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anonymous

  • 5 years ago

x^3-12x^2+32x=0

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  1. anonymous
    • 5 years ago
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    x(x^2 - 12x + 32) = 0 x(x - 4)(x - 8) = 0 x = 0, 4, 8

  2. Regan
    • 5 years ago
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    divide by x, then solve the remaining quadratic. x^2-12x+32=0=(x-8)(x-4) x=8,2

  3. anonymous
    • 5 years ago
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    regan, you have to account for the fact that you divided by x, meaning that x could be zero as well you see this is true if you plug zero back into the equation, you'll get 0 = 0, which is true, hence, x = 0 is also a solution to this equation

  4. Regan
    • 5 years ago
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    noted : ) ah oui, the fundamental theory of algebra

  5. anonymous
    • 5 years ago
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    still dont get it

  6. anonymous
    • 5 years ago
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    how did u get 0 4 8?

  7. anonymous
    • 5 years ago
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    x(x^2 - 12x + 32) = 0 x(x - 4)(x - 8) = 0 do you get this part?

  8. anonymous
    • 5 years ago
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    after that part you just set each part equal to 0 so x = 0 x - 4 = 0 x - 8 = 0 solve for each x x = 0, 4, 8

  9. anonymous
    • 5 years ago
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    No i dont get how u got the x(x-4) and (x-8)?

  10. anonymous
    • 5 years ago
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    x(x^2 - 12x + 32) = 0 this part? i factored out the x to get to this part x(x - 4)(x - 8) = 0 you have to factor do you know how to factor?

  11. anonymous
    • 5 years ago
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    so how would u do 4x^3-x=0

  12. anonymous
    • 5 years ago
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    Ya i know how 2 factor i jus dont get how x factored 12 and got 4?

  13. anonymous
    • 5 years ago
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    factor out the x x(4x^2 - 1) = 0 x = 0 4x^2 - 1 = 0 4x^2 = 1 x^2 = 1/4 \[x = \pm 1/2\]

  14. anonymous
    • 5 years ago
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    i factored x out of x^3-12x^2+32x=0 to get x(x^2 - 12x + 32) = 0 than i factored INSIDE the parenthesis to get x(x - 4)(x - 8) = 0 go to this site http://tutorial.math.lamar.edu/Classes/Alg/Factoring.aspx it teaches you how to factor

  15. anonymous
    • 5 years ago
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    thanks all check it out

  16. anonymous
    • 5 years ago
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    good luck

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