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Math Help - Prime and Factorable Polynomials

  1. #1
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    Prime and Factorable Polynomials

    I'm new to factoring. Don't really understand the definition of "prime". I did 47 questions in my homework, but I want to make sure if some are right or wrong.

    I will answer the questions. Tell me if they are right/ wrong. If they are wrong, then tell me how. At least show me some type of work , so I understand.

    1) P^4-49= (P^2+7)(P^2-7)

    2) X^4-1 = (X^2+1)(X^2-1)
    (X^2-1) can be factored into (X+1) (X-1).
    Total answer: (X^2+1)(X+1)(X-1)

    3) 100x^2+49= PRIME

    4) K^2-18= PRIME

    5) 81W^2+16= Prime

    6) 6t^3-2= Prime

    7) M^2-12= Prime


    I think I have a problem with primes, because it seemed weird I had too many primes.

    Help me , please
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    re: Prime and Factorable Polynomials

    "Prime", hm! Ok, let it be...


    7) M^2-12 = (M+\sqrt{12})(M-\sqrt{12})


    \sqrt{12}=\sqrt{4\cdot 3}=2\sqrt{3}


    M^2-12 = (M+\sqrt{12})(M-\sqrt{12})= (M+2\sqrt{3})(M-2\sqrt{3})
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  3. #3
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    re: Prime and Factorable Polynomials

    So all of them are correct except number 7?
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  4. #4
    MHF Contributor alexmahone's Avatar
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    re: Prime and Factorable Polynomials

    Quote Originally Posted by xivivx View Post
    So all of them are correct except number 7?
    I assume that the problem requires you to factorize into polynomials with integer coefficients, in which case all your answers are correct (including 7).
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    re: Prime and Factorable Polynomials

    Quote Originally Posted by xivivx View Post
    So all of them are correct except number 7?
    No.

    In 1) you can factoring a little bit more, 2) correct, 3) correct, 4) incorrect , 5) correct, 6) incorrect.
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  6. #6
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    re: Prime and Factorable Polynomials

    I get what you are saying. I need to square root them. For example, number 1 and 4 I can solve like this:


    1) P^4-49= (P^2+7)(P^2-7)

    (P+square root of 7) (P-square root of 7) (P^2+7)

    4) K^2-18=

    (K-3 square root of 2) (K+3 square root of 2)

    I,however seriously have no idea on number 6.
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  7. #7
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Prime and Factorable Polynomials

    Quote Originally Posted by xivivx View Post
    I get what you are saying. I need to square root them. For example, number 1 and 4 I can solve like this:


    1) P^4-49= (P^2+7)(P^2-7)

    (P+square root of 7) (P-square root of 7) (P^2+7)



    4) K^2-18=

    (K-3 square root of 2) (K+3 square root of 2)

    I,however seriously have no idea on number 6.



    a^3-b^3= (a-b)(a^2+ab+b^2)
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  8. #8
    MHF Contributor alexmahone's Avatar
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    Re: Prime and Factorable Polynomials

    Quote Originally Posted by Also sprach Zarathustra View Post
    No.

    In 1) you can factoring a little bit more, 2) correct, 3) correct, 4) incorrect , 5) correct, 6) incorrect.
    Out of curiosity, why do you refrain from factoring (X^2+1)=(X+i)(X-i)?
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  9. #9
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    Re: Prime and Factorable Polynomials

    Quote Originally Posted by alexmahone View Post
    I assume that the problem requires you to factorize into polynomials with integer coefficients, in which case all your answers are correct (including 7).
    I'm inclined to agree
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  10. #10
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    Re: Prime and Factorable Polynomials

    (X^2+1)=(X+1)(X-1)

    If you foil them, then you'll know they won't work.

    X (multiplied by) X= X^2
    X multiplied by -1=-x
    1 multiplied by x= x
    1 multiplied by -1= -1

    X^2-x+x-1 = X^2+(-1) instead of what the formula says which is X^2+1
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  11. #11
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Prime and Factorable Polynomials

    Quote Originally Posted by xivivx View Post
    (X^2+1)=(X+1)(X-1)

    If you foil them, then you'll know they won't work.

    X (multiplied by) X= X^2
    X multiplied by -1=-x
    1 multiplied by x= x
    1 multiplied by -1= -1

    X^2-x+x-1 = X^2+(-1) instead of what the formula says which is X^2+1
    Look here:

    Complex number - Wikipedia, the free encyclopedia
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  12. #12
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    Re: Prime and Factorable Polynomials

    xivivx, please answer the question that was implied in allexmahone's response 4: What FIELD is this over? Complex numbers (in which x^2+ 1= (x+ i)(x- i), the real numbers, in which x^2+ 1 is prime but x^2- 18= (x- 3\sqrt{2})(x+ 3\sqrt{2}), or the rational numbers in which both of those are prime? I suspect the rational numbers but you really need to tell us that.
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