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Math Help - Factoring again

  1. #1
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    Factoring again

    Factor:

    27x^2+64

    I am completely lost...
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  2. #2
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    <br />
27x^2+64<br />

    this is a binomial expression- an expression with two terms, the variable, x is only present in the first term and thus you cannot factor it. We are only left with the constants 27 and 64. These numbers do not have a common favtor hence we cannot factor again. Thus the expression canot be factored
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  3. #3
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    Quote Originally Posted by mt_lapin View Post
    Factor:

    27x^2+64

    I am completely lost...
    I'll assume you're required to factorise over the complex number field. Otherwise, as has been previously correctly noted, it can't be done and the question is pointless.

    Recall: A^2 + B^2 = (A + iB)(A - iB).

    In your case, A = \sqrt{27} \, x = 3 \sqrt{3} \, x and B = 8 ....
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  4. #4
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    Oh dear. I meant 27^3+64. Does that make it solvable now? Really sorry :/
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  5. #5
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    Quote Originally Posted by mt_lapin View Post
    Oh dear. I meant 27^3+64. Does that make it solvable now? Really sorry :/
    = (27)^3 + (4)^3.

    Factorise using the sum of two cubes formula.
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  6. #6
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    I keep missing things. It's supposed to be 27x^3+64

    I promise this is correct.
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  7. #7
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    Would I then need (3x)^3+4^3 and to factorise using the sum of two cubes formula?
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  8. #8
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    The answer would be: (3x+4)(9x^2-12x+16)

    Is this correct?
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  9. #9
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    Quote Originally Posted by mt_lapin View Post
    The answer would be: (3x+4)(9x^2-12x+16)

    Is this correct?
    Yes.
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