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Is it 2nd order polynomials, or 2nd order quadratics?

  1. May 14, 2008 #1
    I thought this was rather odd, and wanted to just show it to see what you all thought of it. Well, also, if anyone knows what I should read to exactly understand what I did.

    1: Lets define each answer of a polynomial such as (ax + b)(cx + d) as x1 = (-b/a), x2 = (-d/c).

    2: The polynomial written out is qx^2 + rx + s

    3: Change the polynomial from qx^2 + rx + s to where x = s, and rewrite the problem as qS^2 + rS = S.

    4: Devide S out, which leaves qS + r = 1

    5: subtract r, qS + r - r = 1 - r. qS = 1 - r

    6: Devide q, qS/q = (1 - r)/q. S = (1-r)/q

    7: Now if the equation (q x1 + 1)/q is subtracted from S will solve for x2. Or if x1 is replaced with x2 the equation then solves for x1.
    (1-r)/q - (q x1 + 1)/q = x2
    (1-r)/q - (q x2 + 1)/q = x1

    I did simplify this, but I still think it is rather interesting as to 'how' I did this, rather than the 'simplicity' of it.
    (Darn it, I forgot what these kinds of polynomials are called. Is it 2nd order polynomials, or 2nd order quadratics?)

  2. jcsd
  3. May 14, 2008 #2
    I just ran it through a couple of problems but it seems that in the third order....

    qx3 + rx2+sx+t
    x1 = -b/a
    x2 = -d/c
    x3 = -f/e

    (1-r)/q - (q x1 + 1)/q = x2 + x3
    (1-r)/q - (q x2 + 1)/q = x1 + x3
    (1-r)/q - (q x3 + 1)/q = x1 + x2
    Last edited: May 14, 2008
  4. May 14, 2008 #3


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    The original polynomial is 0 only at its roots. x=s is NOT a root, so your rewrite is wrong. Also you put in a sign change in the S term.
  5. May 14, 2008 #4


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    To what?? Isn't that a crucial part of the problem?

  6. May 14, 2008 #5
    Yikes...I had thought I did a better job at explaining it, I guess not...Well back to the drawing board...

    I just thought it was interesting how I came up with it...

    I'll try one more time, though...

    qx2 + rx = s, and 'Falsely' implying s as the value of x (Actually, I consider this change as 'False' and call the new variable formed as 'False x', however, I use s instead here)

    qx2+rx = s...Making S = x (Yes, I know there is a change in the variable, that is what I find interesting)

    qS2+rS = S...Deviding S out then leaves qS + r = 1.

    qS = 1 - r, then devide q out.

    Thus S is the 'False' x.

    (Which is the reason that I started using the capital letter S instead of the lowercase s when I was expressing that in the first post. Perhaps I should have mentioned that...)

    Then using the 'False' x in the equation

    (FALSE X) - (qx2 + 1)/q = x1
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