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AngelaB97

  • one year ago

how do you factorize x^4+27x

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  1. zepdrix
    • one year ago
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    Hey Angela! They've both got some x's... factor those out first.\[\large\rm x^4+27x=x(x^3+27)\]From there you want to utilize your `Sum of Cubes Formula`: \(\large\rm a^3+b^3=(a+b)(a^2-ab+b^2)\)

  2. zepdrix
    • one year ago
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    Hmm but we only have one thing being cubed... x. If you can find a way to rewrite 27 as a perfect cube, you can apply this formula!

  3. AngelaB97
    • one year ago
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    but how do you get (a+b)(a^2-ab+b^2) from x(x^3+27)? @zepdrix

  4. zepdrix
    • one year ago
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    You get \(\large\rm (a+b)(a^2-ab+b^2)\) from \(\large\rm a^3+b^3\) and you get \(\large\rm a^3+b^3\) from \(\large\rm x^3+27\)

  5. zepdrix
    • one year ago
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    the x^3 matches up with the a^3, it's in the correct format. But we need to rewrite 27 as \(\large\rm something^3\)

  6. zepdrix
    • one year ago
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    Example: \(\large\rm 8=2\cdot2\cdot2\) therefore \(\large\rm 8=2^3\) We found a way to write 8 as a perfect cube! Can you find a way to do that with 27?

  7. AngelaB97
    • one year ago
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    thank you!

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