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Math Help - stuck on some problems

  1. #1
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    stuck on some problems

    I have been out of math for years and I am stuck on 3-4 problems of my first struggle since returning to school

    10x(to the 9th power) = 2x( ) I think it's 5x to the 8th power but am I forgetting some rules or something?

    x(to the 2nd power) + 27 + 28 I believe it's prime but I may be wrong. none of the other answers work out for me or is it (x+28)(x-1)

    how do I work out something like x(to the 4th power) -1
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  2. #2
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    Talking

    Quote Originally Posted by mmaquick View Post
    10x(to the 9th power) = 2x( ) I think it's 5x to the 8th power but am I forgetting some rules or something?
    Is it (10x)^9, or 10x^9?

    Quote Originally Posted by mmaquick View Post
    x(to the 2nd power) + 27 + 28 I believe it's prime but I may be wrong. none of the other answers work out for me or is it (x+28)(x-1)
    What value have they given you for x, that you have concluded that the value of the expression is prime? What "other answers" are you talking about?

    Quote Originally Posted by mmaquick View Post
    how do I work out something like x(to the 4th power) -1
    What are you supposed to be doing with x^4 - 1? What is the "working out" that you want to do?
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  3. #3
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    Red face

    Quote Originally Posted by stapel View Post
    Is it (10x)^9, or 10x^9?


    What value have they given you for x, that you have concluded that the value of the expression is prime? What "other answers" are you talking about?


    What are you supposed to be doing with x^4 - 1? What is the "working out" that you want to do?

    10x^9

    2nd question I am factoring completly, and if the polynomial is prime I choose that answer my other options are
    (x+7)(x-4) - is option A.
    (x+28)(x-1) - opttion B
    prime - option C
    (x+28)(x+1) option D.

    with x^4 - 1 these are my options
    (x^2+1) (x-1)(x+1) option A
    (x+2)(x^2-2x+4) -B
    (x-2)(x^2 +2x +4) -C
    (x+8)(x^2-1) -D
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  4. #4
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    Quote Originally Posted by mmaquick View Post

    with x^4 - 1 these are my options
    (x^2+1) (x-1)(x+1) option A
    (x+2)(x^2-2x+4) -B
    (x-2)(x^2 +2x +4) -C
    (x+8)(x^2-1) -D
    Right, i haven't a clue what your trying to do in the first 2 questions but here's how you do the question above

    Use the difference of two squares formula twice i.e.  x^{4} - 1 = (x^{2})^{2} - (1)^{2} = (x^{2}+1)(x^{2}-1) = (x^{2}+1)((x)^{2} - (1)^{2}) = (x^{2}+1)(x+1)(x-1)

    Difference of two squares formula  x^{2} - y^{2} = (x+y)(x-y)
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  5. #5
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    Quote Originally Posted by piglet View Post
    Right, i haven't a clue what your trying to do in the first 2 questions but here's how you do the question above

    Use the difference of two squares formula twice i.e.  x^{4} - 1 = (x^{2})^{2} - (1)^{2} = (x^{2}+1)(x^{2}-1) = (x^{2}+1)((x)^{2} - (1)^{2}) = (x^{2}+1)(x+1)(x-1)

    Difference of two squares formula  x^{2} - y^{2} = (x+y)(x-y)
    so would (x^2 + 1) (x-1) (x+1) work ? the - & + are switched
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  6. #6
    Junior Member piglet's Avatar
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    Quote Originally Posted by mmaquick View Post
    so would (x^2 + 1) (x-1) (x+1) work ? the - & + are switched
    Yes of course it would. Think of  x as just a number. It's also useful to think of an easy example like the following;  2*3 = 3*2 = 6
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  7. #7
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    Quote Originally Posted by piglet View Post
    Right, i haven't a clue what your trying to do in the first 2 questions but here's how you do the question above

    Use the difference of two squares formula twice i.e.  x^{4} - 1 = (x^{2})^{2} - (1)^{2} = (x^{2}+1)(x^{2}-1) = (x^{2}+1)((x)^{2} - (1)^{2}) = (x^{2}+1)(x+1)(x-1)

    Difference of two squares formula  x^{2} - y^{2} = (x+y)(x-y)
    I worked out that x^2 + 27 + 28 is prime so I no longer need assistance on that one

    I am having more problems on the ones you seem good at
    does x^6+1 work out this way
    (x^3) + (1)^2 = (x^3-1)(x^3+1) = (x^3 +1) (x^3-1) did I do that right or is it something like (x+1)(x-1)(x^4-x^2+1)
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  8. #8
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    Quote Originally Posted by mmaquick View Post
    I worked out that x^2 + 27 + 28 is prime so I no longer need assistance on that one

    I am having more problems on the ones you seem good at
    does x^6+1 work out this way
    (x^3) + (1)^2 = (x^3-1)(x^3+1) = (x^3 +1) (x^3-1) did I do that right or is it something like (x+1)(x-1)(x^4-x^2+1)
    still stuck on this one...
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  9. #9
    Junior Member piglet's Avatar
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    Quote Originally Posted by mmaquick View Post
    still stuck on this one...
    Yes this question is very similar but it has a little twist. First you use the difference of two square formula that i mentioned above so
     x^{6}+1 = (x^{3})^{2} + (1)^{2} = (x^{3}+1)(x^{3}-1)

    Now you need to use the difference of two CUBES on  (x^{3}-1)

    Formula =  x^{3 }- y^{3} = (x-y)(x^{2}+xy+y^{2})

    so  (x^{3}-1) = (x-1)(x^{2} +x + 1)

    You can also tidy up  (x^{3}+1) using the SUM of two CUBES formula.

    Formula =  x^{3}+y^{3} = (x+y)(x^{2}-xy+y^{2})

    so  (x^{3}+1) = (x+1)(x^{2} -x +1)

    => your overall answer  x^{6}+1 = (x-1)(x^{2} +x + 1)(x+1)(x^{2} -x +1)

    Hope that helps
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