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anonymous

  • one year ago

What are the real or imaginary solutions of the polynomial equation. 64x^3+27=0 Please help!!!! medal and fann!!!!

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  1. anonymous
    • one year ago
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    do you know how to factor the sum of two cubes?

  2. anonymous
    • one year ago
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    no i cant say i do.

  3. anonymous
    • one year ago
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    i skipped 3 units so idk how to do any of this?

  4. anonymous
    • one year ago
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    ok i will show you \[a^3+b^3=(a+b)(a^2-ab+b^2)\]

  5. anonymous
    • one year ago
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    in this case you have \[64x^3+27=(4x)^3+3^3\] put \(a=4x,b=3\) and plug them in

  6. anonymous
    • one year ago
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    so 4x^3+3^3=(4x+3)(4x^2-4x*3+3^2) right?

  7. anonymous
    • one year ago
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    is that what you meant by plug them in? cause that was a formula right?

  8. anonymous
    • one year ago
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    yeah

  9. anonymous
    • one year ago
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    oops no

  10. anonymous
    • one year ago
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    but close

  11. anonymous
    • one year ago
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    \[4x^3+3^3=(4x+3)(4x^2-4x*3+3^2)\] is what you wrote but you need to square the 4 as well, should be \[4x^3+3^3=(4x+3)(16x^2-12x+9)\]

  12. anonymous
    • one year ago
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    oh ok that makes more sense

  13. anonymous
    • one year ago
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    what do i do from there?

  14. geerky42
    • one year ago
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    \[0 = (4x)^3+3^3=(4x+3)(16x^2-12x+9)\] So either \(4x+3 = 0\) OR \(0 = 16x^2-12x+9\) So from here, you can easily solve first equation, then for second equation, just use quadratic formula.

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