricks again! Jhevon's Avatar
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    otherwise, just look for things that the numbers have in common. for instance. if the coefficient of x^2 is odd and the constant is even, and for the two numbers you get one odd and one even, then you'd want to pair th odd and even numbers together and so on. don't worry about it, it comes with experience. if you put the numbers the wrong way, you'll realize within 2 seconds when you try to factorize, and then you'd just switch them.
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  4. #4
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    Quote Originally Posted by Jhevon View Post
    otherwise, just look for things that the numbers have in common. for instance. if the coefficient of x^2 is odd and the constant is even, and for the two numbers you get one odd and one even, then you'd want to pair th odd and even numbers together and so on. don't worry about it, it comes with experience. if you put the numbers the wrong way, you'll realize within 2 seconds when you try to factorize, and then you'd just switch them.

    Thanks so much! Do you think it would be better if I just use the quadratic forumla for these, just to be more accurate? It seems like when I use the quadratic forumula things turn out better, but according to my teacher I should know how to use the AC method and that it's much easier for simple polynomials. Oddly enough the AC method seems to be giving me more trouble than the quadratic forumula. Thanks agian for your help and quick response.

    And one more question.................

    When you have

    2x^2-3x+16x-24
    = x(2x - 3) + 8(2x - 3)
    = (2x - 3)(x + 8)...............same thing

    I thought you would have to factor out a - X and a -8 . Could you tell me why that isn't so. Thanks.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathfailure View Post
    Thanks so much! Do you think it would be better if I just use the quadratic forumla for these, just to be more accurate? It seems like when I use the quadratic forumula things turn out better, but according to my teacher I should know how to use the AC method and that it's much easier for simple polynomials. Oddly enough the AC method seems to be giving me more trouble than the quadratic forumula. Thanks agian for your help and quick response.
    there is no "better" here, its just a matter of preference. it is convention to use the quadratic as a last resort, when you can't find any rational roots, but it really doesn't matter. do what you are comfortable with--unless your professor requires you to do a problem using a particular method
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    Sorry for all the questions but I have 1 more.

    2x^2-3x+16x-24
    = x(2x - 3) + 8(2x - 3)
    = (2x - 3)(x + 8)...............same thing

    I thought you would have to factor out a - X and a -8 . Could you tell me why that isn't so?
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathfailure View Post
    Sorry for all the questions but I have 1 more.

    2x^2-3x+16x-24
    = x(2x - 3) + 8(2x - 3)
    = (2x - 3)(x + 8)...............same thing

    I thought you would have to factor out a - X and a -8 . Could you tell me why that isn't so?
    well, i could take out a -x and -8, but what would happen? let's see

    2x^2-3x+16x-24
    = -x(-2x + 3) - 8(-2x + 3)
    = (-x - 8)(-2x + 3) ............this is the same answer, but it doesn't look as pretty. you want to factor out positive numbers as long as possible. becuase having negative signs in front look ugly. there are times when you have to factor out a positive from the first two terms and a negative from the last two or vice versa. again, you just have to match up the signs right

    i was looking at the second pair of numbers the whole time. i realize if i took out positive numbers for both of them, everything would work out nicely (and beautifully)

    don't listen to anyone that tells you otherwise, math is supposed to be aesthetic
    Last edited by Jhevon; March 24th 2007 at 08:35 PM.
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  8. #8
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    Quote Originally Posted by Jhevon View Post
    well, i could take out a -x and -8, but what would happen? let's see

    2x^2-3x+16x-24
    = -x(-2x + 3) - 8(-2x + 3)
    = (-x - 8)(-2x + 3) ............this is the same answer, but it doesn't look as pretty. you want to factor out positive numbers as long as possible. becuase having negative signs in front look ugly. there are times when you have to factor out a positive from the first two terms and a negative from the last two or vice versa. again, you just have to match up the signs right

    i was looking at the second pair of numbers the whole time. i realize if i took out positive numbers fo rboth of them, everything would work out nicely (and beautifully)

    don't listen to anyone that tells you otherwise, math is supposed to be aesthetic

    You rock.. thanks so much for your help on this problem!!!
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathfailure View Post
    You rock.. thanks so much for your help on this problem!!!
    don't mention it, that's what we're here for
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