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sabika13

  • 3 years ago

How do I factor this: x^3 -9x^2 +27x -27

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  1. Mr.Math
    • 3 years ago
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    Write \[x^3 -9x^2 +27x -27=(x^3-27)+(-9x^2+27x)=(x^3-3^3)-9x(x-3)=\dots\] Does this help?

  2. sabika13
    • 3 years ago
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    let me try: (x^3 - 3^3) - 9x (x-3) 1^3 (x-3) -9x (x-3) (x-3) (-9x) lol am i right :S?

  3. Mr.Math
    • 3 years ago
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    Not quite. You first need to know the formula for the difference of two cubes: \[(a^3-b^3)=(a-b)(a^2+ab+b^2).\] http://www.mathsisfun.com/algebra/polynomials-difference-two-cubes.html

  4. Mr.Math
    • 3 years ago
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    Now we will apply this to the problem in our hand. You wanna try?

  5. sabika13
    • 3 years ago
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    im so confused.. i havent learned this formula.. All i really needed from this equation was the zeros.. is there a easier way to find zeros than factoring?

  6. Mr.Math
    • 3 years ago
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    Don't be. It's an easy-to-learn formula. You can find help in understanding it in the link I gave above. Now back to our problem (and by applying that formula), we can write: \[(x^3 - 3^3) - 9x (x-3)=(x-3)(x^2+3x+9)-9x(x-3)\] \[=(x-3)(x^2+3x+9-9x)=(x-3)(x^2-6x+9)\] \[=(x-3)(x-3)^2=(x-3)^3.\]

  7. Mr.Math
    • 3 years ago
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    Makes sense?

  8. sabika13
    • 3 years ago
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    how did you get from: =(x−3)(x^2+3x+9−9x) to =(x−3)(x−3)^2

  9. sabika13
    • 3 years ago
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    oh no not that one sorry.

  10. sabika13
    • 3 years ago
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    (x^3−3^3)−9x(x−3) to (x-3) (x^2+3x+9-9x)

  11. braw
    • 2 years ago
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    (2x+3)^_4

  12. braw
    • 2 years ago
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    ??????///

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