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Math Help - Factoring

  1. #1
    Super Member 11rdc11's Avatar
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    Factoring

    How would I factor

    64x^4-768x^3 + 3392x^2 -6529x +4625 = 0

    to

    (4x^2-25x +37)(16x^2-92x+125) = 0
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  2. #2
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    ok, so, are you asking us to finish factoring? or how to get where its at in factoring the problem?

    <br /> <br />
(4x^2-25x +37)(16x^2-92x+125) = 0<br />

    you can just use the quadratic formula to finish it:

     x= /frac {-b(+-) /sqrt {b^2 - 4ac}}{2a}

    or to see how to get to:

    <br /> <br />
(4x^2-25x +37)(16x^2-92x+125) = 0<br />

    with


    <br /> <br />
64x^4-768x^3 + 3392x^2 -6529x +4625 = 0<br />

    you simply go like this:

     /sqrt {64^4}

    then same thing with:

     /sqrt {3392^2}
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  3. #3
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    Quote Originally Posted by 11rdc11 View Post
    How would I factor

    64x^4-768x^3 + 3392x^2 -6529x +4625 = 0

    to

    (4x^2-25x +37)(16x^2-92x+125) = 0
    This might help:
    Quartic equation - Wikipedia, the free encyclopedia
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  4. #4
    Moo
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    Hello,

    I can see no other obvious way than getting two factors (ax+bx+c)(dx+ex+f), since if there is a complex root, there will be its conjugate as a root too. And (x-1st root)(x-conjugate) will always give an quadratic with rational coefficients.
    If it's real roots, no problem.
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  5. #5
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by log(xy) View Post
    ok, so, are you asking us to finish factoring? or how to get where its at in factoring the problem?

    <br /> <br />
(4x^2-25x +37)(16x^2-92x+125) = 0<br />

    you can just use the quadratic formula to finish it:

     x= /frac {-b(+-) /sqrt {b^2 - 4ac}}{2a}

    or to see how to get to:

    <br /> <br />
(4x^2-25x +37)(16x^2-92x+125) = 0<br />

    with


    <br /> <br />
64x^4-768x^3 + 3392x^2 -6529x +4625 = 0<br />

    you simply go like this:

     /sqrt {64^4}

    then same thing with:

     /sqrt {3392^2}
    Thanks i can find the solutions using the quad formula. I'm having problems figuring out how the TI89 broke the 4th degree polynomial to two 2nd degree polynomial. I'm kind of confused as to what you are doing. Can you finish working it out so i can understand, thanks.
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  6. #6
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    I can see no other obvious way than getting two factors (ax+bx+c)(dx+ex+f), since if there is a complex root, there will be its conjugate as a root too. And (x-1st root)(x-conjugate) will always give an quadratic with rational coefficients.
    If it's real roots, no problem.
    Hi Moo, I'm still a little confused. I'm about to try the quaditic formula chop suey suggested but would still like to understand your input. Thanks Moo and o i did like that techno song lol.
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  7. #7
    Moo
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    Quote Originally Posted by 11rdc11 View Post
    Hi Moo, I'm still a little confused. I'm about to try the quaditic formula chop suey suggested but would still like to understand your input.
    Well, there are 3 or 4 situations.

    - all roots are rational. This is not a problem, you can develop 64(x-a)(x-b)(x-c)(x-d), where a,b,c,d are the roots.
    - at least one root is complex (this is the most general case). This is not a problem either. We know that if the coefficients of a polynomial are rational (and thus integers), then if a complex number z_1=x_1+i y_1 is a root, then its conjugate z_2=x_1-i y_2 is also a root. \boxed{\text{Complex roots go by } \underline{\text{pairs}}}

    So (x-z_1)(x-z_2) divides the polynomial, that is to say (x-x_1-i y_1)(x-x_1+i y_1) this, again, is a product of a complex number and its conjugate : (x-x_1)-i y_1 and (x-x_1)+i y_1
    If you do the product, you'll get a quadratic (x is the variable) : =(x-x_1)^2+y_1^2 (rational coefficients)

    It can be deduced (with a same reasoning for the 3rd and 4th roots) that any quartic with rational coefficients can be factored into a product of 2 quadratics with rational coefficients.


    Since you don't know if there are complex roots or only real roots, it's better to assume there are complex roots (real numbers are complex numbers), and thus to write the original polynomial into the product of 2 quadratics.

    Is it clear ?

    Thanks Moo and o i did like that techno song lol.
    that's cool !
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  8. #8
    Super Member 11rdc11's Avatar
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    O now it makes total sense lol. Thanks Moo!!!
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