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factorising a cubic..leading onto something else. Watch

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    hello everyone, hpe you're all good.

    i was just hoping to be helped in the explanation of factorising a cubic, and then lead me on to factorising a polynomial of order 4.

    basically te question is
    (x^4)-(4x^3)+(35x^2)-62x+130

    and the answer is (x^2-2x+26)(x^2-2x+5)
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    Well assuming you were told that that quartic had two quadratic factors, you could write:

    x^4 -4x^3 +35x^2 -62x + 130 = (Ax^2 + Bx + C)(Dx^2 + Ex + F)

    You can see immediately that A = D = 1 and CF = 130.

    The prime factorisation of 130 = 2 x 5 x 13 so the only possible pairs are 10 and 13, 26 and 5 and 65 and 2 (negative pairs are possible too).

    If you expand the brackets and equate coefficients, you will be able to figure out what to do.
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    (Original post by Mr M)
    Well assuming you were told that that quartic had two quadratic factors, you could write:

    x^4 -4x^3 +35x^2 -62x + 130 = (Ax^2 + Bx + C)(Dx^2 + Ex + F)

    You can see immediately that A = D = 1 and CF = 130.

    The prime factorisation of 130 = 2 x 5 x 13 so the only possible pairs are 10 and 13, 26 and 5 and 65 and 2 (negative pairs are possible too).

    If you expand the brackets and equate coefficients, you will be able to figure out what to do.
    I'll just ask you or someone else here, but is integrating e^(x^2) too difficult for A level or have I missed a simple trick?
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    (Original post by Jeppasloth)
    I'll just ask you or someone else here, but is integrating e^(x^2) too difficult for A level or have I missed a simple trick?
    Too difficult for A Level. You need to know about the error function.
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    (Original post by Mr M)
    Too difficult for A Level. You need to know about the error function.
    Thank god for that. I was puzzling over it all day
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    Actually,my method is a little bit complicate,which means that it is not easy to generalise.First,we know the coefficient of x^4 is 1,and x^3 is -4,so we can assume(x^2-2x+y)*(x^2-2x+z),so,according to simultaneous equation 4+y+z=35 and y*z=130,y=26,z=5.
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    (Original post by Gary Gong)
    Actually,my method is a little bit complicate,which means that it is not easy to generalise.First,we know the coefficient of x^4 is 1,and x^3 is -4,so we can assume(x^2-2x+y)*(x^2-2x+z),so,according to simultaneous equation 4+y+z=35 and y*z=130,y=26,z=5.
    Do you think that is a reasonable assumption?

    Why couldn't it have been (x^2 - x + y)(x^2 - 3x + z) or (x^2 +12x + y)(x^2 -16x +z) for example?
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    (Original post by Jeppasloth)
    integrating e^(x^2)
    on a similar note, what's \displaystyle\int xe^{x^2}\ dx?
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    (Original post by Pheylan)
    on a similar note, what's \displaystyle\int xe^{x^2}\ dx?
    This one is a lot nicer. You should be able to spot that it's the derivative of something nice (chain rule etc).
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    (Original post by Mr M)
    Well assuming you were told that that quartic had two quadratic factors, you could write:

    x^4 -4x^3 +35x^2 -62x + 130 = (Ax^2 + Bx + C)(Dx^2 + Ex + F)

    You can see immediately that A = D = 1 and CF = 130.

    The prime factorisation of 130 = 2 x 5 x 13 so the only possible pairs are 10 and 13, 26 and 5 and 65 and 2 (negative pairs are possible too).

    If you expand the brackets and equate coefficients, you will be able to figure out what to do.
    Prime factors ... certainly narrows factorisation scope.
 
 
 
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