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Math Help - Factoring Help

  1. #1
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    Factoring Help

    Are these questions right?

    1. 32x^5 - 108x^2
    4x^3 (8x^2 - 27)

    2. 8y^6 - 38y^4 - 10y^2
    2y^2 (4y^4 - 19y^2 - 5)
    2y^2 (y^2 - 5)(4y^2 + 1)

    Also, how would I solve:

    z^5 - 32z^4 - 16z + 48

    My teacher did not explain to us how to do this...
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  2. #2
    Super Member Quacky's Avatar
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    Re: Factoring Help

    Quote Originally Posted by callmebea View Post
    Are these questions right?

    1. 32x^5 - 108x^2
    4x^3 (8x^2 - 27)
    No, there's an error here.

    2. 8y^6 - 38y^4 - 10y^2
    2y^2 (4y^4 - 19y^2 - 5)
    2y^2 (y^2 - 5)(4y^2 + 1)
    Looks good.

    Also, how would I solve:

    z^5 - 32z^4 - 16z + 48

    My teacher did not explain to us how to do this...
    z^5 - 32z^4 - 16z + 48
    This doesn't factor. Are you sure this is exactly the polynomial?
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  3. #3
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    Re: Factoring Help

    Thank you for replying!

    Could you give me a clue as to what the error I made is?
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  4. #4
    Junior Member mathbyte's Avatar
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    Post Re: Factoring Help

    If you multiply out the factored expression you arrived at, you'll see that the result does not match the polynomial you were given in the problem. That will show you where the error is.
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  5. #5
    MHF Contributor
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    Re: Factoring Help

    Quote Originally Posted by callmebea View Post
    Are these questions right?

    1. 32x^5 - 108x^2
    4x^3 (8x^2 - 27)

    2. 8y^6 - 38y^4 - 10y^2
    2y^2 (4y^4 - 19y^2 - 5)
    2y^2 (y^2 - 5)(4y^2 + 1)

    Also, how would I solve:

    z^5 - 32z^4 - 16z + 48

    My teacher did not explain to us how to do this...
    1. The HCF is 4x^2, not 4x^3...

    2. Correct

    3. You are expected to know that if you can find a z = a such that f(a) = 0, then (z - a) is a factor. You are also expected to know that a can only be factors of 48...
    Once you have found a factor (z - a), long divide, follow the same process again with the remaining factor until it's completely factorised.
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