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Calculus expansion

  1. Mar 11, 2004 #1
    I am posting this for my friend who does not have the internet:

    Yes, I know it's easy but even following an example in the book, I can't come up with the correct answer.

    --------------------------------------------------------------------------------
    The roots and degree of a polynomial function are given. Write the equation in standard form.

    roots: 2, -2i | degree:4

    --------------------------------------------------------------------------------

    I thought to start f(x)=(x-2)(x-2i)(x+2i) and simply expand, but this would leave me with a degree 3 answer.

    The final correct answer is x^4 - 4x^3 + 8x^2 - 16x + 16

    Would anyone be so kind as to show me how to attain this answer step by step?

    Thanks in advance.
     
  2. jcsd
  3. Mar 11, 2004 #2
    to get degree four assume that x=2 is a root with multiplicity 2.
    That is
    [tex]
    f(x)=(x-2)^2(x-2i)(x+2i)
    [/tex]
     
  4. Mar 12, 2004 #3

    matt grime

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    If that is exactly what the question is then it as best ambiguous, and at worst impossible.

    Why are we presuming the coefficients must be real? When they say roots, do they mean the only roots are 2 and -2i? If so then there are 3 answers, none of them the one you wrote. So we must have to include other roots, and there is nothing in the question that tells me what I'm allowed to include.

    standard form isn't a standard phrase for me.
     
  5. Mar 13, 2004 #4

    HallsofIvy

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    As matt grime said, the question is ambiguous.

    IF the problem is to find a fourth degree polynomial with real coefficients having 2 and -2i as roots and no other real roots then the answer is exactly what Philcorp said: use (x+2) twice to get (x-2)2(x-2i)(x+2i). You must have both -2i and 2i to get real coefficients.

    IF the problem is to find a fourth degree polynomial with real coefficients and having -2 and 2i as roots but maybe other roots also, then pick any number you like and write (x-2)(x-a)(x-2i)(x+2i) and multiply it out.

    IF the problem is to find a fourth degree polynomial having exactly -2 and 2i as roots (and no others) then any of
    (x-2)n(x+2i)m (where m+n= 4) will work. That will have complex numbers as coefficients.
     
  6. Mar 14, 2004 #5
    But wouldn't that give roots of 2i AND -2i?
     
  7. Mar 14, 2004 #6

    matt grime

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    That's part of the ambiguity. The question lists 2 of the 4 roots that the poly will have and states nothing about the other roots. Guessing what the others might be is what leads to all the possibilities. Requiring real coeffs means -2i and 2i would be roots. We aren't told we can do this, nor is it explicitly barred.
     
  8. Mar 14, 2004 #7

    HallsofIvy

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    Yes, which is why I said "no other REAL roots".
     
  9. Mar 15, 2004 #8
    Sorry. I can't read.
     
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