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Thread: Factoring...2

  1. #1
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    Factoring...2

    What is the first step in factoring the following problem?

    64 (x - a)^4 - x + a
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  2. #2
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    Re: Factoring...2

    rewrite as
    $64(x-a)^4-1(x-a)$
    and then you can factor out the x - a
    $=(x-a)[64(x-a)^3-1]$ and then complete the factor using the difference of two cubes
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    Re: Factoring...2

    (x - a)[64 (x - a)^3 - 1]

    We use a^3 – b^3 = (a – b)(a^2 + ab + b^2).

    Let a = (x - a)

    Let b = 1

    (x - a - 1)[(x - a)^2 + (x - a)(1) + (1)^2]

    (x - a - 1)[64 (x - a)^2 + (x - a) + 1

    Putting it all together I got the following:

    (x - a)[(x - a - 1)64 (x - a)^2 + (x - a) + 1]

    Is this right?
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    Re: Factoring...2

    Quote Originally Posted by mathdad1965 View Post
    (x - a)[64 (x - a)^3 - 1]

    We use a^3 – b^3 = (a – b)(a^2 + ab + b^2).

    Let a = (x - a)

    Let b = 1

    (x - a - 1)[(x - a)^2 + (x - a)(1) + (1)^2]

    (x - a - 1)[64 (x - a)^2 + (x - a) + 1

    Putting it all together I got the following:

    (x - a)[(x - a - 1)64 (x - a)^2 + (x - a) + 1]

    Is this right?
    Not quite. Note that:
    64 (x - a)^3 - 1 = 4^3*(x-a)^3 -1^3 = [4(x-a)]^3 - 1^3 and now factorise the diff of 2 cubes.
    Thanks from mathdad1965
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    Re: Factoring...2

    Quote Originally Posted by Debsta View Post
    Not quite. Note that:
    64 (x - a)^3 - 1 = 4^3*(x-a)^3 -1^3 = [4(x-a)]^3 - 1^3 and now factorise the diff of 2 cubes.
    Ok. I will work on it later or tomorrow.
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    Re: Factoring...2

    Quote Originally Posted by mathdad1965 View Post
    Ok. I will work on it later or tomorrow.
    [4(x-a)-1]([4 (x-a)]^2 + 4 (x-a) + 1)

    (4x - 4a - 1)[(4x-4a)^2 + (4x - 4a) - 1]

    Yes or no?
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  7. #7
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    Re: Factoring...2

    Quote Originally Posted by Debsta View Post
    Not quite. Note that:
    64 (x - a)^3 - 1 = 4^3*(x-a)^3 -1^3 = [4(x-a)]^3 - 1^3 and now factorise the diff of 2 cubes.
    [4(x-a)-1]([4 (x-a)]^2 + 4 (x-a) + 1]

    (4x - 4a - 1)[(4x-4a)^2 + (4x - 4a) + 1]

    Yes or no?
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  8. #8
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    Re: Factoring...2

    Quote Originally Posted by mathdad1965 View Post
    [4(x-a)-1]([4 (x-a)]^2 + 4 (x-a) + 1]

    (4x - 4a - 1)[(4x-4a)^2 + (4x - 4a) + 1]

    Yes or no?
    Yes
    Thanks from mathdad1965
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  9. #9
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    Re: Factoring...2

    Quote Originally Posted by Debsta View Post
    Yes
    Good.
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  10. #10
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    Re: Factoring...2

    It feels good to be right.
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