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

    $\displaystyle 16y^2-8y+1 = 0 $

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

    Hence the entire expression is

    $\displaystyle -2n(4n^2 - 1)^2$

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

    3) Notice that this can be written as $\displaystyle (4m)^2+40mn+(5n)^2 = (4m)^2+4\times 5 \times 2mn+(5n)^2 $. This can be written $\displaystyle (4m+5n)^2$
    Last edited by Mush; Jan 15th 2009 at 08:55 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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