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Homework Help: Create a polynomial with desired characteristics, factoring

  1. Aug 3, 2015 #1
    1. The problem statement, all variables and given/known data

    I understand that this is a very simple thing, but somehow I can't find the key :)
    Please, take a look a pictures attached with a problem and an answer. The task is to create a polynomial f with real number coefficients which has all of the desired characteristics.
    I can't figure out how we got factor x+2.

    2. Relevant equations
    if f(o) = -16, it means that according to the Remainder Theorem the remainder from the division by (x - 0) of polynomial f should equal - 16. Well, dividing by (x-0) means dividing by x. Hence, I thought that factoring should be of the form
    -2x (x - 2i) (x + 2i)

    How did they arrive at x+2?
    Thank you!

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Aug 3, 2015 #2


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    If you assume the third factor is of the form (a*x + b), then what must a and b equal in order to satisfy the conditions that f(0) = -16 and that the leading term of the polynomial is -2x3 ?

    Note: the third factor is (x + 2) only if a = 1 and b = 2.
  4. Aug 3, 2015 #3

    Ray Vickson

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    Please type out the problem (short and easy in this case) and the solution (also pretty short in this case). The desired PF standard is to type out your work and not just post images, with exceptions made when there are diagrams or lengthy tables involved.
  5. Aug 5, 2015 #4


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    I don't think this is right. You might want to revisit what the Remainder Theorem says.
    If f(0) = -16, and you divide the function by (x-0), you are dividing by 0 at x=0. Essentially imposing the idea that 0*g(x) = -16. That can't be right.
    If you know that f(0) = -16 and the leading term has a coefficient of -2, you should be able to say that the factor has the form (-2x +a)(x+b)(x+c) with abc = -16, or as in the answer you posted -2(x+a)(x+b)(x+c) where abc = 8. Since when x=0, the only remaining term in the expanded polynomial is abc.
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