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Thread: can some one help me with factoring

  1. #1
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    can some one help me with factoring

    factoring completely

    250x3-128y3

    x2+100
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by dblued View Post
    factoring completely

    250x3-128y3
    $\displaystyle 250x^3 - 128y^3 = 2(125x^3 - 64y^3) = 2[(5x)^3 - (4y)^3]$

    now use the difference of two cubes formula: $\displaystyle a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

    here of course, your $\displaystyle a$ is $\displaystyle 5x$ and your $\displaystyle b$ is $\displaystyle 4y$
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  3. #3
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    Quote Originally Posted by dblued View Post
    factoring completely

    250x3-128y3

    x2+100
    $\displaystyle 250x^3 - 128y^3 = 2(125 x^3 - 64 y^3) = 2([5x]^3 - [4x]^3)$. Now apply the formula for the difference of two perfect cubes.

    $\displaystyle x^2 + 100 = (x + 10i)(x - 10i)$ where i is the square root of -1.
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  4. #4
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    Quote Originally Posted by Jhevon View Post
    $\displaystyle 250x^3 - 128y^3 = 2(125x^3 - 64y^3) = 2[(5x)^3 - (4y)^3]$

    now use the difference of two cubes formula: $\displaystyle a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

    here of course, your $\displaystyle a$ is $\displaystyle 5x$ and your $\displaystyle b$ is $\displaystyle 4y$
    Snap
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by dblued View Post
    x2+100
    did you mean $\displaystyle x^2 ~{\color{red}-}~ 100$?

    maybe that's too easy
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mr fantastic View Post
    $\displaystyle x^2 + 100 = (x + 10i)(x - 10i)$ where i is the square root of -1.
    i thought about doing that, but complex numbers seemed, well, too advanced here...
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