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Math Help - Factoring Imperfect squares

  1. #1
    Junior Member Freaky-Person's Avatar
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    Factoring Imperfect squares

    Well, my question is how.

    Here's one from the sheet, I have the answers already, but everyone knows the answers don't matter in math >.>

    1) x^4 + x^2 + 25


    2) 2x^4 + 8


    If you think these are both done with a pretty similar method, then you don't have to show me both. They just look different.
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  2. #2
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    Quote Originally Posted by Freaky-Person View Post
    1) x^4 + x^2 + 25
    x^4+x^2+25=(x^2+5)^2-9x^2, now factorise the difference of two perfect squares.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Freaky-Person View Post
    Well, my question is how.

    Here's one from the sheet, I have the answers already, but everyone knows the answers don't matter in math >.>

    1) x^4 + x^2 + 25


    2) 2x^4 + 8


    If you think these are both done with a pretty similar method, then you don't have to show me both. They just look different.
    I'm not sure what you mean by "factoring imperfect squares." Neither of these expressions factor over the integers, nor do they factor over the rationals. In fact, neither of them factor over the reals, either.

    By solving for the zeros of the polynomials I can tell you that
    x^4 + x^2 + 25 = \left ( x - \left ( \frac{3}{2} + i \cdot \frac{\sqrt{11}}{2} \right ) \right ) \left ( x - \left ( \frac{3}{2} - i \cdot \frac{\sqrt{11}}{2} \right ) \right )  \left ( x - \left ( -\frac{3}{2} + i \cdot \frac{\sqrt{11}}{2} \right ) \right ) \left ( x - \left ( -\frac{3}{2} - i \cdot \frac{\sqrt{11}}{2} \right ) \right )

    You can get this by setting x^4 + x^2 + 25 = 0 and putting the factors in terms of (x - r_1)(x - r_2).... where r_1, r_2, ... are the zeros of the polynomial.

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Krizalid View Post
    x^4+x^2+25=(x^2+5)^2-9x^2, now factorise the difference of two perfect squares.
    Cool! Way to show me up! (Again. )

    -Dan
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  5. #5
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    Quote Originally Posted by Freaky-Person View Post
    2) 2x^4 + 8
    2(x^4+4)

    It remains to factorise x^4+4=(x^2+2)^2-4x^2, it's the same as above.

    Quote Originally Posted by topsquark View Post
    I'm not sure what you mean by "factoring imperfect squares."
    I get it like "complete the square".

    Cheers,
    K.
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  6. #6
    Junior Member Freaky-Person's Avatar
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    miss me?

    okay, you've all been a TON of help, but then I stumbled onto a grevious problem! (oh noez)

    In the following equation

    4b^4 - 13b^2 + 1

    it factors to

    (2b^2 - 3b - 1)(2b^2 + 3b - 1)

    and I'm wondering why it's MINUS 1 and not PLUS 1
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  7. #7
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    Quote Originally Posted by Freaky-Person View Post
    okay, you've all been a TON of help, but then I stumbled onto a grevious problem! (oh noez)

    In the following equation

    4b^4 - 13b^2 + 1

    it factors to

    (2b^2 - 3b - 1)(2b^2 + 3b - 1)

    and I'm wondering why it's MINUS 1 and not PLUS 1
    Because -1 X -1 = 1.

    I know 1 X 1 also = 1, but if you multiply out (2b^2 - 3b + 1)(2b^2 + 3b + 1) it does not equal 4b^4 - 13b^2 + 1
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  8. #8
    Junior Member Freaky-Person's Avatar
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    Quote Originally Posted by WWTL@WHL View Post
    Because -1 X -1 = 1.

    I know 1 X 1 also = 1, but if you multiply out (2b^2 - 3b + 1)(2b^2 + 3b + 1) it does not equal 4b^4 - 13b^2 + 1

    I shall now try and STUMP YOU!! RAWR!!

    m^4 - 19m^2 + 9

    factors to

    (m^2 - 5m + 3)(m^2 + 5m + 3)

    LOOK!! There be a PLUS now!!

    It's magic!! The wizard did it!!
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  9. #9
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    I'm scared.
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  10. #10
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    Quote Originally Posted by Freaky-Person View Post
    m^4 - 19m^2 + 9
    Mmm... what's the problem with this?

    m^4-19m^2+9=(m^2+3)^2-25m^2, the rest is trivial.
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  11. #11
    Junior Member Freaky-Person's Avatar
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    Quote Originally Posted by Krizalid View Post
    Mmm... what's the problem with this?

    m^4-19m^2+9=(m^2+3)^2-25m^2, the rest is trivial.
    look up up uuuuup!

    Cause I wrote a question like this one, but it ended up with the 3rd term being NEGATIVE!! So I was like ZOMG! WHY!?!

    =3
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  12. #12
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    Quote Originally Posted by Freaky-Person View Post
    Cause I wrote a question like this one, but it ended up with the 3rd term being NEGATIVE!! So I was like ZOMG! WHY!?!
    Why not?

    What's so surprising about the 3rd term being negative? Is it the fact that you haven't seen it negative before?
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