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Math Help - Three Factoring Problems

  1. #1
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    Three Factoring Problems

    1) r^8 - 1
    = (r^4)^2 - (1)^2
    = (r^4 - 1)(r^16 + 1r ^4 + 1)
    = Not sure what to do from here

    2) 4x^2 - 4x - 3 = (x+2)(2x - 1)(2x + 3)

    Why do you have to make it 2x and not just have one as 4x?

    3) (x^2 + 1) ^ 1/2 + 3(x^2 + 1) ^ -1/2

    How do you do this???
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  2. #2
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    are you solving for the values of the variables?
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  3. #3
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    Quote Originally Posted by qzno View Post
    are you solving for the values of the variables?
    Oh sorry no

    For number 1 I just need to finish factoring.

    #2 I don't understand why when factored 4x^2 - 4x - 3 = (2x - 1)(2x + 3) and not (4x - 2)(x + 3)

    #3 I just don't understand how to factor it down.

    For 1 I used difference of squares and for 3 I'm supposed to use Sum of Cubes I think.
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  4. #4
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    #2:

    4x^2 - 4x - 3 = 0
    where there is a 4 in front of the x squared, you take this and multiply it by the last number
    so 4 x -3 = -12 meaning
    what multiplies to give you -12 and adds to give you -4?
    (-6 and +2)

    4x^2 + 2x - 6x - 3 = 0
    2x(2x + 1) - 3(2x + 1)

    Factors:
    (2x - 3)(2x + 1)

    hope this helps with #2 : )
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  5. #5
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    Quote Originally Posted by qzno View Post
    #2:

    4x^2 - 4x - 3 = 0
    where there is a 4 in front of the x squared, you take this and multiply it by the last number
    so 4 x -3 = -12 meaning
    what multiplies to give you -12 and adds to give you -4?
    (-6 and +2)

    4x^2 + 2x - 6x - 3 = 0
    2x(2x + 1) - 3(2x + 1)

    Factors:
    (2x - 3)(2x + 1)

    hope this helps with #2 : )
    Oh geez yeah I forgot to put the -6 and 2 back into the equation...thanks a lot.
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  6. #6
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    #3:
    If this is how you had it written:

    (x^2 + 1)^{\frac{1}{2}} + 3(x^2 + 1)^{\frac{-1}{2}} = 0

    Then we know that anything raised to the power of a half is the same as the square root
    and anything raised to the negative a half is the same as one over the square root

    \sqrt{x^2 +1} + \frac{3}{\sqrt{x^2 + 1}} = 0

    Then make it a common denominator

    \frac{x^2 + 1 + 3}{\sqrt{x^2 + 1}} = 0

    \frac{x^2 + 4}{\sqrt{x^2 + 1}} = 0

    I don't know if you personally want the denominator out of the square root or not but this is it in a nutshell.
    If you want to get rid of the square root in the denominator just multiply it by:

    \frac{\sqrt{x^2 +1}}{\sqrt{x^2 +1}}
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  7. #7
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    Quote Originally Posted by Dickson View Post
    1) r^8 - 1
    = (r^4)^2 - (1)^2
    = (r^4 - 1)(r^16 + 1r ^4 + 1)
    = Not sure what to do from here

    2) 4x^2 - 4x - 3 = (x+2)(2x - 1)(2x + 3)

    Why do you have to make it 2x and not just have one as 4x?

    3) (x^2 + 1) ^ 1/2 + 3(x^2 + 1) ^ -1/2

    How do you do this???
    Nevermind about #1 just made sense to me...not sure why I didn't get it haha
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  8. #8
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    #1:

    r^8 - 1 = 0

    (r^4)^2 - 1 = 0

    (r^4 + 1)(r^4 - 1)

    (r^4 + 1)((r^2)^2 -1)

    (r^4 + 1)(r^2 + 1)(r^2 - 1)

    (r^4 + 1)(r^2 + 1)(r + 1)(r - 1)

    : )
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  9. #9
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    Quote Originally Posted by qzno View Post
    #1:

    r^8 - 1 = 0

    (r^4)^2 - 1 = 0

    (r^4 + 1)(r^4 - 1)

    (r^4 + 1)((r^2)^2 -1)

    (r^4 + 1)(r^2 + 1)(r^2 - 1)

    (r^4 + 1)(r^2 + 1)(r + 1)(r - 1)

    : )
    Haha thanks...yeah I got it, don't know how I didn't before
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