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Math Help - Having trouble w/ a few Factoring Polynomials and Radical Expression problems

  1. #1
    Newbie Connor's Avatar
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    [Solved]Having trouble w/ a few Factoring Polynomials and Radical Expression problems

    I'm having a tough time understanding these:

    X^6 - 1





    And if you can, please give me a little explanation on how you got the answer.
    Last edited by Connor; January 12th 2011 at 03:19 PM.
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  2. #2
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    48=3*16

    75=3*25

    \displaystyle\frac{5\sqrt{16}\sqrt{3}+\sqrt{25}\sq  rt{3}}{5\sqrt{3}}

    Does this help for the first one?
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  3. #3
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    x^6-1=0\Rightarrow x^6=1 \ \ \ x=\pm 1

    Do you want all the solutions or just the two real solutions?
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    (4\sqrt{2}-3\sqrt{5})2\sqrt{25}=4\sqrt{2}*2\sqrt{25}-3\sqrt{5}*2\sqrt{25}=\cdots
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    Quote Originally Posted by Connor View Post
    I'm having a tough time understanding these:

    X^6 - 1
    Do you need to factor this guy?

    \displaystyle x^6-1 = (x^3)^2-1^2

    Now apply the difference of two squares. \displaystyle a^2-b^2 = (a-b)(a+b)
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  6. #6
    Newbie Connor's Avatar
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    Quote Originally Posted by dwsmith View Post
    48=3*16

    75=3*25

    \displaystyle\frac{5\sqrt{16}\sqrt{3}+\sqrt{25}\sq  rt{3}}{5\sqrt{3}}

    Does this help for the first one?
    No not really. You can't have a radical as a denominator and I have to simplify it more.

    Quote Originally Posted by dwsmith View Post
    x^6-1=0\Rightarrow x^6=1 \ \ \ x=\pm 1

    Do you want all the solutions or just the two real solutions?
    X^6 - 1 isn't equal to anything in this problem. I'm supposed to factor a polynomial here.
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  7. #7
    Newbie Connor's Avatar
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    Quote Originally Posted by pickslides View Post
    Do you need to factor this guy?

    \displaystyle x^6-1 = (x^3)^2-1^2

    Now apply the difference of two squares. \displaystyle a^2-b^2 = (a-b)(a+b)
    Are you sure? I don't have to use this formula?:
    (a \pm b)(a^2 \pm ab + b^2)
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  8. #8
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    Quote Originally Posted by Connor View Post
    No not really. You can't have a radical as a denominator and I have to simplify it more.
    You need to factor.

    \displaystyle\frac{5\sqrt{16}\sqrt{3}+\sqrt{25}\sq  rt{3}}{5\sqrt{3}}\Rightarrow \left(\frac{\sqrt{3}}{\sqrt{3}}\right)\left(\frac{  5\sqrt{16}+\sqrt{25}}{5}\right)

    Then you can further simplify.

    Refer to pickslides post then.
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    Quote Originally Posted by Connor View Post
    Are you sure? I don't have to use this formula?:
    (a \pm b)(a^2 \pm ab + b^2)
    That is for cubics.
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  10. #10
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    Once you've used the difference of two squares (as Pickslides suggests) then you can use the sum/difference of two cubes

    (a \pm b)^3 = (a \pm b)(a^2 \mp ab + b^2)


    Since x^6 and 1 are both square numbers and cube numbers you can start with either formula
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  11. #11
    Newbie Connor's Avatar
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    Quote Originally Posted by dwsmith View Post
    You need to factor.

    \displaystyle\frac{5\sqrt{16}\sqrt{3}+\sqrt{25}\sq  rt{3}}{5\sqrt{3}}\Rightarrow \left(\frac{\sqrt{3}}{\sqrt{3}}\right)\left(\frac{  5\sqrt{16}+\sqrt{25}}{5}\right)

    Then you can further simplify.

    Refer to pickslides post then.
    Three of my friends + my mom got an answer of 5 for this problem. I still don't understand how they got that...
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  12. #12
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    Quote Originally Posted by Connor View Post
    Three of my friends + my mom got an answer of 5 for this problem. I still don't understand how they got that...
    \displaystyle\frac{5\sqrt{16}\sqrt{3}+\sqrt{25}\sq  rt{3}}{5\sqrt{3}}\Rightarrow \left(\frac{\sqrt{3}}{\sqrt{3}}\right)\left(\frac{  5\sqrt{16}+\sqrt{25}}{5}\right)\Rightarrow\frac{5\  sqrt{16}+5}{5}\Rightarrow\left(\frac{5}{5}\right)(  \sqrt{16}+1)=4+1=5

    It should probably really be \pm 5 but you may only need the positive solution.
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  13. #13
    Newbie Connor's Avatar
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    Quote Originally Posted by dwsmith View Post
    \displaystyle\frac{5\sqrt{16}\sqrt{3}+\sqrt{25}\sq  rt{3}}{5\sqrt{3}}\Rightarrow \left(\frac{\sqrt{3}}{\sqrt{3}}\right)\left(\frac{  5\sqrt{16}+\sqrt{25}}{5}\right)\Rightarrow\frac{5\  sqrt{16}+5}{5}\Rightarrow\left(\frac{5}{5}\right)(  \sqrt{16}+1)=4+1=5
    Well I certainly feel retarded for not being able to figure that out.
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  14. #14
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    Do the last problem in the same method. I have already completed one step for you, see post 4.
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  15. #15
    Newbie Connor's Avatar
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    Quote Originally Posted by dwsmith View Post
    Do the last problem in the same method. I have already completed one step for you, see post 4.
    Ok. And for the first problem. Is this the answer?
    (x + 1)(x - 1)(x^2 - x + 1)(x^2 + x + 1)

    And for the last one is this the answer?
    40\sqrt{2} - 30\sqrt{5}
    Last edited by Connor; January 12th 2011 at 03:07 PM.
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