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Math Help - How would I factor these polynomials completely?

  1. #1
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    How would I factor these polynomials completely?

    I was supposed to factor 16p^2-64, and following the steps I ended up with 8(2p^2-8) I don't think that's factored COMPLETELY though?

    Then I was supposed to factor 4n^2-32n+48 , and I got 4(n^2-8n+12) . I don't think that's factored completely either?! Were there different steps I was supposed to follow or something?


    your input would be greatly appreciated!
    thanks!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by deathtolife04 View Post
    16p^2-64
    Hint: this is the difference of two squares, there's a formula for factoring such expressions, do you know it?

    4n^2-32n+48 , and I got 4(n^2-8n+12)
    well, you are right for feeling that way, it isn't complete. have you ever heard of "foiling"?
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    Quote Originally Posted by Jhevon View Post
    Hint: this is the difference of two squares, there's a formula for factoring such expressions, do you know it?

    well, you are right for feeling that way, it isn't complete. have you ever heard of "foiling"?
    I don't think I'm familiar with a formula for factoring expressions like the first one.


    Yes, I've heard of foiling. I think it may be starting to come back to me, how to turn them into 'foil' problems. Is that the thing where I would need to find two numbers whose product is 48, and the sum would have to be -32? And then you somehow turn that into a 'foil' problem? I can't remember exactly...
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    Quote Originally Posted by deathtolife04 View Post
    I don't think I'm familiar with a formula for factoring expressions like the first one.
    here's the formula. try it and tell me what you get.

    x^2 - y^2 = (x + y)(x - y)


    Yes, I've heard of foiling. I think it may be starting to come back to me, how to turn them into 'foil' problems. Is that the thing where I would need to find two numbers whose product is 48, and the sum would have to be -32? And then you somehow turn that into a 'foil' problem? I can't remember exactly...[/quote]you already made it simpler didn't you? by factoring out the 4? don't worry about 48 and 32, now all you have to worry about is 8 and 12. (By the way, if you are factoring the original expression, it would be even harder, since the coeffcient of n^2 is not 1. with experience you should be able to foil it, but beginners have this whole process they have to go through to factor by groups and stuff.

    anyway, you were right so far, now you just have to put it together. let's say the two numbers you found were a and b (and they can be positive or negative), then the answer would be in the form:

    (n + a)(n + b)

    contracting the original expression into that form is called "foiling." try it


    This is my 31th post!!!!
    Last edited by Jhevon; August 29th 2007 at 10:33 PM.
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    ok, so for the first one would that be: 16(p-2)(p+2) ?

    I'm still working on the second one...
    Last edited by deathtolife04; August 29th 2007 at 08:59 PM.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by deathtolife04 View Post
    ok, so for the first one would that be: 16(p-2)(p+2) ?
    that's fine. i would probably say (4p + 8)(4p - 8) though. it looks neater that way, there's no constant hanging out in front, but to each his own
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    Quote Originally Posted by Jhevon View Post
    that's fine. i would probably say (4p + 8)(4p - 8) though. it looks neater that way, there's no constant hanging out in front, but to each his own




    ok, yeah, that sounds better

    so for the second one I figured out that -6 and -2 have a product of 12 and a sum of -8


    so, how would i write my final answer? i think if i just write it 4(n-6)(n-2) that would be wrong? cuz then it would look like you just had to distribute the 4 to the n-6?????
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by deathtolife04 View Post
    ok, yeah, that sounds better

    so for the second one I figured out that -6 and -2 have a product of 12 and a sum of -8


    so, how would i write my final answer? i think if i just write it 4(n-6)(n-2) that would be wrong? cuz then it would look like you just had to distribute the 4 to the n-6?????
    write (2n - 12)(2n - 4)

    i split the 4 into 2 times 2, distributed one 2 in the first set of brackets and the other in the second. it is what you would get if you foiled directly


    on second thought, i think it may be better if you keep the constants out in front, it looks messier, but it is factored "completely" when in that form
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    ok, thanks!
    Last edited by deathtolife04; August 29th 2007 at 09:48 PM.
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    Quote Originally Posted by Jhevon View Post
    write (2n - 12)(2n - 4)

    i split the 4 into 2 times 2, distributed one 2 in the first set of brackets and the other in the second. it is what you would get if you foiled directly


    on second thought, i think it may be better if you keep the constants out in front, it looks messier, but it is factored "completely" when in that form
    actually, so would i need to put a 4 in front of each set of parenthesis? or keep it as 4(n-6)(n-2)
    ?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by deathtolife04 View Post
    6p^2 + p - 12
    foil

    Quote Originally Posted by deathtolife04 View Post
    actually, so would i need to put a 4 in front of each set of parenthesis? or keep it as 4(n-6)(n-2)
    ?
    yes, do what you did before

    you got the other one ok?
    Last edited by ThePerfectHacker; September 2nd 2007 at 03:20 PM.
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    Quote Originally Posted by Jhevon View Post
    yes, do what you did before

    you got the other one ok?

    yeah, i think i'm good now! thanks


    let me clarify, though. do what i did before as in, keep it with ONE 4, right?


    so: 4(n-6)(n-2)
    but then wouldn't they think that the 4 should only be distributed through the FIRST set of parenthesis? or does it matter?
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    Quote Originally Posted by deathtolife04 View Post
    yeah, i think i'm good now! thanks


    let me clarify, though. do what i did before as in, keep it with ONE 4, right?


    so: 4(n-6)(n-2)
    but then wouldn't they think that the 4 should only be distributed through the FIRST set of parenthesis? or does it matter?
    to "factor completely" means we should be able to "pull anything else out." if you distribute the 4, then it will become something you can "pull out"

    so leave it as it was the first time, which is what i believe you have here
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  14. #14
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    Quote Originally Posted by Jhevon View Post
    to "factor completely" means we should be able to "pull anything else out." if you distribute the 4, then it will become something you can "pull out"

    so leave it as it was the first time, which is what i believe you have here

    ok. thanks.
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