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Math Help - factoring

  1. #1
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    factoring

    I have three I'm having problems with:

    1.) 16n^3 - 32n^5 - 2n


    2.) x^2 - 2/3x + 1/9


    3.) 16m^2 + 40mn +25n^2
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  2. #2
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    Quote Originally Posted by MathMack View Post
    I have three I'm having problems with:

    1.) 16n^3 - 32n^5 - 2n


    2.) x^2 - 2/3x + 1/9


    3.) 16m^2 + 40mn +25n^2
    1) Each term has two things in common! They're all divisible by -2, and they're all divisible by n. Hence they're all divisible by -2n. So that's your factor!

    This gives  -2n(16n^4-8n^2+1) . This can be further factorised by letting  y=n^2 and considering the quadratic!

     16y^2-8y+1 = 0

    By the quadratic equation  y = \frac{1}{4} hence  4y - 1 is the only factor! Hence 16y^2-8y+1 = (4y - 1)^2 = (4n^2 - 1)^2

    Hence the entire expression is

    -2n(4n^2 - 1)^2

    2) Use the quadratic equation to solve the equation  x^2+\frac{2}{3}x+\frac{1}{9} = 0 . You should get a single root  \frac{1}{3} . This means that  (x - \frac{1}{3}) is a factor. And this the discriminant is zero, in fact TWO of these are factors. Hence the answer is  (x - \frac{1}{3})(x - \frac{1}{3})  =(x - \frac{1}{3})  ^2

    3) Notice that this can be written as  (4m)^2+40mn+(5n)^2 =  (4m)^2+4\times 5 \times 2mn+(5n)^2 . This can be written (4m+5n)^2
    Last edited by Mush; January 15th 2009 at 09:55 PM.
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  3. #3
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    on number 1, I understand that I take out the two and go from there, but I'm confused how to factor it after that.
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  4. #4
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    Quote Originally Posted by MathMack View Post
    on number 1, I understand that I take out the two and go from there, but I'm confused how to factor it after that.
    Please review my editted post!
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